GRAVITATION 8/95

WITH
m SOLUTIONS
1*1

Q. 1. Answer the followings ;

(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you
shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere

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or by some other means ?
(b) An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the
space station orbiting around the earth has a large size, can he hope to detect gravity ?
(c) If you compare the gravitational force on the earth due to the Sun to that due to the moon, you
would find that the Sun’s pull is greater than the moon’s pull. However, the tidal effect of the moon’s

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pull is greater than the tidal effect of Sun. Why ?

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Sol. (a) We cannot shield a body from the gravitational influence of nearby matter, because the gravitational

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force on a body due to nearby matter is independent of the intervening medium or the presence of other

rF
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matter in their way, whereas it is not so in case of electrical forces, it means the gravitational screens are
not possible.

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(b) Yes, if the size of the spaceship orbiting around the earth is large enough, an astronaut inside the
spaceship can detect the variation in g.
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(c) Tidal effect depends inversely on the cube of the distance, unlike force which depends inversely on the
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square of the distance. Since the distance of Moon from the ocean water is very small as compared to the
distance of Sun from the ocean water on Earth. Therefore, the tidal effect of Moon s pull is greater than
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the tidal effect of the Sun.
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Q. 2. Choose the correct alternative

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(a) Acceleration due to gravity increases/decreases with increasing altitude.

(*) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a
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sphere of uniform density),

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(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
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(d) The formula - CM m — is more/less accurate than the formula mg (r2 - rj) for the
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difference of potential energy between two points T2 and rj distance away from the centre of earth.
(b) decreases
F
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Sol. (a) decreases

(c) mass of the body (d) more.
Q. 3. Suppose there existed a planet that went around the sun twice as fast as the Earth. What would be
its orbital size as compared to that of the Earth ?
T 1
Sol. Here, T. = 1 year \ T = —
P i
= _ year;r
2
= lA.U.;r^ = ?
Using Kepler’s third law, we have
x2/3
(T 1/2
r
,
= r
_P_ = 1 = 0*63 AU
p e j 1
\ e )
Q. 4. One of the satellites of Jupiter, has an orbital period of 1*769 days and the radius of the orbit is
4*22 X 10* m. Show that mass of Jupiter is about one thousandth times that of the sun. (Take 1 year
= 365*25 mean solar day).
8/96
^Mdee^ Fundamental Physics (XI) VOL.II

Sol. For a satellite of Jupiter, orbital period, = 1-769 days = 1-769 x 24 x 60 x 60 s

Radius of the orbit of satellite, rj = 4-22 x 10® m
Mass of the Jupiter, Mj is given by M
4tc^ ^ 47t2x (4-22x10®)® (1)
1 “
GT^ ~ Gx(1-769 x 24 x 60 x 60)2
We know that the orbital period of Earth around the Sun, T = 1 year = 365-25 x 24 x 60 x 60 s ;
Orbital radius, r = 1 A.U. = 1-496 x 10** m

Mass of the Sun is given by M =

4jc2;-3 47t2 X(1-496x10**)®
.(2)
Gr2 "gx(365-25x24x60x60)2
Dividing (2) by (1), we get
M _ 4tc2x (1-496x10**)® Gx (1-769 x 24 x 60 x 60)2
Mj Gx(365-25x24x60x60)2 ^ 4tc2 x(4-22x10®)® = 1046

w
M 1 1
1 _ 1
or or Af. w M, which was to be proved.
M 1046 1000 1
1000

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Q. 5. Let us consider that our galaxy consists of 2*5 x 10^^ stars each of one solar mass. How long will this
star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Ihke the

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diameter of the Milky way to be 10® ly. G = 6*67 x lO-^^ Nm2 kg2.
Sol. Here, r = 50,000 ly = 50,000 x 9-46 x 10*® m = 4-73 x 102**

Fr
m

M = 2-5 X 10** solar mass = 2-5 x 10** x (2 x 10®**) kg = 5-0 x 10'** kg

\l/2
4jc2 ^3
for
^47c2r® -\M2
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We know that, M - or 7’ = 4 X (22/7)2 x(4-73x1Q20)3 = 112xl0^®s
GF2 . GM ) [ (6-67X10"**)x (5-0x104*)
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Q. 6. Choose the correct alternative :
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(a) If the zero of the potential energy is at infinity, the total energy of an orbiting satellite is negative
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of its kinetic/potential energy,

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(b) The energy required to rocket an orbiting satellite out of Earth’s gravitational influence is
more/less than the energy required to project a stationary object at the same height (as the satellite)
out of Earth’s influence.
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Sol. (fl) kinetic energy (h) less.

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Q. 7. Does the escape speed of a body from the Earth depend on (a) the mass of the body Q>) the location
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from where it is projected (c) the direction of projection (d) the height of the location from where the
body is launched ? Explain your answer.
Sol. The escape velocity is independent of the mass of body and the direction of projection. It depends upon
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the gravitational potential at a point from where the body is launched. Since, this potential depends slightly
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on the latitude and height of the point, therefore, the escape velocity depends sUghtly on these factors.
Q. 8. A comet orbits the Sun in a highly elliptical orbit Does the comet have a constant (a) linear speed
(h) angular speed (c) angular momentum (d) kinetic energy (e) potential energy (/) total energy ..
throughout its orbit ? Neglect any mass loss of the comet when it comes very close to the Sun.
Sol. A comet while going on elliptical orbit around the Sun has constant angular momentum and total energy at
all locations but other quantities vary with locations.
Q. 9. Which of the following symptoms is likely to affect an astronaut in space (a) swollen feet (b) swollen
face, (c) headache, (d) orientational problem.
Sol. (a) We know that the legs cany the weight of the body in the normal position due to gravity pull. The
astronaut in space is in weightlessness state. Hence, swollen feet may not affect his working.
(A) In the conditions of weightlessness, the face of the astronaut is expected to get more supply. Due to it,
the astronaut may develop swoUen face. As eyes, ears, nose, mouth etc. are all embedded in the face!
hence, swollen face may affect to great extent the seeing/hearing/smelling/eating capabilities of the astronaut
in space.
GRAVITATION 8/97

(c) Headache is due to mental strain. It will persist whether a person is an astronaut in space or he is on
earth. It means headache will have the same effect on the astronaut in space as on a person on earth.
{d) Space also has orientation. We also have the frames of reference in space. Hence, orientational problem
will affect the astronaut in space.
In the following two exercises 10 and 11 choose the correct answer among the given ones.
Q. 10. The gravitational intensity at the centre Q of the drumhead defined by a hemispherical shell of
uniform mass density has the direction indicated by the arrow (8(N).l), (i) a, (ii) b, (iii) c, (iv) zero.
Sol. We know that the gravitational potential is constant
at all points inside a spherical shell. Therefore, the
gravitational potential gradient at all points inside the
dV
spherical shell is zero [i.e., as Vis constant. = 01.
dr
Since, gravitational intensity is equal to negative of

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the gravitational potential gradient, hence the
gravitational intensity is zero at all points inside a
hollow spherical shell.
This indicates that the gravitational forces acting on

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a particle at any point inside a spherical shell, will be

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symmetrically placed.

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Therefore, if we remove the upper hemispherical shell, the net gravitational force acting on the particle at
the centre Q or at some other point P will be acting downwards which will also be the direction of

Fr
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gravitational intensity. It is so because, the gravitational intensity at a point is the gravitational force per
at the centre Q will be along c, i.e., option (iii) is
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unit mass at that point. Hence, the gravitational intensity
correct.

Q. 11. For the above problem, the direction of the gravitational

by the arrow (i) d, (ii) e, (Hi)/, (iv) g.
s for
intensity at any arbitrary point P is indicated
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Sol. As per explanation given in the answer of Q. 10, the direction of gravitational intensity at P will be along
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e. i.e., the option (ii) is correct.

Q. 12. A rocket is fired from the earth towards the sun. At what point on its path is the gravitational force
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on the rocket zero ? Mass of sun = 2 x lO^O kg, Mass of earth = 6x10^ kg. Neglect the effect of other
planets. Orbital radius of earth = 1-5 x 10*^ m.
Sol. Here, = 2 x 10^° kg ; = 6 x 10^‘* kg ; r = 1-5 x 10** m.
ur

Let X be the distance of a point from the earth where gravitational forces on the rocket due to sun and earth
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become equal and opposite. Then, distance of rocket from the sun = (r - x). If m is the mass of rocket then
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GM ^ m GMg in (r-jc)2 M or
r-x

“^6x102^
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or

{r-x9 .x2 ^2 M
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in

10^
^ = 1+
ic)2 Vs + io^
^-1 =
F

or or

■S V3
rV3 1-5x10** xV3 1-5 X 1-732x10** = 2-59 X 10** m
or
>^+icP 1-732+1000

Q. 13. How will you ‘weigh the Sun’, that is estimate its mass ? You will need to know the period of one of
its planets and the radius of the planetary orbit. The mean orbital radius of the Earth around the
Sun is 1*5 X 10® km. Estimate the mass of the Sun.
Sol. To estimatethe mass of the Sun, we require, the time period of revolution T of one of its planets (say the
Earth) Let M, A/ be the masses of Sun and Earth respectively and r be the mean orbital radius ot the
" GA/^.A/^
Earth around the Sun. The gravitational force acting on Earth due to Sun is F =
r2
Let, the Earth be moving in circular orbit around the Sun, with a uniform angular velocity co, the centripetal
4jt2
force acting on Earth is F' = Af^ r 0)2 = Af^ r j2
 8/98
^ Fundamental Physics (XI) VOL.II

As this centripetal force is provided by the gravitational pull of Sun on ,Faith, so

GM M 47i2 4jc^r^
S i
= r or M
r2 j2
Knowing r and 7*, mass of the Sun can be estimated.
In this question, we are given, r = 1-5 x 10* km = 1-5 x 10^* m; T= 365 days = 365 x 24 x 60 x 60 s
4x(22/7)2x(1-5x10^^)3
M »2xl0*®kg
^ (6-67 X10-11) X (365 X 24 X 60 X 60)2
Q. 14. A Saturn year is 29*5 times the earth year. How far is the Saturn from the sun if the earth is 1*5 x 10*
km away from the sun ?

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Sol. Here,r, = 29-5T^;/?^=l-5xlO*km;/e^=?
'pl <t>2 (T \2/3
s _^ r29-5T
Using the relation. or R=R = 1-5x10* = l^SxlO’km
R] Rl s ^ T T

e
V ey

Q. 15. A body weighs 63 N on the surface of Earth. What is the gravitational force on it due to the Earth at

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a height equal to half the radius of the Earth ?

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Sol. Weight of body = mg = 63 N

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At height h, the value of g' is given by. /_ gR^ _ gR^ 4

(R + h)^ (R+R/2)'^ 9^
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sor rF
4 4
Gravitational force on body at height h is F = mg' = mx -g=-mg =
kffo |x63 = 28N
Q. 16. Assuming the Earth to be a sphere of uniform mean density, how much would a body weigh half way
down to the centre of Earth if it weighed 250 N on the surface ?g on the surface of Earth is 9*8 mS-2.
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Sol. Weight of body at a depth d = mg' = mxg = 250 f1 - = 125 N

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V /?; ^ j
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Q. 17. A rocket is fired verticaUy with a speed of 5 km s"* from the Earth’s surface. How far from the Earth
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does the rocket go before returning to the Earth ? Mass of the Earth = 6 x 1()24 kg, mean radius of
Earth = 6-4 x 10® m, G = 6*67 x 10r« Nm 2 kg-2 ^
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Sol. Let the rocket be fired with velocity v from the surface of Earth and it reaches a height h from the surface
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of Earth where its velocity becomes zero.

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d

Total energy of rocket at the surface of Earth = K.E. 1 2

+ RE. = -mv^+ f-GMm^
2 { R
nidn
Re

GMm
At the highest point, u = 0, K.E. = 0 and RE. = -
F
Fi

(R + h)
f-GMm'\ GMm
Total energy = K.E. + RE. = 0 +
^ R+h ) R+ h
According to law of conservation of energy.
1
—mv
2 GMm GMm
2 R (R + h)

1 GM _gR^ gR^ /

= gR 1-
R f u \
or
= gR
2 R (R + h) R {R + h) V R+h) [R + h)
or
u2(/? + /,) = 2gRh or
Rv^=2gRh-v^h = {2gR-v^)h
or Rv^ (6-4x10®)x(5x10*)2 = l*6xl0®m
2gR-u2 2 X 9-8 X (6-4x10®)-(5x103)2
GRAVITATION 8/99

Q. 18. The escape velocity of projectile on the surface of Earth is 11*2 km s"^. If a body is projected out
with thrice of this speed, find the speed of the body far away from the Earth. Ignore the presence of
other planets and sun.
Sol. Here, = 11 -2 km s“* ; Velocity of projection of the body v = 3 Let m be the mass of the projectile and
Vq be the velocity of the projectile when far away from the Earth (i.e., out of gravitational field of Earth).
Then, from the law of conservation of energy
1 1 9 1 1
-mvl=-mv ^ —mvt
2 ^

or V
0 f = .^/(3v^?^ = V8t;^ = ^8x11-2 =31-68 km
Q.19. A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be
expended to rocket the satellite out of the gravitational influenceof Earth ? Mass of the satellite ef
200 kg, mass of the Earth = 6-0 x 10^ kg, radius of the Earth = 6*4 x 10® m, G = 6-67 x 10^^^ N m^ kg" .

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Sol. Total energy of orbiting satellite at a height h
GMm 1i 9
+ -mv^ = - GMm I GM ^ -GMm
(R + h)'^ 2^(R + h) 2{R + h)

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iR + h) 2
Energy expended to rocket the satellite out of the Earth’s gravitational field

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= - (total energy of the orbiting satellite)

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GM m (6-67 X 10-^^)x(6xl0^)x200
= 5-9x10^1
2{R+h) 2(64x10®+4x1()5)
for
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Q. 20. Two stars each of 1 solar mass (= 2 x 10^ kg) are approaching each other for a head on collision.
When they are at a distance 10® km, their speeds are negligible. What is the speed with which they
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collide ? The radius of each star is lO'* km. Assume the stars to remain undisturbed until they collide.
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Use the known value of G.
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Sol. Here, mass of each star, M = 2 x 10^® kg.

Initial distance between two stars, r = 10® km = 10^^
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m.

GMM
Initial potential energy of the system = -
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r
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Total K.E. of the stars = -M + —M =Mv^

2 2

where V is the speed of stars with which they collide. When the stars are about to collide, the distance
d
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between their centres, / = 2 R.

in

GMM
F

Final potential energy of two stars = - 2R

Since, gain in K.E. is at the cost of loss in P.E.

Mv^ =- GMM f GMM^ -GMM ^ GMM

r
. 2R J
2R r

2 X 10^® =-
6-67x10
-11
x(2xl0^°)^ 6-67xl0~*^ x(2x10^Q)^
or
10*2 ^ 2x10*^
= - 2-668 X 1()38 + 1-334 x 10^^ = 1-334 x 10^^ J
1-334x10^2
= 2-583x10® ms-1
i 2x1020
V =

k
8/100
^ftadee^'^ Fundamental Physics (XI) VOL.II

Q. 21. IWo heavy spheres each of mass 100 kg and radius 0*10 m are placed 1*0 m apart on a horizontal
table. What is the gravitational field and potential at the mid point of the line joining the centres of
the spheres ? Is an object placed at that point in equilibrium ? If so, is the equilibrium stable or
unstable.

Sol. Gravitational field at the mid-point of the line joining the centres of the two spheres
GM GM A _
(-P) + jr = 0
(rll9 (r/2)
Gravitational potential at the mid point of the line joining the centres of the two spheres is

V = - —+{1 - -4x6-67X10"^ ^ X100 = -2*7xlO-*J/kg

r/2\ rl2 ^ r 10

As the effective force on the body placed at mid-point is zero, so the body is in equilibrium. If the body is
displaced a little towards either mass body from its equilibrium position, it will not return back to its initial

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position of equilibrium. Hence, the body is in unstable equilibrium.

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e
rree
r FF
uurr
for
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ad
Yo
dY
Re
innd
Fi

i
GRAVITATION 8/101

1 WITH ANSWERS,
PJ HINTS AND SOLUTIONS
f ■i

MULTIPLE CHOICE QUESTIONS-I

1. The earth is an approximate sphere. If the 4. Satellites orbiting the earth have finite life and

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interior contained matter which is not of the sometimes debris of satellites fall to the earth.
same density everywhere, then on the surface This is because,
of the earth, the acceleration due to gravity {a) the solar cells and batteries in satellites run
out.
(fl) will be directed towards the centre but not
the same everywhere. (b) the laws of gravitation predict a trajectory

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(b) will have the same value everywhere but not spiralling inwards,

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rFl
directed towards the centre, (c) of various forces causing the speed of

Fre
satellite and hence height to gradually
(c) will be same everywhere in magnitude
decrease.

rr F
directed towards the centre.
(d) of collisions with other satellites.
(d) cannot be zero at any point.
5. Both earth and moon are subject to the
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2. As observed from earth, the sun appears to
move in an approximate circular orbit. For
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gravitational force of the sun. As observed
from the sun, the orbit of the moon
the motion of another planet like mercury as
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(a) will be elliptical.
observed from earth, this would
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(b) will not be strictly elliptical because the total

(a) be similarly true.
gravitational force on it is not central,
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(b) not be true because the force between earth (c) is not elliptical but will necessarily be a
and mercury is not inverse square law. closed curve.
(c) not be true because the major gravitational (d) deviates considerably from being elliptical
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force on mercury is due to sun. due to influence of planets other than earth.
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(d) not be true because mercury is influenced 6. In our solar system, the inter-planetary region
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by forces other than gravitational forces. has chunks of matter (much smaller in size
3. Different points in earth are at slightly compared to planets) called asteroids. They
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different distances from the sun and hence {a) will not move around the sun since they have
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experience different forces due to gravitation. very small masses compared to sun.
For a rigid body, we know that if various (b) will move in an irregular way because of
F

forces act at various points in it, the resultant their small masses and will drift away into
motion is as if a net force acts on the c.m. outer space,
(centre of mass) causing translation and a net (c) will move around the sun in closed orbits
torque at the c.m. causing rotation around an but not obey Kepler’s laws.
axis through the c.m. For the earth-sun system (d) will move in orbits like planets and obey
(approximating the earth as a uniform density Kepler’s laws.
sphere) 7. Choose the wrong option,
(fl) the torque is zero. (a) Inertial mass is a measure of difficulty of
(b) the torque causes the earth to spin, accelerating a body by an external force
(c) the irgid body result is not applicable since whereas the gravitational mass is relevant
the earth is not even approximately a irgid in determining the gravitational force on it
body. by an external mass.
(cf) the torque causes the earth to move around (b) That the gravitational mass and inertial mass
the sun. are equal is an experimental result.

%
8/102
Fundamental Physics (XDSEIMD

(c) That the acceleration due to gravity on earth FIGURE 8(N).2

is the same for all bodies is due to the equa A B C
lity of gravitationalmass and inertial mass.
2M m M
id) Gravitational mass of a particle like proton
can depend on the presence of neighbouring At subsequent times before any collision takes
heavy objects but the inertial mass cannot. place :
8. Particles of masses 2 A/, m and M are respec (a) m will remain at rest.
tively at pointsyl, B and C with Afi = 1/2 (BQ, (b) m will move towards M.
m is much-much smaller than M and at time
(c) m will move towards 2 A/.
t = 0, they are all at rest (Fig. 8(N).2).
(d) m will have oscillatory motion.

MULTIPLE CHOICE QUESTIONS-II

9. Which of the following options are correct ? ib) we will become hotter after billions of years,

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{a) Acceleration due to gravity decreases with (c) we will be going around but not strictly in
increasing altitude. closed orbits.
{b) Acceleration due to gravity increases with id) after sufficiently long time we will leave the

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increasing depth (assume the earth to be a solar system.
sphere of uniform density), 14. Supposing Newton’s law of gravitation for

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(c) Acceleration due to gravity increases with
—>

gravitation forces and between two

increasing latitude.

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{d) Acceleration due to gravity is independent masses and ^2 at positions and read
of the mass of the earth.
10. If the law of gravitation, instead of being inverse- for
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square law, bec'omes an inverse-cube law ^1 =
hi Ml
{a) planets will not have elliptic orbits. 0 ;
kkss
ib) circular orbits of planets is not possible, where Mq is a constant of dimension of mass,
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^2 = ^ Is 3 number. In such a case,

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(c) projectile motion of a stone thrown by hand

on the surface of the earth will be
(a) the acceleration due to gravity on earth will
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approximately parabolic. be different for different objects.

{d) there will be no gravitational force inside a ib) none of the three laws of Kepler will be valid,
spherical shell of uniform density. (c) only the third law will become invalid.
r
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11. If the mass of sun were ten times smaller and

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id) for n negative, an object lighter than water

gravitational constant G were ten itmes larger will sink in water.
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in magnitudes 15. Which of the following are true ?

{a) walking on ground would became more (a) A polar satellite goes around the earth’s pole
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difficult.
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in north-south direction.
ib) the acceleration due to gravity on earth will ib) A geostationary satellite goes around the
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not change, earth in east-west direction,

(c) raindrops will fall much faster. (c) A geostationary satellite goes around the
earth in west-east direction.
id) airplanes will have to travel much faster.
12. If the sun and the planets carried huge id) A polar statellite goes around the earth in
east-west direction.
amounts of opposite charges
(fl) all three of Kepler’s laws would still be valid.
16. The centre of mass of an extended body on the
surface of the earth and its centre of gravity
{b) only the third law will be valid.
(a) are always at the same point for any size of
id) the second law will not change. the body.
id) the first law will still be valid. ib) are always at the same point only for
13. There have been suggestions that the value spherical bodies,
of the gravitational constant G becomes (c) can never be at the same point.
smaller when considered over very large time id) is close to each other for objects, say of sizes
period (in billions of years) in the future. If less than 100 m.
that happens, for our earth ie) both can change if the object is taken deep
id) nothing will change. inside the earth.
GRAVITATION 8/103

ANSWERS

!.(</) 2.{c) 3. (a) 4. (c) 5. (b) 6. id) 7. W 8. (c)

9. (a,c) 10. (a,b,c) 11. ia,c,d) 12. (a,c,d) 13. (c,d) 14. (a.c.i/) 15. (a,c) 16. (^0

HINTS FOR DIFFICULT MULTIPLE CHOICE QUESTIONS

Multiple Choice Questions -1

1. If the earth is an approximate sphere of non-unifonn density, then the centre of gravity of earth will not be
situated at the centre of earth. The distance of different points on earth will be at different distances from
GM I
the centre of gravity of earth. As, g = OT g
oc
—r-, so g is different on different points on the surface
r

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of earth but never zero.

2. The sun appears to move in an approximate circular orbit as observed from earth. The gravitational force of
attraction between earth and sun always followed inverse square law.

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Due to relative motion between earth and mercury, the orbit of mercury as observed from earth will not be
approximately circular, since the major gravitational force on mercury is due to sun.

e
reree
3. The earth is revolving on circular orbit around sun due to gravitational force (F) which acts along the

r FF
j ^
radius of circular path, towards the sun i.e. angle between r andF is zero. As
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—*

Torque = !xl = lrxFI=rFsin0° = 0

Therefore, torque is zero.
foor
GM
ks s
4. Orbital velocity of satellite at distance r from the centre of earth is, v =
y r
Yoo
ooook

GMm 1 GMm 1 GM GMm

2 = - + -m =
eBB

+ — mv
Total energy of satellite = PE + KE 2 r 2 r 2r

The viscous force acting on satellite decreases the energy of satellite. As a result of it, the total energy of
satellite becomes more negative. Therefore, the value of r gradually decreases, consequently the height of
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ad

satellite gradually decreases.

Yo

5. Moon is revolving around earth in almost circular orbit. Sun exerts gravitational pull on both, earth and
moon. When observed from sun, the orbit of the moon will not be strictly elliptical because the total
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gravitationalforce {i.e. force due to earth on moon and force due to sun on moon) is not central.
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innd

6. Asteroids move in circular orbits like planets under the action of central forces.
7. The gravitational mass of proton is equivalent to its inertial mass. Hence option (d) is wrong.
FFi

G2Mxm GMm
8. Resultant force on mass m due to masses at A and B is F = towards BA.
(A5)2 {BC9
Therefore, m will move towards 2 M.

Multiple Choice Questions - II

g
9. At altitude h, g/, -
(R + h)^ {\ + h/Rf
As h increases, g/j decrea.ses.
/
d ^
At depths/, ; As d increases, g^ decreases

At latitude X, gy^ = g - R (£>^ cos^ AsX increases, the value of cos X decreases so gy^ increases
8/104 Fundamental Physics (XI) VOL.II

GM
On surface of earth, Z -
/?2 i.e. g depends on mass M of the earth.
10. If the law of gravitation becomes an inverse cube law then

F =
GM m rmp-
or v =
‘Jgm
r3 r r

Hme period of revolution of a planet, T =

2nr _ 2nr^ or oc r It means a planet will not have an
V "Vcw
elliptical orbit The circular orbit of a planet may not be possible as the gravitational attractive force obeys
inverse square law and not inverse cute law. A stone thrown by hand on the surface of the earth will follow
nearly parabolic path, under the gravitational force. There will be some gravitational force inside a spherical
shell of uniform density.

ww
(lOG)Af
11. When G = 10 G, then g' = = 10g.

Weight of person = mg' = m x 10 g = 10 mg i.e. gravity pull on person will increase. Due to it, walking

Flo
_= 2r^(p-g)g

e
on ground would become more difficult. As critical velocity of rain drop,

eree
9^
where letters have the usual meanings, so v^.« g. Since g increases, hence increases.

FFr
To overcome the increased gravitational pull of earth, the aeroplanes will have to travel much faster.
oorr
uur r
12. Due to huge amounts of opposite charges on sun and earth there will be a large force of electrostatic
attraction as well as gravitational attraction. Both the forces obey inverse square law and are central forces.
sf
Due to it, the distance between sun and earth will decrease. In this situation, all the three Kepler’s laws will
sk
be valid.
Yoo
oook

13. When G decreases with time, the gravitational force F = will become weaker with time. Then r
eBB

y.1
increases with time. Due to it, earth will be going around the sun not strictly in closed orbit and after long
time it will leave the solar system.
uurr
ad

fm, m, 1"
14. F = fGMl L^
Yo

'll "o J 'll

dY
Re
innd

Acceleration due to gravity, g =

r,2
FFi

mass mass
12

Thus acceleration due to gravity on earth is different for different objects. In this situation Kepler’s third
law will not be valid.

When n is negative, then

8 =
rl
12
mass

which is positive as Mq > mj or

Therefore, object lighter than water will sink in water.
15. Knowledge based question.
16. Centre of mass is very close with the centre of gravity of a body if size of the body is very very small as
compared to size of the earth. Thus only option id) is correct.
GRAVITATION 8/105

VERY SHORT ANSWER QUESTIONS

17. Molecules in air in the atmosphere are attracted by gravitational force of the earth. Explain why all
of them do not fall into the earth just like an apple falling from a tree.
Ans. Air molecules in the atmosphere experience vertically downward force due to gravity, just like an apple
falling vertically from a tree. Due to temperature, the air molecules have random motion. Due to it, the
velocity of air molecules is not exactly in the vertical downward direction. The downward gravity pull on
air molecules increases the density of air in the atmosphere close to earth. It is due to this reason, the
density of air decreases as we go up.
18. Give one example each of central force and non-central force.
Ans. Examples of central forces. Gravitational force of a point mass and electric force due to a point charge.

ooww
Examples of non-central force. Nuclear forces which hold the nucleons together in the nucleus and
magnetic forces between two current carrying loops.
19. Draw areal velocity versus time graph for Mars.
FIGURE 8(N).3
Ans. We know that areal velocity of a planet revolving around the sun
in an orbit is constant with time. Therefore, the variation of areal Areal Velocity

e
velocity versus time for Mars is as shown in Fig. 8(N).3 by a

re
straight line AB.

rFFl
ree
F
20. What is the direction of areal velocity of the earth around the A B
sun ?

rF
Ans. Areal velocity of earth around the sun is given by dA^L ■►Time

fsfoor
ouur
dt 2m 0

where l is the angular momentum.

kosk
—»
—»
dA^\ rxv
Yo
As, L = r X p = r xm V ● (rxmv) =
oo

dt 2m 2
Y
BB

-*

The direction of areal velocity dA

is the same as that of ("rx^) i.e.
rre

dt J
oYuu

perpendicular to the plane containing r and v and is directed as given by

ad

Right Hand rule.

dY

21. How is the gravitational force between two point masses affected when they are dipped in water
keeping the separation between them the same ?
innd
Re

Ans. Since, f - G mj m2 , is independent of the nature of medium between two

point masses /M] and m2-
Fi
F

r2
Therefore, the gravitational force between two point masses will not be affected, when they are dipped in
water.

22. Is it possible for a body to have inertia but no weight ?

Ans. Yes, a body can have inertia (i.e. mass) but not weight. In fact a body always has mass (i.e. inertia) but the
gravitational force on it can be zero. When a body is taken at the centre of the earth, it has mass m, but its
weight (= mg) is zero.
23. We can shield a charge from electric fields by putting it inside a hollow conductor. Can we shield a
body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by
some other means ?
Ans. No, because gravitational force between two point mass bodies is independent of the intervening medium
between them. It is due to the above reason, we cannot shield a body from the gravitational influence of
nearby matter by putting it either inside a hollow sphere or by some other means.
8/106
“Pnadeefi. d- Fundamental Physics (XI) tvjnm
24. An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space
station orbiting around the earth has a large size, can he hope to detect gravity ?
Ans. Yes, if the size of space-station orbiting around the earth is very large, then the astronaut inside the spaceship
will experience gravitational force and hence can detect gravity.
25. The gravitational force betvt'een a hollow spherical shell (of radius F FIGURE 8(N).5
K and uniform density p) and a point mass is F. Show the nature of
F vs r graph where r is the distance of the point from the centre of
the hollow spherical shell of uniform density.

Ans. ^ _ GM m where M =
3 ^

F = m i.e. F OC r
O
r )

Inside the hollow spherical shell the gravitational force is zero.

w
The variation of F with r is shown in Fig. 8(N).5.
26. Out of aphelion and perihelion, where is the speed of the earth

Flo
more and why ?
Ans. Refer to Fig. 8(N).6, Aphelion A is the location at which the earth is at

reee
the greatest distance from the sun and perihelion P is the location at
which the earth is at the smallest distance from the sun. Since areal

FFr
velocity of the earth around the sun is constant, therefore, the speed of
the earth at the perihelion is more than that at the aphelion.
27.
for
What is the angle between the equatorial plane and the orbital plane of
ur
(a) Polar satellite ?
{b) Geostationary satellite ?
kkss
Ans. (a) Angle between the equatorial plane and orbital plane of polar satellite is 90°.
Yo
ooo

{b) Angle between equatorial plane and orbital plane of the geostationary satellite is 0°.
eB

SHORT ANSWER QUESTIONS

r

28. Mean solar day is the time interval between two succesive noon when sun passes through zenith
ou
ad

point (meridian).
YY

Sidereal day is the time interval between two successive transit

of a distant star through the zenith point (meridian).
nndd
Re

By drawing appropriate diagram showing earth’s spin and

orbital motion, show that mean solar day is four minutes
Fi

longer than the sidereal day. In other words, distant stars

would rise 4 minutes early every successive day.
Ans. The earth’s spin and orbital motion of earth around the sun is
shown for one solar day and sidereal day in Fig. 6(CF).7(a), (b).
We know that every day the earth advances in the orbit by 1°.
Therefore, the earth will have to rotate 361° (which we define as
1 day) to have sun at the zenith point again. It means
361°=lday = 24h
24
1° = X1 = 0-066 h = 3-99 min « 4 min.
361

29. Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed
at the mid point of the line joining their centres be in stable equilibrium or unstable equilibrium ?
Give reason for your answer.
GRAVITATION 8/107

Ans. Refer to Fig. 8(N).8, when an object of mass m is placed

at P, the mid point of AB, then each sphere will attract
GMm
this object with a force F = . Due to it, the
(5«)2
resultant force on object at P is zero and the object at O
is in equilibrium. If the object is displaced a small
distance x from P towards sphere B, then the force on
GM m
object due to sphere A, = acting towards sphere A. The force on object due to sphere B,
(5/? + x)2

oww
GMm
^2 = acting towards sphere B. As F2 > Fj, so a net force will be acting on object towards
{5R-x)^
sphere B and object will move towards B.
Therefore, tlie object will be in unstable equilibrium.

e
30. Show the nature of the following graph for a satellite orbiting the earth.

re
(a) KE vs orbital radius R (b) PE V.S orbital radius R (c) TE vs orbital radius R.

FFrllo
GM
Ans. Orbital speed of a satellite in an orbit of radius R is, Vq =

reF
R

e
1 2 1 GM
ouru
.●. KE of satellite, E
^
= —mVn = —m
2 ^ 2 R

osrF
GM m
PE of satellite, Ep = ~ R

TE of satellite, Ej. = Ef^ + Ep = —

ffor
k
I GM m GM m GM m
>R
kso
R R 2R
ooo
Yo
Y
In Fig- 8(N).9,
BB

Variation of E^ with R is shown by curve (a)

Variation of Ep with R is shown by curve (b)
Variation of Ep with R is shown by curve (c).
r ree
Y
uu

31. Shown are several curves (Fig. 8(N).10). Explain with reason, which ones amongst them can be
possible trajectories traced by a projectile (nelgect air friction).
ad
Ydoo
nidn
Re
F
Fi
8/108 “P>iadeefr U Fundamental Physics (XI) VOL.II

Ans. The nature of trajectory of a particle under the action of gravitational force of earth will be a parabolic
rather a conic section for motion outside the earth with focus lying at the centre of the earth. Only option
(c) meets this erquirement.
32. An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth,
that is, taken from a distance R to 2 R from the centre of the earth. What is the gain in its potential
energy ?

^ GAfm'\_GMm gR^m _
\mgR
f GMm
Ans. Gain in PE = final PE - initial PE = -
2R K R 2R IR

33. A mass m is placed at P a distance h along

the normal through the centre O of a thin
circular ring of mass M and radius r
(Fig. 8(N).ll(a)).

ww
If the mass is removed further away such
that OP becomes 2 /t, by what factor the
force of gravitation will decrease, if A = r ?

Flo
Ans. Refer to Fig. 8(N). 11(b), mass per unit length

e
of ring.

reree
M

r FF
dM =
2nr
uurr
Take an element of length dl of the ring at A.
The mass of this element of the ring.
foor
ks s
M
Yoo
dM^ dl
ooook

2nr
eBB

Here, AP = Vr2+/i2
Gravitational force on the particle of mass m at P due to element at A is
uurr
ad

GmdM Gm(M/2nr)dl
Yo

dF=^ acting along PA

(r^ + h^) (r^+h^)
dY
Re

Resolving dF into two rectangular components, we get dF cos 0 acts along the axis directed towards O
innd

and dF sin 6 acts perpendicular to axis OP.

FFi

Now we calculate the force on the particle at P due to all the elements of the ring and ersolve these forces
into two rectangular components dong the axis and perpendicular to the axis of ring. The components
forces perpendicular to the axis of ring due to all the elements will cancel out each other and component
forces along the axis will add up. So the resultant gravitational force on particle at P due to whole ring is
GmM h GmMh
F = ^dF COS0 = 2nr(r^
\di
GmMh GmM h
x2nr =
2iir(r2+/i2)3/2 (r2+/,2)3/2
GM mx2h
When h = 2h, then F' =
(r2+4/i2)2/2
GRAVITATION 8/109

GM mr GM m GM m X 2 r 2GMm
If /i = r, then F = and F' —
(r2 + r2)3/2 “ (^2^4^2)3/2 5^^
F' _ 4V2 or F'A
5 Vs

LONG ANSWER QUESTIONS

34. A star like the sun has several bodies moving around it at different distances. Consider that all of
them are moving In circular orbits. Let r be the distance of the body from the centre of the star and
let its linear velocity be t;, angular velocity co, kinetic energy K, gravitational potential energy V, total
energy E and angular momentum L. As the radius r of the orbit increases, determine which of the

ooww
above quantities increase and which ones decrease.

Ans. Linear velocity of a body orbiting a star is, v - , when r increases, v decreases.

ee
Angular velocity, co = —= When r increases, co decreases.
T

rFl
re
Fre
_ 1 GM
Kinetic energy, K = -mv 2 = —m When r increases, KE decreases.
2 2

rrF
r

GM m
Gravitational PE, U = - When rincreases, PE becomes less negative i.e. increases.
sffoo
ouur
r

GMm (-GMm\ -GM m

kosk
Total energy, £ = KE + PE = ; When r increases, £ increases.
2r r 2r
Yo
oo
Y

GM
yjGMr ;
BB

Angular momentumL~mv r =mr = m

When r increases, L increases.
V r
35. Six point masses of mass m each are at the vertices of a regular hexagon of side /. Calculate the force
rre

on any of the masses.

ouu

-llSn = Sl
Y

Ans. InFig. 8(N).I2, /iC = AM + MC= 2 AM = 2 /cos 30°

ad

1 1
dY

AD = AG+ G// + //Z) = /sin 30° + /+/sin 30° =/x- + / + /x = 2/

2 2
AE = AC=^Sl,
innd

AB = AF=l
Re

Force on mass m at A due to mass m at £ is,

Fi
F

Gmm
/l = /2
along AB.

Force on mass mat A due to mass m at C is,

Gmxm _Gm^
/2 = ~ 3/2
along AC.

Force on mass m at A due to mass m aXD is,

Gmxm _GnP-
/3 = along AD.
”(2/)2 “ 4/2
Force on mass m at A due to mass m at £ is,

GmXm _Gm^
4 = along A£
(-n^/)2 " 3/2
8/110 ^ Fundamental Physics (XI)

Gm X m Gm^
Force on mass m at A due to mass m at F is, = along AF.
/2 /2

Gm^
Resultant force due to /, and/5 is, '"i=i//i^ + /5" + 2/,/5™s120° = /2
along AD.

■^Gm^ _ Gm^
Resultant force due to/2 and/j is. ^2=l/-^2+/4+2/2/4 cos 600 = 3/2 ~V3/2
along AD.

Gm2 Gm2 Gm2 _ Gw2 ^

oww
Net force along AD = Fj + F2 +f^ =
"7^^>^/2 ^ 4/2 /2 L Vs 4
36. A satellite is to be placed in equatorial geostationary orbit around earth for communication,
(a) Calculate height of such a satellite.
(/>) Find out the minimum number of satellites that are needed to cover entire earth so that at least

e
one satellites is visible from any point on the equator.

re
FFrlo
[M = 6xl0^kg,R = 6400 km, F= 24 h, G = 6-67 x ir“ SI units]
Ans. Here, G = 6-67 x 10“'' SI units, M = 6 x 102“* kg ,

rF
ee
R = 6400 km = 64 x 10^ m, r = 24 h = 24 X 60 X 60 s

rF
ouru
-|i/3
G/W7'2
(fl) h = -R
4 k-

fosor
skf
nl/3
(6-67 X10“") X (6 X 1Q24 ) X (24x 60 X 60)2
ooko
-64x10^
Yo
4x(3-142)2
Y
Bo

= 4-23 X 10^ - 64 X 10^ = 3-59 x lO’ m

reeB

(/?)From Fig. 8(N).13,

R I
ooY
uur

COS0 =
R +h
1+A
ad

R
dY

Now
h _ 3-59x1Q2 = 5-61
nind

R ~ 64x10^
Re
F
Fi

1
Hence, cos 6 = = 0-1513 = COS 81“I8'
1 + 5-61
0 = 81° 18' and 2 6 = 81° 18'x 2

360° 360°
If n be the number of satellites that are needed to cover entire earth, then n = = 2-31 = 3
20 81°18'x2

37. Earth’s orbit is an ellipse with eccentricity 0*0167. Thus, earth’s distance from the sun and speed as
it moves around the sun varies from day to day. This means that the length of the solar day is not
constant through the year. Assume that earth’s spin axis is normal to its orbital plane and find out
the length of the shortest and the longest day. A day should be taken from noon to noon. Does this
explain variation of length of the day during the year ?
Ans. As earth orbits the sun, the angular momentum is conserved and areal velocity is constant. Let m be the
mass of earth, Vp, be the velocity of earth at perihelion and and aphelion respectively co^ and O)^ are the
angular velocity of earth at perihelion and aphelion respectively.
GRAVITATION 8/111

According to law of conservation of angular momentum

mvprp=^mv„r^
or

2 2
or a or co„
p p
- 0).a ra

0)
P — a
or ..(0
2
(0 r
a
P

If a is the semimajor axis of earth’s orbit, then

rp = a(l-e) and r^ = a(l+e)

oww
^p (1 + e)^ (1 + 0-0167)^ = 1-0691
From (i).
CO.
a (l-e)2 "(1-00167)2
If Tp , are the time period of spinning motion of earth at perigee and apogee respectively, then

ee
2ti 271
and CO

FFrlo
CO a
p T T

r
a
P

rF
ee
-2-= 1-0691 ...(«)
ouru
rF
CO T
a p

Let T be the mean solar day (= 24 h). It is the geometric mean of and Tp. Therefore,
7’^r^ = r2 = 24x24 ffosor ...(«i)
os k
Solving (») and {Hi), we have Tp = 23-211 h and
Tp = 24-815 h
This does not represent actual variation of the length of the day during the year.
ook
Yo
Y

38. A satellite is in an elliptical orbit around the earth with aphelion of 6 and perihelion of 2 /f where
Bo

R - 6400 km is the radius of the earth. Find eccentricity of the orbit. Find the velocity of the satellite
reeB

at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of
radius 6 1? ? [G = 6-67 x 10"*^ SI units and = 6 x kg]
oouY
ur

Ans. Here, = 2 /? and r^ = 6R

ad

Weknow, Pp = a (\-e) = a-aer^ = a {\ + e) = a+ae

dY

’’a ~ ^P _ 6R-2R ^ -1 = 0-5

nidn

e =
6R+2R 2
Re
F
Fi

Let m be the mass of satellite.

According to law of conservation of angular momentum ;

angular momentum at perigee = angular momentum at apogee

^p rp =
or ^ = !s. = ^ = 3 or v.=3v a
P
...(0
V 2R
a rp

Suppose M is the mass of earth.

1 2 ^
Total energy of satellite at perihelion — 2 r
P

1 2 GM m
Total energy of satellite at aphelion = —mv a r
a
8/112
'P'uuieefr'^ Fundamental Physics (XI) VOL.II

According to law of conservation of eneigy, we have

1 2 -
GMm 1
-~mv“ -
2 GM m
2 P 2 «

or
vl-vl=2GM —-1 =2GM !a t
y ** it»
L'i- I
r r
^ P J

or jP
r_ r
L o P J

w
or 8v2=2GAf
L 'i'i. J

Flo
or ^:2_2GM Pa-Pp _2GMf6/?-2/?l GAf
% “
8 8 6Rx2R

ee
r_ r \2R
L o P j

Fr
nl/2
^ ^ (6-67xlO»^)x(6xlO^Q) = 2-28 X 10^ ms-' = 2*28 km/s
"1 12x6-4x10®
for
ur
From equation (/). Vp = 3 x 2-28 = 6-84km/s
ks
Velocity of satellite in an orbit of radius r is given by
Yo
v =
oo
eB

Forr = 6/?,
M-l (6-67xl0““)x(6xl024) 6x6-4x10® = 3-23 X 1Q3 m/s = 3-23 km/s
ur

Hence to transfer the satellite to a circular orbit at aphelion, the additional speed to be given to satell'^e ii
ad
Yo

= 3-23 - 2-28 = 0*95 km/s. This can be done by firing rockets from the satellite.
d
Re
in
F
GRAVITATION 8/113

!^F;o.r£
* MCQs in Physics for NEET
m^■*‘' >4|Pradeep*s Stellar Series....

ww
● MCQs in Physics for JEE (Main)
V#'

FF loo
Multiple Choice Questions (with One Correct
Anyvv^if)|

ree
4. Two small satellites move in a circular orbits
I. Kepler’s laws around the earth, at distance r and (r + dr) from

reFe
the centre of the earth. Their time periods of
1. The largest and shortest distance of earth from
oroFr
rur
rotation are T and T + dT (A r << r, AT < < T).
the sun are r, and r2- its distance from the sun
s ff
Then
when it is perpendicular to the major axis of the
orbit drawn from the sun is
k
ib) AT = ~7' —
YYouo
(a) A7’=-r —
2 r
okso

'1+^2 n+t-2 2 r
(a) ib)
4
BBoo

Ar

2r, .r (c) Ar = |r— 3 r

id) AT=T
r
r ee

1 '2 rj+rj
(c) (^0
'i+^2 2 5. A system of binary stars of masses and mg are
moving in circular orbits of radii and rg
ad
ouur

2. A satellite is launched into a circular orbit of

respectively. If and Tg are the time periods of
Yo

radius R around the earth. A second satellite is masses and respectively, then
launched into an orbit of radius 1 -Ol R. The period
xl/2
of the second satellite is larger than the first one
d

(«) ^ =
idnY
Re

by approximately
Tb rs}
ib) 10%
FFin

(a) 0-5%
(c) 1-5% id) 3-0%. ib) T^>TgiifrA>rg)
3. The time period of a satellite revolving around (f) Ta > Tg (if > mg)
earth in given orbit is 7 hours. If the radius of the id) T^ = Tg
orbit is increased to three time its previous value, 6. A satellite moves in a circle around the earth. The
then approximate new time period of the satellite radius of this circle is equal to one half of the
will be; radius of the moon’s orbit. The satellite completes
(a) 40 hours ib) 36 hours one revolution is :

(c) 30 hours id) 25 hours (a) 1/2 lunar month ib) 2/3 lunar month
(JEE Main 2022) (c) 2“^^ lunar month id) 2^^ lunar month
ANSWERS

1. (c) 2. (c) 3. ib) 4. (a) 5. id) 6. (c)

8/114
^ieidecfr'4. Fundamental Physics (Xl)EZsl9D
II. Newton’s law of Gravitation with cavity now applies a gravitational force
on the same particle at P. The ratio F2/FJ is
7. If the mass of the Sun were ten times smaller and
the universal gravitational constant were ten times
larger in magnitude, which of the following is not
correct ?

(a) Time period of a simple pendulum on the

Earth would decrease
(b) Walking on the ground would become more -P
difficult
(c) Raindrops will fall faster
(d) ‘g’ on the Earth will not change
(NEET 2018)
8. Two masses, 800 kg and 600 kg are at a distance

ww
0-25 m apart. The magnitude of total force (a) 3/25 (b) 9/50
experienced by a body of mass 1 kg placed at a (c) 22/25 (d) 41/50
point distance 0-2 m from the 800 kg mass and 11. In the above question, if the solid sphere is a large

FF loo
0-15 m from the 600 kg mass is :
rock, what is the gravitational acceleration at a
(a)34x 10“^ N (b) 2-22 X 10-6 N
point on the surface of the rock at a point just

ree
(c) 3-22 X 10"^ N (d) 2-22 X 10-^ N above the cavity ?
9. A uniform sphere of mass M and radius R exerts

rFee
GM GM
a force F on a small mass m situated at a distance (a) (b)
of 2 /? from the centre O of the sphere. A spherical R- 2r2

F
oor r
rur
portion of diameter R is cut from the sphere as GM IGM
(c)
s ff
shown in Fig. 8(CF).l. The force of attraction id)
8F^ 8/?^
between the remaining part of the sphere and the
12.
Suppose, the acceleration due to gravity at the
k
mass m will be
YYoou

Earth’s surface is 10 ms-^ and at the surface of

ookos

Mars it is 4-0 ms'^. A 60 kg passenger goes from

BBo

the Earth to the Mars in a spaceship moving with

a constant velocity. Neglect all other objects in
re

the sky. Which part of Fig. 8(CF).3 best represents

the weight (net gravitational force) of the
ouur
ad

passenger as a function of time ?

Yo

8(CF).3
dY
Re
idn
FFin

2F
(a) - (b)
3 3
4F IF
(c) id)
3 9
10. A solid sphere of uniform mass M and radius R
■►Tlrne
applies a gravitational force of attaction equal to
F| on a particle of mass m placed at point P
distance 3 R from the centre of the sphere. A
spherical cavity of radius R/2 is now made in the id) A ib) B
sphere as shown in the Fig. 8(CF).2. The sphere ic) C id) D
ANSWERS

7. (d) 8. (h) 9.(d) 10. (rf) !!.(/,) 12. (c)

GRAVITATION 8/115

13. A point mass M is placed at a distance L from one 16. Two spherical bodies of mass M and 5 M and radii
end of a uniform rod of mass M and length L as R and 2 R aie released in free space with initial
shown in Fig. 8(CF).4. The gravitational force separation between their centres equal to 12 R. If
experienced by point mass due to rod is they attract each other due to gravitational force
only, then the distance covered by the smaller
FIGURE 8(CF).4
body before collision is
M M (a) 2-5 R {b) 4-5 R
(c) 7-5 R id) 1-5 R
rt- -m- M
L L (AIPMT 2015)
17. Four identical particles of each mass M are located
GM^ GM^ at the corners of a square of side a. What should
(.a) ib)
I? 21? be their speed, if each of them revolves under the
influence of other’s gravitational field in a circular
2GM^ agm'^

ww
id)
orbit, circumscribing the square ?
(c)
3l2 91?
14, A large spherical mass M is fixed at one position

FF loo
and two identical point masses m are kept on a
line passing through the centre of M (See Fig.

ree
8(CF).5). The point masses are connected by a
rigid massless rod of length / and this assembly is
free to move along line connecting them. All three

reFe
masses interact only through their mutual
gravitational interaction. When the point mass
oor rF
rur
nearer to A/ is at a distance r = 3 I from M, the
s ff
M
tension in the rod is zero for m = k . The
k
288 GM GM
(b) 116
YYoou
value of k is (a) 1-35
okso

V a a
BBoo

FIGURE 8(CF).5
GM GM
(c) 141 id) 1-21
r ee

M m a a

r
*\4-l (JEE Main 2019)
ouur
ad

18. A body weighs 72N on the surface of earth. What

Yo

io)3 ib)^ is the gravitational froce on it, at a height equal

to half the radius of the earth?
ic)l id) 9
d
idnY
Re

ia) 32 N {b) 30 N
(JEE Advanced 2015)
(c) 24 N (J)48N (NEET2020)
FFin

15. Four particles, each of mass m, are moving along

a circle of radius r under the influence of mutual 19. The ratio of the weights of the Earth’s surface to
gravitational attraction. The speed of each particle that on the surface of planet is 9 : 4. The mass of
will be

GM 2V2GM
the planet is ^ th of that of the Earth. If R is the
ia) ib) radius of the Earth, what is the radius of the
y r r
planet? [Take the planet and Earth to have the
same mass density]
ic)
GM
2(V2 + 1) id) GM (2V2 + I) R R R
R
y r r 4 (.)- ib) - ic) - id) -
(JEE Main 2014)
(JEE Main 2019)

ANSWERS

\3.(b) 14. (c) 15. (r/) 16.(0 17.(0 18.(0 19.(0

I
8/116
'P’uteUe^'^ Fundamental Physics fXltroSTTl
20. Two objects of equal masses placed at certain 1
distance from each other, attracts each other with 24. A planet of radius /? = — (radius of earth) has
10
a force F. If one third mass of one object is the same mass density as Earth. Scientists dig a
transferred to the other object, then the new force well of depth /?/5 on it and lower a wire of same
will be:
length and of linear mass density 10”^ kg m -1
into it. If the wire is not touching anywhere, the
(a)|p («-
force applied at the top of the wire by a person
holding in place is (take the radius of the earth
(d) F (JEE Main 2022) = 6 X 10^ m and acceleration due to gravity on
earth is 10 ms“^).
(a) 96 N ib) 108 N
m. Acceleration due
(c) 120 N (d) 150 N
to gravity and its variation
(JEE Advanced 2014)

ww
21, If the radius of earth were to increase by 1%, its 25. A particle hanging from a spring stretches it by
mass remaining the same, the acceleration due to 1 cm at earth’s surface. How much the same
gravity on the surface of earth will
particle stretches the spring at a place 1600 km

FF loo
(a) increase by 1 % (b) decrease by 2 % above the surface of earth (R = 6400 km)
(c) decrease by 1 % (d) increase by 2 %. (a) 16/50 cm (b) 16/25 cm

ree
22. The dependence of acceleration due to gravity g (c) 25/16 cm (d) 50/16 cm.
on the distance r from the centre of the ea^, 26. What is the percentage change in the value of g

reFe
assumed to be a sphere of radius R of uniform as we shift from equator to pole on the surface of
density is as shown in Fig. 8(CF).7 below :
earth ? (Given equatorial ra^us of earth is greater
oor rF
rur
than polar radius by 21 km and mean radius of
s ff
earth is 6300 km).
(a) 0-52% ib) 0-61%
k

(c) 1-67% id) 6-7%

YYoou
okso

27. The reading of a spring balance corresponds to

BBoo

100 N while situated at the north pole and a body

is kept on it. The weight record on the same scale
r ee

if it is shifted to the equation is (lake g = 10 m/s^,

and radius of earth, /? = 6-4 x 10^ km)
ouur
ad

(a) 99-66 N (/?) 110 N

Yo

(c) 97-66 N id) 106 N

(AIIMS 2015)
d
idnY
Re

28. The ratio of radii of earth to another planet is 2/3

and the ratio of their mean densities is 4/5. If an
FFin

The correct figure is astronaut can jump to a maximum height of 1-5 m

ia) (4) ib) (1) on the earth, with the same effort, the maximum
height he can jump on the planet is
(c) (2) (d) (3)
id) 1 m ib) 0-8 m
(JEE Main 2017 ; AIPMT Main 2010)
(c) 0-5 m id) 1-23 m
23. A body weighs 200 N on the surface of the earth.
ie) 2 m (Kerala PET 2009)
How much will it weigh half way down to the
centre of the earth ? 29. A body is weighed with a spring balance in a train
ia) 150 N (b) 200 N at rest, shows a weight W. When the train begins
(c) 250 N id) lOON
to move with a velocity v around the equator from
west to east and if the angular velocity of the train
(NEET 2019) is (0 then the weight shown by spring balance is
ANSWERS

20. (c) 21, ib) 22. ia) 23.id) 24. ib) 25.ib) 26. ib) 27.ia) 28. ib)

t
GRAVITATION 8/117

2i;co (a) 48 kg wt (b) 48-83 kg wt

(a) W (b) W 1 + (c) 47-84 kg wt (d) 47 kg wt
8
(JEE Main 2021)
2v(d
(c) W l- (d) w{i + v^/r)- rv. Gravitational Intensity,
8
Gravitational potential
30. What is the depth at which the value of
and Gravitationalenergy
acceleration due to gravity becomes 1/n times the
value at the surface of earth? (radius of earth 34. A spherically symmetric gravitational system of
= R) particles has a mass density
R for r<R
p=r.“0
R
(a) - (b) —
n n for r>R

ww
R{n-\) Rn where Po is a constant. A test mass can undergo
(.c) (d) circular motion under the influence of the
n (n-1) gravitational field of particles. Its speed v as a

FF loo
(NEET 2020) function of distance r (0 < r < <») from the centre
of the system is represented by
31. The value of the acceleration due to gravity is

ree
FIGURE 8(CF).8
at a height b = — {R = radius of the earth) from

reFe
the surface of the earth. It is again equal to gj at a
depth d below the surface of the earth. The ratio
oor rF
rur
(d/R) equals :
s ff
4 1

(«) 5 (i) 3
k
■►r
R
YYoou
okso

5 7
© o
(c) 9
BBoo

(JEE Main 2020)

r ee

32. The acceleration due to gravity on the earth’s

surface at the poles is g and angular velocity of
ouur
ad

the earth about the axis passing through the pole

Yo

is o>. An object is weighed at the equator and at a

height h above the poles by using spring balance.
d
idnY
Re

If the weights are found to be the same, then h

is: (A « /?, where R is the radius of the earth)
35. The gravitational field due to a mass distribution
FFin

R^oP' R'^(o^ is / = K/r^ in the X-direction. (K is a constant).

(a) (b) Taking the gravitational potential to be zero at
8 ^8
infinity, its value at a distance x is
R?-(i?- R^isP- (a) Klx (b) K/2x
ic) id)
(c) KI:P id) K/2x^.
(JEE Main 2020) 36. Two concentric spherical shells A and B of radii
and 2 /? and mases 4 M, and M, respectivelyare
33. The weight of a person on pole is 48 kg wt then
placed in space as shown in Fig. 8(CF).9. The
the weight of the person on equator is : [Given
redius of earth - MOO km; R oP = 0-034 ms“^ ; gravitational potential at P at a distance
r(R<r<2R) from the centre of shells is
g = 10 ms'^]
ANSWERS

29.(c) 30. (c) 31. (c) 32. id) 33.(c) 34. (c) 35. id)
8/118
^ Fundamental Physics (XI) IV»7WM

FIGURE 8(CF).11
m

I -P
d

4Gm 2Gm
(a) (b)
d-^i^+Ad^

ww
4GM 9GM
(a) - (b) -
R 2R
(c)
2^G. m

(d) none of the above

d-^l^+4d^

Floo
4GM
(c) - (d) none of these
3R 39. A point P lies on the axis of a irng of mass M and

ree
radius a, at a distance a from its centre C. A small
37. From a solid sphere of mass M and radius R, a particle starts from P and reaches C under

rFee
R gravitational attraction only, Fig. 8(CF).12. Its
spherical portion of radius — is removed, as speed at C will be

F
2

oor r
rur
shown in Fig. 8{CF).10. Taking gravitational FIGURE 8(CF).12
s ff
potential V= 0 at r = «>, the potential at the centre
of cavity thus formed is (G = gravitational
osk

constant)
YYoou
P
oook

a-
eBB
uur r

'2GM
ad

2GM 1
(a) ib) 1-
Yo

a
a

'2GM
iS-\) id)
dY

(c) zero
Re

a
idn

40. Infinite number of masses, each of 1 kg, are placed

FFin

along the x-axis at x = ± 1 m, ± 2 m, ± 4 m, ± 8 m,

2GM GM
ia) ~ ib) - ± 16 m The magnitude of the resultant
3R R gravitational potential in terms of gravitational
constant G at the origin (x = 0) is
GM -2GM
ic) - id) ia) G/2 ib) G
2R R
(c) 2G id) 4 G
(JEE Main 2015) ie) 8G (Kerala PET 2008)
38. A uniform rod of mass m and length I is taken. 41. The energy required to take a satellite to a height
Find the gravitational field intensity at point P at h above Earth surface (radius of Earth = 6-4 x
distance d which is on the peipendicular bisector 10^ km) is £j and kinetic energy required for the
of the rod as shown in Fig. 8(CF).I1. satellite to be in a circular orbit at this height be
ANSWERS

36. (</) 37. ib) 38. {b) 39. {b) 40. {d)

i
GRAVITATION 8/119

£2- The value of h for which Ej and £2 equal ~2Gm 2 / 1

IS : (i’) 2+
/ V2
(a) 1-6 X lO^km (fc) 3-2 X 10^ km
(c) 64 X 10^ km (f/) 1-28 X lO'^km -V2G m 2 /
1
(c) V2-
(JEE Main 2019) / V2
42. A satellite is moving with a constant speed v in a
1 ^
circular orbit around the earth. An object of mass id) V24-
I
m is ejected from the satellite such that it just
escapes from the gravitational pull of the earth. 47. Energy required to transfer a 400 kg satellite in a
At the time of ejection, the kinetic energy of the circular orbit of radius 2 /? to a circular orbit of
object is : radius 4 R, where R is the radius of the earth is
(fl) Imv^ (b) m (g = 9-80 ms-2, /; = 6 4 X 10^ m)
{b) 3-13 X 10^ J

ww
(c) mv^l {d) 3 m v2/2 (a) 1-65 X lO^J
(JEE Main 2019) (c) 6-26 X 10^ J id) 4-80 X 10^ J
43. A particle of mass M is situated at the centre of a 48. There is a crater of depth (/?/100) on the surface

FF loo
spherical shell of same mass and radius a. The of moon, where R is the radius of moon. A particle
magnitude of the gravitational potential at a point is projected vertically upwards from the crater

ree
situated at a/2 distance from the centre, will be with the velocity which is equal to the escape
velocity v from the surface of the moon. The

reeF
2GM 3GM maximum height attained by the projectile is
(a) (b)
a a (a) R (b) 85 E

oroFr
r ur
(c) 99 R (d) \00R
AGM GM
s ff
ic) id)
a
49. A body is released at a distance r(r>R) from the
centre of earth. The velocity of the body when it
a
k
(AIPMT Main 2011, AIPMT 2010) strikes the surface of the earth is
YYouo
kos o

44. A person brings a mass of 1 kg from infinity to a

point P. Initially the mass was at rest but it moves ia) ^2gr ib) ^2giR + r)
BBoo

at a speed of 2 ms“’ as it reaches at P. The work -1I/2 1/2

r ee

1 1
done by the person on the mass is - 3 J. The (c) R 2s
potential at P is /J
ad
ouur

ia) -2J/kg ib) -3 J/kg 50. A rocket is launched vertically from the surface
Yo

(c) - 5 J/kg id) -7 J/kg. of earth with an initial velocity u the height up to
45. The work done to raise a mass m from the surface which the rocket can go from the surface of earth,
before falling back is
d
Re

of the earth to a height h, which is equal to the

dinY

radius of the earth (R) is : ia) -JtJr ib) u^/2g

FFin

ia) mgR ib) 2mgR -1

24' 1
25 1
ic) id)
id) ^ mg R
1 2 R
ic) -mgR u
u 2 R]
51. The earth is assumed to be a sphere of radius R. A
(NEET 2019)
platform is arranged at a height R from the surface
46. Four particles each of mass m are placed at the of the earth. The escape velocity of a body from
vertices of a square of side 1. What is the potential this platform is/v, where v is its escape velocity
energy of the system ? from the surface of the earth. The value of/is

-j2Gm- 1 ia) 1/2 ib) V2

ia)
I
(<^) U^l2 id) 1/3

ANSWERS

41. ih) 42. ib) 43. (b) 44. (c) 45. (c) 46. ib) 47. ib) 48. ic) 49. ic) 50. (d) 51. ic)

\
8/120
^ Fundamental Physics (XI) OTWfl
52. What is the minimum energy required to launch rotation plane. In order to escape from the
a satellite of mass m from the surface of a planet gravitational field of this double star, the
of mass M and radius /? in a circular orbit at an
altitude of 2 /? ?
minimum speed that meteorite should have at 0
is : (Given Gravitational constant = 6-67 x
(a)
SGmM
(b)
IGmM 10“’* Nm^ kg~^)
6R 3R
(a) 2-8 X 1()5 ms -1 ib) 14 X 10^ ms"*
CmM GmM (c) 24 X 10“ ms"* (d) 3-8 X i0“ m.s -1
(c) (d)
2R 3R (JEE Main 2019)
(JEE Main 2013} 56. Consider two solid spheres of radii = 1 m,
53. A thin uniform annular disc (see Fig. 8(CF).13) /?2 = 2 m and masses A/j and M2 respectively.
of mass M has outer radius 4 R and inner radius The gravitational field due to two spheres 1 and
3 R. The work required to take a unit mass from 2 are shown in Fig. 8(CF).14. The value of

ww
point P on its axis to infinity is (M1/M2) is

FF loo
ree
reFe
oroFr
rur
k s ff
YYouo
okso

2GM
(a) (4-^-5) (b) -
2GM
(4V2-5)
BBoo

7R 1 I
IR (a)~
r ee

GM 2GM
(c) 4R (d) 5R
(V2-I) 1

(0^
ad
ouur

(IIT 2010)
Yo

(JEE Main 2020)

54. A particle of mass m is projected with a velocity
V = k Vg (k < 1) from the surface of Earth (v^ is 57. A body of mass 60 g experiences a gravitational
force of 3*0 N when placed at a particular point.
d

the escape velocity). The maximum height above

idnY
Re

the surface of earth reached by the particle is The magnitude of the gravitational field intensity
at that point is:
FFin

Rk^ / If \2
id) ib) R (a) 0-05 N/kg ib) 50 N/kg
l-k (c) 20 N/kg id^ 180 N/kg
{ ic
\2 R^k (NEET 2022)
(c) R id)
l +k (1 + *) V. Satellite and Escape velocity
(NEET 2021) 58.
Two particles A and B are moving in uniform
55, Two stars of masses 3 x 10^* kg each, and at circular motion in concentric circles of radii
distance 2 x 10*’ m rotate in a plane about their and rg with speed and respectively. Their
common centre of mass 0. A meteorite passes lime period of rotation is the same. The ratio of
through O moving perpendicular to the stars angular speed of A to that of B will be :

ANSWERS

52.(«) S3.(«) 54.(a) 55.(«) 56.{«) 57. (b)

GRAVITATION 8/121

(a) (b) : Vq 62. A satellite in a force-free space sweeps stationary-

id) 1: I dM
(c) rg: interplanetary dust at a rate = pu , where v
(NEET 2019) dt

59. Kepler’s third law states that square of period of is the speed of escaping dust w.r.t. satellite and M
is the mass of satellite at that instant. The
revolution (T) of a planet around the sun, is
acceleration of satellite is
proportional to third power of average distance r
between sun and planet i.e. = K where K is (a) (b) - P v^/2 M
constant.
(c) - p v^m id) -M p/u-
If the masses of sun and planet are M and m 63. A satellite is moving around the earth with speed
respectively then as per Newton’s law of gravi V in a circular orbit of radius r. If the orbital radius
tation force of attraction between them is is decreased by 1 % its speed will
(a) increase by I % ib) increase by 0-5%
F= .9^^, where G is gravitational constant. The

ww
(c) decrease by 1% id) decrease by 0-5%.
relation between G and K is described as : 64. The radii of circular orbits of two satellites A and
B of the earth, are 4 /? and R, respectively. If the
ia)GMK=A%^ ib)K-G
speed of satellite A is 3 V, then the speed of

Flo
1 satellite B will be
id) GK=4nr^

e
(c) K = -

e
G (a) 3V/4 ib) 6V

reer
(AIPMT 2015) (c) 12 V id) 3 V/2

rFF
60. A particle is moving with a uniform speed in a (AIPMT 2010)
uur r
circular orbit of radius R in a central force 65. Two bodies each of mass M, are kept fixed with a
inversely propertional to the nth power of R. If
ffoor
separation 2L A particle of mass m is projected
period of rotation of the particle is T, then for from the mid point of the line joining their centres,
sks
any value of n perpendicular to the line. The gravitational
YYoo

constant is G. The correct statement(s) is (are);

ooko

-+l
(a) the minimum initial velocity of the mass m to
eBB

(a) 7’«/?3/2 ib) TocR2

escape the gravitational field of the two bodies
(d) TocR'^'^ is AylGM/L
uurr

(JEE Main 2018)

(b) the minimum initial velocity of the mass m to
ad

61. A rocket is launched normal to the surface of

Yo

escape the gravitational field of the two bodies

the earth, away from the sun, along the line
joining the sun and the earth. The sun is 3 x 10^ is 2^GM/L
dY

times heavier than the earth and is at a distance

Re

(c) the minimum initial velocity of the mass m to

innd

2-5 X 10“^ times larger than the radius of the escape the gravitational field of the two bodies
earth. The escape velocity from earth’s gravi
FFi

tational field is = 11-2 ms“^ The minimum is ^2GM/L

initial velocity (u^) = 11-2 ms"^ The minimum (d) the energy of mass m remains constant.
initial velocity (f^) required for the rocket to (JEE Advanced 2013)
be able to leave the sun-earth system is closest 66. A satellite is moving with a constant speed V in a
to (Ignore the rotation of the earth and the circular orbit about the earth. An object of mass
presence of any other planet m is ejected from the satellite such that it just
-1
(a) = 22 km s“’ (b) = 72 km s escapes from the gravitational pull of the earth.
(c) = 42 km s“* id) Uj = 62 km s“* At the time of its ejection, the kinetic energy of
(JEE Advanced 2017) the object is

ANSWERS

58.id) 59.(a) 60.(c) 61.(c) 62.(c) 63.(b) 64.ib) 65.ib)

8/122
“P%Ade€^ '4. Fundamental Physics (XI) LV«TII
1
- mV2 70. A bullet is fired vertically upwards with a velocity
(a) ^ {b) mV^ V from the surface of a spherical planet when it
reaches its maximum height, its acceleration due
-mV^ (d) 2mV^ (nT2011)
w 2 1
to the planet’s gravity is — th of its value at the
4
67. A satellite is revolving in a circular orbit at a height
It from the earth’s surface (radius of earth R ; surface of the planet. If the escape velocity from
h « R). The minimum increase in its orbital the planet is Vescape = V yjN, then the value of N
velocity required, so that the satellite could escape
from the earth’s gravitational field, is close to is : (ignore energy loss due to atmosphere).
(Neglect the effect of atmosphere) (o) 1 (b)2

(a) -^gR ih)

(c)3 (d}4
(JEE Advanced 2015)

ww
(c) id) 4gRirl2~l) 71. The escape velocity from the earth’s surface is v.
(JEE Main 2016) The escape velocity from the surface of another
planet having a radius fourtimes that of earth and
68. A particle is fired vertically upward from the

Flo
same mass density is
surface of earth with a velocity ofkv^, where

e
(a) 4 V (b)v
is the escape velocity and k < 1. Neglecting air

e
resistance, the maximum height to which it will

reer
ic)2v (d)3v (NEET 2021)

rFF
rise, measured from the centre of the earth, is 72. A planet A has a mass M and radius R. Planet B
{R - radius of earth)
has half the mass and half the radius of planet A.
uur r
R R If the escape velocities from the planets A and B
ffoor
(fl) V
l-k^ are and Vg respectively, then — =
n
sks
V
B
4’
i-k^
YYoo
the value of n is
(C)
ooko

(d)
R R (a) 3 (b)2
eBB

(AMU PET 2011) (c) 4 id) 5

69. A black hole is an object whose gravitational field (JEE Main 2020)
is so strong that even light cannot escape from it.
uurr

To what approximate radius would earth (mass = 73. Two satellites are revolving in the orbit around a
ad

planet. The ratio of their time period is 1 : 8. Find

5-98 X 10^ kg) have to be compressed to be a
Yo

black hole ? the ratio of their angular velocities,

(a) 1/2 (b) 8/1
(a) 10“^ m (b) 10-^ m
dY
Re

(c) 10"^ m (d) 100 m (c)2/l (d) 2/3

innd

(AIPMT 2014) (JEE Main 2021)

FFi

EH Multiple Choice Questions {with One or More than One Correct Answers)
74. In case of earth
75. Which of the following statements are true about
(a) potential is minimum at the centre acceleration due to gravity
(a) ‘g’ is zero at the centre of earth
(b) potential is zero, both at centre and infinity
(b) 'g' decreases if earth stops rotating on its axis
(c) field is zero both at centre and infinity (c) ‘g’ decreases in moving away from centre if
(d) potential is same, both at centre and infinity r>R
but not zero
(d) ‘g’ decreases in moving away from centre if
r<R
ANSWERS

66. (ft) 67. (d) 68. (fl) 69. (c) 70. (ft) 71. {a)
72. (c) 73. (ft) 74. («,c) 75. (a.c)
GRAVITATION 8/123

76. If two satellites of different masses are revolving

in the same orbit, they have same
(a) speed (b) energy
(c) time period
{d) angular momentum
77. The escape velocity of an object projected from
the surface of a given planet is independent of
(a) mass of the planet
{b) the mass of the object
(c) the radius of the planet
(d) the direction of projection
78. Consider a planet moving in an elliptical orbit

ww
(a) the gravitational force due to this object at the
round the sun. The work done on the planet by
origin is zero
the gravitational force of the sun
(b) the gravitational force at the point B (2, 0, 0)
(a) is zero in some part of the orbit

Flo
is zero
(b) is zero in no part of the motion
(c) the gravitational potential is the same at all

ee
(c) is zero in any small part of the orbit
points of the circle y^ + = 36

rere
(d) is zero in one complete revolution
id) the gravitational potential is the same at all

r FF
79. Three point masses are at the corners of an
points on the circle = 4. (IIT)
equilateral triangle of side r. Their separations do
uurr
82. A ring has a total mass M but non-uniformly
not change when the system rotates about the
centre of the triangle. For this, the time period of
foor
distributed over its circumference. The radius of

rotation must be proportional to the irng is R. A point mass m is placed at the centre
ks s
of the irng. Workdone in taking away this point
Yoo
(a) (b)r
ooook

-1/2 mass from centre to infinity is

(c)m id) m
GM m GMm
eBB

80. Two objects of masses m and 4 m are at rest at an id) - ib)

R R
infinite separation. They move towards each other
under mutual gravitational attraction. If G is the GM m GM m
uurr

universal gravitational constant, then at separation r (c) - id)

ad

2R IR
(a) the total energy of the two objects is zero
Yo

83. Suppose universal gravitational constant starts to

(lb) net angular momentum of both the objects is
dY

decrease, then
zero about any point
(a) length of the day, on earth, will decrease
Re

(c) the total K.E. of the objects is 4 G rn^lr

innd

ib) length of the year will decrea.se

id) their relative velocity of approach is
FFi

nI/2
(c) earth will follow a spiral path of increasing
(SGm radius

id) kinetic energy of earth will decrease

81. A .solid sphere of uniform density and radius 4 84. Two satellites of same mass of a planet in circular
units is located with its centre at the origin O of orbits have periods of revolution 32 days and 256
coordinates. Two spheres of equal radii 1 unit, days. If the radius of the orbit of the first is R,
with their centres at A (- 2, 0, 0) and B (2, 0, 0) then the
respectively are taken out of the solid leaving ia) radius of the orbit of the second is 4 /?
behind spherical cavities as shown in the Fig.
ib) radius of the orbit of the second is 8 /?
8(CF).15, then

ANSWERS
76. (rt.c) 77. {h,d) 78. (c.d) 79. ia.d) 80. ia.b,c) 81. (a,c,d)
82.(b) 83. ia.c.d)
8/124 "P^iidee^ ’4. Fundamental Physics (XI) kviwii

(c) total mechanical energy of the second is also has uniform density p and its mass is {Mp +
greater than that of the first Mq). The escape velocities from the planets P, Q
(</) kinetic energy of the second is more than that
am
R, are Vp, Vq, Vp, respectively, then
of the first
(a)VQ>Vp>Vp
85. Two spherical planets P and Q have the same uni {b)Vp>VQ>Vp
form density p, masses Mp and Mq , and surface {c)Vp/Vp = 3
areas A and 4A, respectively. A spherical planet R
{d)Vp/VQ=m (HT 2012)

DQ Multiple Choice Questions (Based on the given Passage/Compreh ensipri).

Each comprehension given below is followed by some multiple choice questions. Each question has one
correct option. Choose the correct option.

ww
D There are three identical
(c)
3Gm^
(d)
■y/SGm^
point mass bodies each of mass m located at r2 r2
the vertices of an equilateral triangle with

Flo
side r. They are exerting gravitational force 87. At what speed must they move if they all revolve

e
of attraction on each other, which can be under the influence of one another’s gravitation

e
reer
given by Newton’s law of gravitation. Each in a circular orbit circumscribing the triangle still

rFF
mass body producesits gravitationalfield in preserving the equilateral triangle
the surrounding region. The magnitude of
uur r
gravitational field at a point due to a point ffoor
Gm 2Gm
mass body is the measure of gravitational {a) ib)
r V r
intensity at that point. The gravitational
sks
potential at a point in a gravitational field Is
YYoo
ooko

the amount of workdone in bringing a unit (c)

Gm
(d)
mass body from infinity to the given point r
eBB

without acceleration.
88. Work done in taking one body far away from the
other two bodies is
uurr
ad

Gm^
Yo

Gm^
(a) - (b)
r r
dY
Re

2 Cm2 2Gm2
innd

{c) (d)~
r
FFi

89. Magnitude of gravitational field at the mid point

D of arm BC of triangle ABC is
Answer the following questions :
Gm2 Gm
(a) (b)
86. The magnitude of the gravitational force on one r2 3r2
body due to other two bodies is
3Gm 4Gm
Gm2 2Gm2 (c) id)
(a) ib) r2 3r2
r2 r2

ANSWERS

84. (a.c) 85. (b,d) 86. id) 87. (c) 88. (c) 89. (d)
GRAVITATION 8/125

Time period of revolution of satellite around the

n A rocket is fired vertically
91.
earth is

upwards with a speed of t; (= 5 km s”^) from (a) 3550 s (b) 7100 s

the surface of earth. It goes up to a height h
before returning to earth. At height h a body (c) 5330 s (d) 8880 s
is thrown from the rocket with speed Vq in 92. The energy to be spent in taking the satellite out
such a way so that the body becomes a of the gravitational field of the earth is (mass of
satellite of earth. Let the mass of the earth, the satellite is 200 kg)
Af = 6 X 10^ kg ; mean radius of the earth,
ft = 6-4 X 10® m; G = 6*67 x 10"“ Nm^ kg"2 ; (a) 5-0 X 10^ J (b) lO-OxlO^J
g = 9*8 ms"^. (c) 2-5 X 10*0 J id) 5-Ox 10*0J
Answer the following questions : 93. If this satellite is to be taken at double of the

present height from the surface of the earth, tlien

ww
90. The value of h is
the new lime period of revolution is
(a) 1-5 X 10^ m ib) 3-2 X 10® m
(a) 9330 s (i>) 20080 s
(c) 3-2 X 10® m (d) l-6x 10®m

Flo
(c) 11000 s id) 29400 s

e e
m

reer
Matching Type Questions

rFF
uur r
DIRECTIONS. In each of the following questions, match column I and column II and select the correct
match out of the four given choices.
ffoor
sks
94. A satellite of mass m is orbiting around the earth of mass M and radius ft.
YYoo
ooko

Column I Column n
eBB

GMm
(A) Gravitational potential energy of satellite orbiting close to earth is (ft)
2ft
uurr
ad

GMm
(B) Kinetic energy of orbiting satellite close to earth is (?)
Yo

2ft
dY

GMm
Re

(C) Kinetic energy of orbiting satellite at height ft from the surface of earth (r)
innd

4ft
FFi

GMm
(D) Total mechanical energy of satellite orbiting close to earth is)
ft

(fl) A-p ; : C-r ; D-j ib) A-q ; B-r; C-s ; D-p

(c) A-r; B-p ; C-s ; D-^ id) As; h-p; C-r; D-^

95. A body of mass m is to be escaped from the different positions w.r.t. earth as given below, M is the mass of
earth, ft is the radius of earth and G is the universal gravitational constant.

ANSWERS

90. (d) 91. ib) 92. (n) 93. ((/) 94. id)
8/126
‘Pnadee^ '4- Fundamental Physics (XI)

Column I Column II

%GM
(A) Escape speed of a body from the surface of earth. (P)
■y 5R

'GM
(B) Escape speed of a body from the atmosphere at height R .i4)
2R
from the surface of earth.

GM
(C) Escape speed of a body from the satellite orbiting at a height R (r)
from the surface of earth. R

2GM
(D) Escape speed of a body from the atmosphere at height /?/4 from (s)

ww
R
the surface of earth.

(a) A-q ; B-p ; C-r ; D-i' (b) A-p ; ; C-s ; D-r (c) A-r ; B-j ; C-p ; D-^ (d) As ; B-r; C-q ; D-p

Floo
96. A planet of mass M, has two natural satellites with masses nty and m2. The radii of their circular orbits are R^

ee
and /?2 respectively. Ignore the gravitational force between the satellites. Define, Vj, Lj, ATj and Tj to be,
respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite

eer
1 and V2, L2, K2 and T2 to be the corresponding quantities of satellite 2. Given m,/m2 = 2 and P|//?2 = 1/4,

FFr
Match the ratios in List-I to the numbers in List-II

oorr
uur r
List-I List-II
s ff
(A) U1/U2 ip) 1/8
sk
YYoo

(B) L1/L2 i<i) 1

ooko

(C) K^!K2 ir) 2

eBB

(D) 7'i/7’2 is) 8

(a) As ; B-q ; C-p ; D-r (b) A-r; B-q; C-^ ; D-p (c) A-q ; B-r; C-p ; D-5 {d) A-q; B-r; Cs; D-p
uurr
ad

(JEE Advanced 2018)

Yo

p q
IVIatrix-Match Type Questions
r s
I—11—I T r

A I© © ©I
dY

rr
Re

DIRECTIONS. Each of the following questions contains statements given in

ind

two columns, which have to be matched. The answers to these questions have B
O©
FFin

to be appropriately bubbled. If the correct matches are A-r; As ; B-p ; B-q ;

C“p and D-^ ; D-s, then the correctly bubbled matrix will look like the one c 01© 0 0
shown here.
D
©I© 0 ©
97. Match the following columns :
Column I Column II

(A) Parking orbit ip) Polar satellite

(B) Polar orbit (?) Geostationary satellite

(C) Circular orbit ir) Communication satellite
(D) Elliptical orbit is) Remote sensing satellite

ANSWERS

95. (d) 96. ih) 97. A-p. q. .r s : B-p. s ; C-p. q. r, s : D-r, s

GRAVITATION 8/111

According to law of conservation of angular momentum

mvprp = mv^r^ or Vpr^ = v^r^
or
(.% ^p) ^p = (“a O r a

r2
or
a
.(i)
0)
a
r2
P

If a is the semimajor axis of earth’s orbit, then

oww
rp = a{\-e) and r^ = a(l+e)

% _(l + g)^ (1 + 0-0167)2 = 1-0691

From (/),
(1-0-0167)2

e
If Tp , are the time period of spinning motion of earth at perigee and apogee respectively, then

re
2n 2%
and

FFrllo
CO (0
a
p T T

reF
a
P

e
CO T
uoru £- = -«-= 1.0691
T
...(«■)

osFr
CO
a p

Let T be the mean solar day (= 24 h). It is the geometric mean of and Tp. Therefore,
T^rp = 7'2 = 24x24fkfor ...(m)
okso
Solving (li) and (Hi), we have Tp = 23-211 h and
7), = 24-815 h
This does not represent actual variation of the length of the day during the year.
Y
Yo
oo

38. A satellite is in an elliptical orbit around the earth with aphelion of 6 i? and perihelion of 2 R where
BB

R = 6400 km is the radius of the earth. Find eccentricity of the orbit. Find the velocity of the satellite
at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of
Y
r ree

radius 6 ? [G = 6*67 x 10"” SI units and 3/ = 6 x 102^ kg]

ouu

Ans. Here, rp = 2R and r^ = 6 R

ad
Ydo

We know, rp = a - e) = a - ae \ r^ = a (I + e) = a + ae
6R-2R_ -1
nidn

p _ = 0*5
e =
6R+2R 2
Re
F
Fi

Let nt be the mass of satellite.

According to law of conservation of angular momentum ;

angular momentum at perigee = angular momentum at apogee
V

or ^=3 or
Vp = 3v ...(0
^p '‘p = ^
a
V r 2R
a P

Suppose M is the mass of earth.

1 2 GM m

Total energy of satellite at perihelion - 2^^p r

p

1 2 GM m
Total energy of satellite at aphelion =-mv a r
a
8/112
Fundamental Physics (XI) VOL.II

According to law of conservation of energy, we have

1
—mv
2 GMm 1 2 GM m
2 .-— = 2-:-

or V
l-vl^2GM /pf-1 = 2GM
r. r
^
I o P J

or j.E t
r_ r
I ° P J

ww
or
Svl=2GM -r_ r
L « /> J

Flo
^.2_2Gitf _2GAff6i?-2^1 GA/

e
or

eree
8 r_ r 8 6/?x2/? 12R
L « P J

FFr
nl/2
^ ^ (6.67x10^^)x(6x1q20)
uurr
= 2-28 X 1()3 ms"‘ = 2*28 km/s

orr
"’’L 12x64x10^ sfo
From equation (0, Vp s: 3 X 2-28 = 6*84 km/s
kks
Yoo
Velocity of satellite in an orbit of radius r is given by v =
oooo
eBB

Forr = 6/f, u =
(6»67xlQ-»)x(6xlO^)
6x64x10® = 3-23 X 1Q3 m/s = 3-23 km/s
urr

Hence to transfer the satellite ot a circular orbit at aphelion, the additional speed to be given to satell'ie ii
ad
YYo

= 3*23 - 2*28 = 0*95 km/s. This can be done by firing rockets fix)m the satellite.
dd
Re
inn
F
 8/113
GRAVITATION

NEET/JEE
SPECIAL

-riv 'H
● MCQs In Physics for Nl

ww
Pradeep's Stellar Series...♦ ● MCQs In Physics for JE
sassateiga^iisafeB^yi&cjnir*^Sl^ ‘JZ

FF loo
n Multiple Choice Questions (with one Correct Answej)j

ree
4. Two small satellites move in a circular orbits
I. Kepler’s laws

reFe
around the earth, at distance r and (r + dr) from
the centre of the earth. Their time periods of
1. The largest and shortest distance of earth from
oor rF
rur
rotation are T and T + dT r < < r, AT < < T).
the sun are rj and r2- Its distance from the sun
s ff
Then
when it is perpendicular to the major axis of the
orbit drawn from the sun is
k
(a) aT = ~T — (b) AI=-
YYoou
okso

n + ^2 2 r r

(a) ib)
4
BBoo

Ar
(c) Ar = ^r— 3 r
id) AT=T
r
r ee

2rr
1 '2 rj + rj
ic) id)
n+''2 2 5. A system of binary stars of masses and nig are
moving in circular orbits of radii and rg
ouur
ad

2. A satellite is launched into a circular orbit of

respectively. If and Tg are the time periods of
Yo

radius R around the earth. A second satellite is masses and mg respectively, then
launched into an orbit of radius 1-01 /?. The period
n1/2
of the second satellite is larger than the first one
d

T (r
idnY
Re

by approximately (a) ii= ^

T^b >'b
FFin

(a) 0-5% ib) 10%

ic) 1-5% id) 3-0%. (b) T^>Tg(}f r^> Kg)
3. The time period of a satellite revolving around (c) T^>Tg(ifm^>mg)
earth in given orbit is 7 hours. If the radius of the (d) T^ = Tg
orbit is increased to three time its previous value, 6. A satellite moves in a circle around the earth. The
then approximate new time period of the satellite radius of this circle is equal to one half of the
will be : radius of the moon’s orbit. The satellite completes
id) 40 hours ib) 36 hours one revolution is :

(c) 30 hours id) 25 hours (a) 1/2 lunar month ib) 2/3 lunar month
(JEE Main 2022) (c) 2"^^ lunar month (cf) 2^^ lunar month
ANSWERS

l.(c) 2. (c) 3. (fc) 4. (fl) 5. id) 6. (c)

8/114
“Pn^uiUefi. '4. Fundamental Physics (XI) VOL.II

II. Newton’s law of Gravitation with cavity now applies a gravitational force
on the same particle at P. The ratio F2lF^^ is
7. If the mass of the Sun were ten times smaller and
the universal gravitational constant were ten times
larger in magnitude, which of the following is not
correct ?

(<3) Time period of a simple pendulum on the

Earth would decrease
{b) Walking on the ground would become more P
difficult

(c) Raindrops will fall faster

{d) ‘g’ on the Earth will not change
(NEET 2018) >

8. Two masses, 800 kg and 600 kg are at a distance

ww
0-25 m apart. The magnitude of total force (a) 3/25 (b) 9/50
experienced by a body of mass 1 kg placed at a (c) 22/25 (d) 41/50
point distance 0-2 m from the 800 kg mass and II. In the above question, if the solid sphere is a large
0-15 m from the 600 kg mass is:

Flo
rock, what is the gravitational acceleration at a
{a) 34 X 10-^ N (b) 2-22 X 10-^ N
point on the surface of the rock at a point just

ee
(c) 3-22 X 10-^ N (d) 2-22 X 10-«N above the cavity ?

eer
9. A uniform sphere of mass M and radius R exerts GM GM
a force P on a small mass m situated at a distance

FFr
(a) (b)
R^
of 2 /? from the centre O of the sphere. A spherical 2R-

oorr
uur r
portion of diameter R is cut from the sphere as GM IGM
s ff
shown in Fig. 8(CF).I. The force of attraction (c) id)
8P^ SR-
between the remaining part of the sphere and the
12.
Suppose, the acceleration due to gravity at the
sk
mass m will be
YYoo

Earth’s surface is 10 ms“^ and at the surface of

ooko

Mars it is 4 0 ms“^. A 60 kg passenger goes from

eBB

the Earth to the Mars in a spaceship moving with

a constant velocity. Neglect all other objects in
the sky. Which part of Fig. 8(CF).3 best represents
uurr

the weight (net gravitational force) of the

ad

passenger as a function of time ?

Yo

8(CF).3
dY
Re
innd
FFi

IF
(a) ~ ib)
3 3
4F IF
(c)
3

10. A solid sphere of uniform mass M and radius R

■►Time
applies a gravitational force of attaction equal to
P] on a particle of mass m placed at point P
distance 3 R from the centre of the sphere. A
spherical cavity of radius RH is now made in the id) A ib) B
sphere as shown in the Fig. 8(CF).2. The sphere ic) C id) D

ANSWERS

7. id) 8. (h) 9. (d) 10. (d) 11. (b) 12.(c)

GRAVITATION 8/115

13. A point mass M is placed at a distance L from one 16. Two spherical bodies of mass M and 5 M and radii
end of a uniform rod of mass M and length L as R and 2 R ait released in free space with initial
shown in Fig. 8(CF).4. The gravitational force separation between their centres equal to 12 R. If
experienced by point mass due to rod is they attract each other due to gravitational force
only, then the distance covered by the smaller
FIGURE 8(CF).4
body before collision is
M M
(fl) 2-5 R (/?) 4-5 R
(c) 1-5 R (d) 1-5 R
w-
L L {AIPMT 2015)
17. Four identical particles of each mass M are located
GM^ GA/2 at the comers of a square of side a. What should
(.0) ib)
1? 21? be their speed, if each of them revolves under the
influence of other’s gravitational field in a circular
2GAf2 agm'^

ww
orbit, circumscribing the square ?
(c) id)
31? 9l2
14. A large spherical mass M is fixed at one position

FF loo
and two identical point masses m are kept on a
line passing through the centre of M (See Fig.

ree
8(CF).5). The point masses are connected by a
rigid massless rod of length / and this assembly is
free to move along line connecting them. All three

rFee
masses interact only through their mutual

F
oor r
rur
gravitational interaction. When the point mass
nearer to A/ is at a distance r = 3 / from M, the
s ff
( M
tension in the rod is zero for m = k . The
k
288 GM GM
YYoou

value of k is (a) 1-35 (b) 116

ookos

a a

FIGURE B(CF).5
BBo

'GM GM
(c) 1-41 id) 1-21
re

M m m a a

k (JEE Main 2019)

ouur
ad

18. A body weighs 72N on the surface of earth. What

Yo

(a) 3 ib)6
is the gravitational froce on it, at a height equal
to half the radius of the earth?
{c)l (d)9
dY
Re

(a) 32 N (b) 30 N
idn

(JEE Advanced 2015)

(c) 24 N (d) 48 N (NEF.T 2020)
FFin

15. Four particles, each of mass m, are moving along

a circle of radius r under the influence of mutual 19. The ratio of the weights of the Earth’s surface to
gravitational attraction. The speed of each particle that on the surface of planet is 9 : 4. The mass of
will be 1
the planet is - th of that of the Earth. If R is the
GM 2V2GM radius of the Earth, what is the radius of the
(a) (b)
y r r
planet? [Take the planet and Earth to have the
same mass density]
(c) +1) (d) 5/ (2V2 + 1) R R R R
r 4 (a)- (i) _ (c) _ W -
(JEE Main 2014)
(JEE Main 2019)

ANSWERS

13. (b) 14. (c) 15. (d) 16. (c) V.(b) 18. (a) 19. (rf)

I
8/116
^ Fundamental Physics (XI) voL.n

20. Two objects of equal masses placed at certain 1

distance from each other, attracts each other with 24. A planet of radius ^ ~ (radius of earth) has
a force F. If one third mass of one object is
transferred to the other object, then the new force
the same mass density as Earth. Scientists dig a
well of depth /?/5 on it and lower a wire of same
will be :
length and of linear mass density 10“^ kg m
-1

into it. If the wire is not touching anywhere, the

w|f (.)^F force applied at the top of the wire by a person
holding in place is (take the radius of the earth
(O^F id) F (JEE Main 2022) = 6 X 10® m and acceleration due to gravity
earth is 10 ms"^).
on

m. Acceleration due id) 96 N {b) 108 N

(c) 120 N id) 150 N
to gravity and its variation
(JEE Advanced 2014)

ww
21. If the radius of earth were to increase by 1%,its
25. A particle hanging from a spring stretches it by
mass remaining the same, the acceleration due to 1 cm at earth’s surface. How much the same
gravity on the surface of earth will
particle stretches the spring at a place 1600 km

FF loo
(a) increase by 1 % Qj) decrease by 2 % above the surface of earth {R = 6400 km)
(c) decrease by 1 % id) increase by 2 %. (a) 16/50 cm ib) 16/25 cm

ree
22. The dependence of acceleration due to gravity g (c) 25/16 cm id) 50/16 cm.
on the distance r from the centre of the eai^, 26. What is the percentage change in the value of g

rFee
assumed to be a sphere of radius R of uniform
as we shift from equator to pole on the surface of
density is as shown in Fig. 8(CF).7 below :

F
earth ? (Given equatorial radius of earth is greater
oor r
rur
than polar radius by 21 km and mean radius of
s ff
earth is 6300 km).
id) 0-52% ib) Q-61%
k
YYoou

(c) 1-67% id) 6-7%

ookos

27. The reading of a spring balance corresponds to

BBo

100 N while situated at the north pole and a body

is kept on it. The weight record on the same scale
re

if it is shifted to the equation is (take g = 10 m/s^,

and radius of earth, F = 6-4 x 10^ km)
ouur
ad

id) 99-66 N ib) ilON

Yo

(c) 97-66 N id) 106 N

(AIIMS 2015)
dY
Re

28. The ratio of radii of earth to another planet is 2/3

idn

and the ratio of their mean densities is 4/5. If an

FFin

The correct figure is astronaut can jump to a maximum height of 1 -5 m

ia) (4) ib) (1) on the earth, with the same effort, the maximum
(c) (2) height he can jump on the planet is
id) (3)
ia) 1 m ib) 0-8 m
(JEE Main 2017 ; AIPMT Main 2010)
(c) 0-5 m id) 1-25 m
23. A body weighs 200 N on the surface of the earth.
ie) 2 m (Kerala PET 2009)
How much will it weigh half way down to the
centre of the earth ? 29. A body is weighed with a spring balance in a train
ia) 150 N ib) 200 N at rest, shows a weight W. When the train begins
(c) 250 N (d) lOON to move with a velocity v around the equator from
west to east and if the angular velocity of the train
(NEET 2019) is 0) then the weight shown by spring balance is
ANSWERS

20. (c) 21. ib) 22. ia) 23. id) 24. ib) 25. ib) 26. ib) 27. (a) 28. ib)

I
 8/117
GRAVITATION

2v(a (a) 48 kg wt (b) 48-83 kg wt

(a) W {b) W \ + (c) 47-84 kg wt (d) 47 kg wt
S ) (JEE Main 2021)
2u0)
(c) W 1- (d) w{\ + v'^/r). IV. Gravitational Intensity,
« )
Gravitational potential
30. What is the depth at which the value of and Gravitationalenergy
acceleration due to gravity becomes l/n times the
value at the surface of earth? (radius of earth 34. A spherically symmetric gravitational system of
= R) particles has a mass density
R for r<R
R Po
(a) - (b) — P =
n n 0 for r>R

ww
R{n-\) Rn where po is a constant. A lest mass can undergo
(c) id) circular motion under the influence of the
n in-l) gravitational field of particles. Its speed v as a
function of distance r (0 < r <«») from the centre

Flo
(NEET 2020)
of the system is represented by
31. The value of the acceleration due to gravity is

e
R FIGURE 8(CF).8

rere
at a height ^ (^ = radius of the earth) from

r FF
the surface of the earth. It is again equal to g, at a
depth d below the surface of the earth. The ratio
uurr
id/R) equals: foor
4 1
kss
w 5 ib) 3 ♦t
Yoo
■►r
ooook

R
5 7
e o
(c) 9 id) 9
eBB

(JEE Main 2020)

32. The acceleration due to gravity on the earth’s
uurr

surface at the poles is g and angular velocity of

ad

the earth about the axis passing through the pole

Yo

is CO. An object is weighed at the equator and at a

dY

height h above the poles by using spring balance.

If the weights are found to be the same, then h
Re
innd

is: (/i« R, where R is the radius of the earth)

35. The gravitational field due to a mass distribution
Fi

is / = Kli^ in the X-direction. (AT is a constant).

(a) ib) Taking the gravitational potential to be zero at
g 8^
infinity, its value at a distance x is
R?-(i? rW ia) K/x ib) K!2x
(c) id)
4g 2g ic) Klx^ {d) KITjP-.
(JEE Main 2020) 36. Two concentric spherical shells A and B of radii
and 2 /? and mases 4 A/, and M, respectively are
33. The weight of a person on pole is 48 kg wt then
placed in space as shown in Fig. 8(CF).9. The
the weight of the person on equator is : [Given
redius of earth = WOO km; /? co^ = 0-034 ms"^ ; gravitational potential at F at a distance
riR<r<lR) from the centre of shells is
g = 10 ins“^]
ANSWERS

29. (c) 30. (c) 31. (c) 32. (d) 33. (c) 34. (c) 35. (d)

i
8/118
’a Fundamental Physics (XI) VOL.II

FIGURE 8(CF).11
m

I -P
d

4Gm 2Gm
(a) (.b)
d4^d^+l'^ dl^'^ +Ad'^

ww
AGM 9GM
{a) - {b) -
R 2R
(c)
2-JlGm
(d) none of the above
d^f+Ad^

Floo
AGM
ic) - {d) none of these
2R 39. A point P lies on the axis of a irng of mass M and

ee
radius a, at a distance a from its centre C. A small
37. From a solid sphere of mass M and radius R, a

eer
particle starts from P and reaches C under
gravitational attraction only, Fig. 8(CF).I2. Its
spherical portion of radius — is removed, as

FFr
2 speed at C will be

oor r
uur r
shown in Fig. 8(CF).10. Taking gravitational FIGURE 8(CF).12
s ff
potential V = 0 at r = <», the potential at the centre
of cavity thus formed is (G = gravitational
sk
constant)
YYoo
P
oooko

d'
eBB
uurr

2GM
ad

2GM 1
(fl) (b) 1-
Yo

a
a

2GM
dY

(c) (V2-1) (d) zero

V a
Re
ind

40. Infinite number of masses, each of I kg, are placed

FFin

along the x-axis atx = ± 1 m, ±2 m, ± 4 m, ± 8 m,

2GM GM
(a) - (b) - ± 16 m The magnitude of the resultant
3R R gravitational potential in terms of gravitational
constant G at the origin (x = 0) is
GM -2GM
(c) - (d) (a) G/2 (b) G
2R R
(c) 2C (d) AG
{JEE Main 2015) (e) 8G (Kerala PET 2008)
38. A uniform rod of mass m and length I is taken. 41. The energy required to take a satellite to a height
Find the gravitational field intensity at point P at h above Earth surface (radius of Earth = 6-4 x
distance d which is on the perpendicular bisector 10^ km) is £j and kinetic energy required for the
of the rod as shown in Fig. 8{CF). 11. satellite to be in a circular orbit at this height be
ANSWERS

36. (d) 37. (h) 38. (b) 39. (b) 40. (d)
 8/119
GRAVITATION

£2- The value of h for which £j and £2 are equal —2Gm 2 /'
2+
I
IS : (b)
I V2
id) 1-6 X lO^km (b) 3-2 X 10^ km
(c) 6-4 X 10^ km (d) 1-28 X 10*^1011 -^Gm 2 f 1
(c) V2-
(JEE Main 2019) / V2
42. A satellite is moving with a constant speed u in a 2 /
I
-2Gm
circular orbit around the earth. An object of mass id) V2 +
m is ejected from the satellite such that it just / V2
escapes from the gravitational pull of the earth. 4tl. Energy required to transfer a 400 kg satellite in a
At the time of ejection, the kinetic energy of the circular orbit of radius 2 /? to a circular orbit of
object is : radius 4 R, where R is the radius of the earth is
(a) 2mv^ ib) mv^ ig = 9-80 ms-^ £ = 6-4 X 10^ m)
(a) 1-65 X 10^ J ib) 3-13 X 10^ J

ww
(c) m v^!2 id) 3 m V-/2
(JEE Main 2019) (c) 6-26 X 10^ J id) 4-80 X 10^ J
43. A particle of mass M is situated at the centre of a 48. There is a crater of depth (£/100) on the surface
of moon, where R is the radius of moon. A particle

Flo
spherical shell of same mass and radius a. The
magnitude of the gravitational potential at a point is projected vertically upwards from the crater

e
situated at a/2 distance from the centre, will be with the velocity which is equal to the escape

eree
velocity v from the surface of the moon. The
2GM 3GM maximum height attained by the projectile is

FFr
(a) ib)
a a («) R ib) 85 £

oorr
uur r
(c) 99 £ id) 100 £
4GM GM
49. A body is released at a distance r (r > £) from the
sf
ic) id)
a
a
centre of earth. The velocity of the body when it
sk
(AIPMT Main 2011, AIPMT 2010) strikes the surface of the earth is
Yoo
oook

44. A person brings a mass of 1 kg from infinity to a

point P. Initially the mass was at rest but it moves (a) ^2gr ib) ^2g(£+7j
eBB

at a speed of 2 ms"* as it reaches at P. The work 1/2 nl/2

1 1
done by the person on the mass is - 3 J. The (0 id) 2g --
uurr

potential at P is /j R
ad

(a) -2J/kg ib) - 3 J/kg 50. A rocket is launched vertically from the surface
Yo

ic) 5 J/kg id) -1 J/kg. of earth with an initial velocity u the height up to
which the rocket can go from the surface of earth,
dY

45. The work done to raise a mass m from the surface

before falling back is
of the earth to a height h, which is equal to the
Re
innd

radius of the earth (R) is : (a) -JI^R ib) u^!2g

FFi

(a) mgR ib) 2 mgR -1

r25_i id) 2|._1
ic)
id) I mg £
1 2 £
ic) -mgR u u 2 £

51. The earth is assumed to be a sphere of radius £. A

(NEET 2019)
platform is arranged at a height £ from the surface
46. Four particles each of mass m are placed at the of the earth. The escape velocity of a body from
vertices of a square of side 1. What is the potential this platform isfv, where v is its escape velocity
energy of the system ? from the surface of the earth. The value of/is

-^|2Gm^( 1 (a) 1/2 ib) V2

ia)
I (cO 1/3
ic) I/V2
ANSWERS

41. ib) 42. ih) 43. {b) 44. (c) 45. ic) 46. ib) 47. (b) 48. (c) 49. (c) 50. (d) 51. (c)

A
8/120
'P'Ktdeefr'4. Fundamental Physics (XI) VOL.iT

52. What is the minimum energy required to launch rotation plane. In order to escape from the
a satellite of mass m from the surface of a planet gravitational field of this double star, the
of mass M and radius /? in a circular orbit at an

altitude of 2 /? ? minimum speed that meteorite should have at 0

is : (Given Gravitational constant = 6-67 x
(a)
5GmM
ib)
2GmM Nm^ kg~^)
6R 3R -1
(a) 2-8 X ms (b) 14x10^ ms -1
GmM GmAf (c) 24 X 10*^ ms
-1
(d) 3-8 X 10^ ms''
(c) {d)
2R 3R (JEE Main 2019)
(JEE Main 2013) 56. Consider two solid spheres of radii /?j = 1 m,
53. A thin uniform annular disc (see Fig. 8(CF).13) /?2 = 2 m and masses A/j and M2 respectively.
of mass M has outer radius 4 R and inner radius The gravitational field due to two spheres 1 and
3 R. The work required to take a unit mass from 2 are shown in Fig. 8(CF).14. The value of

ww
is
point P on its axis to infinity is

Flo
e
eree
FFr
oorr
uur r
sk sf
Yoo
oook

2GM
(a) (4>^-5) (b) -
2GM
(4>^-5)
eBB

IR I 1
IR

CM 2 CM
(c) id) (■v/2-1) I
uurr

4R 5R
ad

(IIT 2010)
Yo

(JEE Main 2020)

54. A particle of mass m is projected with a velocity
57. A body of mass 60 g experiences a gravitational
dY

v = kv^{k< 1) from the surface of Earth {v^ is

the escape velocity). The maximum height above force of 3-0 N when placed at a particular point.
Re
innd

the surface of earth reached by the particle is The magnitude of the gravitational field intensity
at that point is :
FFi

Rk^ ( k f
id) ib) R (a) 0-05 N/kg ib) 50 N/kg
i\-k^) 1-* (c) 20 N/kg {d^ 180 N/kg
( k R^k (NEET 2022)
ic) R — id)
I +* H + k) V. Satellite and Escape velocity
(NEET 2021) 58. Two particles A and B are moving in uniform
55. Two stars of masses 3 x 10^' kg each, and at circular motion in concentric circles of radii
distance 2 x lO" m rotate in a plane about their and rg with speed and respectively. Their
common centre of mass 0. A meteorite passes time period of rotation is the same. The ratio of
through O moving perpendicular to the stars angular speed of A to that of B will be :

ANSWERS

52. (ti) 53. (a) 54. (a) 55. (a) 56. (a) 57. (b)

k
 8/121
GRAVITATION

(^) (b) : Vq 62. A satellite in a force-free space sweeps stationary

(^1:1 dM
(c) rg: interplanetary dust at a rate = pi;, where v
(NEET 2019) dt

59. Kepler’s third law states that square of period of is the speed of escaping dust w.r.t. satellite and M
is the mass of satellite at that instant. The
revolution (T) of a planet around the sun, is
acceleration of satellite is
proportional to third power of average distance r
between sun and planet i.e. = Kr’ where K is (a) -Pu2 {b) - P i>^/2 M
constant.
(c) - P v^tM {d) ~M p/u-
If the masses of sun and planet are M and m 63. A satellite is moving around the earth with speed
respectively then as per Newton’s law of gravi u in a circular orbit of radius r. If the orbital radius
tation force of attraction between them is is decreased by 1 % its speed will
(a) increase by 1 % {b) increase by 0-5%
where G is gravitational constant. The

looww
(c) decrease by 1% {d) decrease by 0-5%.
relation between G and K is described as: 64. The radii of circular orbits of two satellites A and
B of the earth, are 4 and R, respectively. If the
{d)GMK = AT0- {b)K=G
speed of satellite A is 3 V, then the speed of
1 satellite B will be
{d)GK = ATiP-

ree
(c)K=-
G (a) 3 V/4 ib) 6 V
(c) 12 V (d) 3V/2

ree F
(AIPMT 2015)
60. A particle is moving with a uniform speed in a
circular orbit of radius R in a central force
r FF (AIPMT 2010)
65. Two bodies each of mass M, are kept fixed with a
inversely propertional to the nth power of R. If
fofroF
separation 2L. A particle of mass m is projected
u
period of rotation of the particle is T, then for from the mid point of the line joining their centres,
ks
any value of n perpendicular to the line. The gravitational
constant is G. The correct statement(s) is (are):
os o
YYouor

-+l
(a) the minimum initial velocity of the mass m to
BBook

(a) 7'<=ci?3/2 {b)TocR2

escape the gravitational field of the two bodies
r ee

(c)7’o=^«+*V2 (d) Tec

is AyjGM/L
(JEE Main 2018)
ouru

(b) the minimum initial velocity of the mass m to

ad

61. A rocket is launched normal to the surface of

escape the gravitational field of the two bodies
Yo

the earth, away from the sun, along the line

joining the sun and the earth. The sun is 3 x 10^ is 2^GM^
d

times heavier than the earth and is at a distance

Re

(c) the minimum initial velocity of the mass m to

inY

2-5 X times larger than the radius of the escape the gravitational field of the two bodies
FFind

earth. The escape velocity from earth’s gravi

tational field is Vg = 11*2 ms“^ The minimum is ^2GM/L
initial velocity (v^) = 11 -2 ms“'. The minimum id) the energy of mass m remains constant.
initial velocity (u^) required for the rocket to (JEE Advanced 2013)
be able to leave the sun-earth system is closest 66. A satellite is moving with a constant speed V in a
to (Ignore the rotation of the earth and the circular orbit about the earth. An object of mass
presence of any other planet m is ejected from the satellite such that it just
(a) = 22 km s
-1
(b) = 72 km s * escapes from the gravitational pull of the earth.
(c) = 42 kra s~* id) = 62 km s"^ At the time of its ejection, the kinetic energy of
(JEE Advanced 2017) the object is

ANSWERS

58. id) 59. (a) 60. (c) 61. (c) 62. (c) 63. ib) 64. (/j) 65. (i»)
8/122
‘P'KuUe^'^i. Fundamental Physics (XI) VOL.II

1
70. A bullet is fired vertically upwards with a velocity
(.) 3 (b) mV^ V from the surface of a spherical planet when it
reaches its maximum height, its acceleration due
-mV^
(0 2 (d) 2mV^ (IIT2011) 1
to the planet’s gravity is -th of its value at the
4
67. A satellite is revolving in a circular orbit at a height
h from the earth’s surface (radius of earth R ; surface of the planet. If the escape velocity from
h « R). The minimum increase in its orbital the planet is Vescape = V then the value of N
velocity required, so that the satellite could escape
from the earth’s gravitational field, is close to is : (ignore energy loss due to atmosphere).
(Neglect the effect of atmosphere) ia) 1 ib)2
(c)3
(a) ^IgR (b) 4sR W4
(JEE Advanced 2015)

ww
(c) 4imi (d) VFs(V2-1) 71. The escape velocity from the earth’s surface is v.
(JEE Main 2016) The escape velocity from the surface of another
planet having a radius fourtimes that of earth and

Flo
68. A particle is fired vertically upward from the
same mass density is
surface of earth with a velocity ofkv^, where

e
{a)Av ib)v
is the escape velocity and jt < 1. Neglecting air

reree
resistance, the maximum height to which it will (c) 2 V {d)3v (NEET 2021)

r FF
rise, measured from the centre of the earth, is 72. A planet A has a mass M and radius R. Planet B
{R = radius of eanh)
has half the mass and half the radius of planet A.
uurr
{a)
\~k^
R R foor
If the escape velocities from the planets A and B
V n
are and Vg respectively, then — =
ks s
V
B
4’
l-k^ k^
Yoo
the value of n is
ooook

(c) (d) —
R R (fl)3 (b)2
eBB

(AMU PET 2011) (c)4 (d)5

69. A black hole is an object whose gravitational field (JEE Main 2020)
is so strong that even light cannot escape from it.
uurr

73. T\vo satellites are revolving in the orbit around a

To what approximate radius would earth (mass =
ad

5-98 X 10^ kg) have to be compressed to be a planet. The ratio of their time period is 1 : 8. Find
Yo

black hole ? the ratio of their angular velocities.

dY

(a) 1/2 (b) 8/1

(a) 10-9 m {b) 10-^ m
Re

(c) 10-2 m (c) 2/1 (cf) 2/3

innd

(d) 100 m
(AIPMT 2014) (JEE Main 2021)
FFi

m Multiple Choice Questions (with One or More than One Correct Answers)
74. In case of earth 75. Which of the following statements are true about
(a) potential is minimum at the centre acceleration due to gravity
(a) 'g' is zero at the centre of earth
(b) potential is zero, both at centre and infinity
(b) ‘g’ decreases if earth stops rotating on its axis
(c) field is zero both at centre and infinity (c) 'g' decreases in moving away from centre if
(d) potential is same, both at centre and infinity r>R
but not zero
(d) 'g' decreases in moving away from centre if
r<R
ANSWERS
66.(h) 67. (d) 68. id) 69. (c) 70. ih) 71. (a)
72. (c) 73. (b) 74. {a,c) 75. («.c)
GRAVITATION 8/123

76. If two satellites of different masses are revolving

in the same orbit, they have same
(a) speed {b) energy
(c) time period
{d) angular momentum
77. The escape velocity of an object projected from
the surface of a given planet is independentof
(a) mass of the planet
(b) the mass of the object
(c) the radius of the planet
(d) the direction of projection
78. Consider a planet moving in an elliptical orbit (a) the gravitational force due to this object at the

ww
round the sun. The work done on the planet by
origin is zero
the gravitational force of the sun
(b) the gravitational force at the point B (2, 0,0)
(a) is zero in some part of the orbit
is zero

Flo
(b) is zero in no part of the motion
(c) the gravitational potential is the same at all

e
(c) is zero in any small part of the orbit
points of the circle ~ 36

eree
{d) is zero in one complete revolution
{d) the gravitational potential is the same at all

FFr
79. Three point masses are at the corners of an points on the circle = 4. (IIT)
equilateral triangle of side r. Their separations do
82. A ring has a total mass M but non-uniformly
oorr
uur r
not change when the system rotates about the
distributed over its circumference. The radius of
centre of the triangle. For this, the time period of
sf
rotation must be proportional to the irng is R. A point mass m is placed at the centre
of the ring. Workdone in taking away this point
sk
Yoo
(^) ib)r
mass from centre to infinity is
oook

-1/2
(c) m id) m
GM m GMm
eBB

80. TWo objects of masses m and 4 m are at rest at an (a) ib)

R R
infinite separation. They move towards each other
under mutual gravitational attraction. If G is the GM m GM m
uurr

universal gravitational constant, then at separation r (c) - id) -

2R
ad

2R
(a) the total energy of the two objects is zero
Yo

83. Suppose universal gravitational constant starts to

ib) net angular momentum of both the objects is decrease, then
dY

zero about any point

(a) length of the day, on earth, will decrease
Re

(c) the total K.E. of the objects is 4 G nr^lr

innd

(b) length of the year will decrease

(d) their relative velocity of approach is
FFi

(c) earth will follow a spiral path of increasing

nI/2
8Gm radius

r (d) kinetic energy of earth will decrease

81. A solid sphere of uniform density and radius 4 84. Two satellites of same mass of a planet in circular
units is located with its centre at the origin O of orbits have periods of revolution 32 days and 256
coordinates. Two spheres of equal radii 1 unit, days. If the radius of the orbit of the first is R,
with their centres at A (- 2, 0, 0) and B (2, 0, 0) then the

respectively are taken out of the solid leaving (a) radius of the orbit of the second is 4 /?
behind spherical cavities as shown in the Fig. (b) radius of the orbit of the second is 8 /?
8(CF).15, then
ANSWERS

76. (a,c) 77. (b.d) 78. {c.d) 79. (a.d) 80. {a.b.c) 81. ia.c.d)
82. (b) 83. (a,c.d)
8/124
‘P’uzdeefo.'^^ Fundamental Physics (XI)EEIHD
(c) total mechanical energy of the second is also has uniform density p and its mass Is {Mp +
greater than that of the first Mq). The escape velocities from the planets P, Q
{d) kinetic energy of the second is more than that and R, are Vp, Vq, Vp, respectively, then
of the first
(a) Vq>Vp> Vp
85. Two spherical planets P and Q have the same uni (b)Vp>V^>Vp
form density p, masses Mp and Mq , and surface (c)Vp/Vp = 3
areas A and 4A, respectively. A spherical planet R
(d)VplVQ=m (IIT 2012)

m Multiple Choice Questions (Based on the given Passage/Comprehehs ibh)'

Each comprehension given below is followed by some multiple choice questions. Each question has one

correct option. Choose the correct option.

ww
HgdmpreRSn^SnlC There are three identical 3Gm^ ■y/3Gm^
point mass bodies each of mass m located at ic) id)
r2 r2
the vertices of an equilateral triangle with

Flo
side r. They are exerting gravitational force 87. At what speed must they move if they all revolve

e
of attraction on each other, which can be under the influence of one another’s gravitation

e
given by Newton’s law of gravitation. Each

reer
in a circular orbit circumscribing the triangle still

rFF
mass body produces its gravitational field in preserving the equilateral triangle
the surrounding region. The magnitude of
uur r
gravitational field at a point due to a point ffoor
Gm IGm
mass body is the measure of gravitational {a) ib)
intensity at that point. The gravitational
r V r
sks
potential at a point in a gravitational field Is
YYoo

'Gm
ooko

the amount of workdone in bringing a unit (c) (d)

3Gm

mass body from infinity to the given point r y r

eBB

without acceleration.
88. Work done in taking one body far away from the
other two bodies is
uurr
ad

Gm^
Yo

Gnx^
(a) - (b)
r r
dY
Re

2Gm^ IGm^
innd

(c) (d)-
r r
FFi

89. Magnitude of gravitational field at the mid point

D of arm BC of triangle ABC is
Answer the following questions :
Gm^ Gm
id) ib)
86. The magnitude of the gravitational force on one r2 3r2
body due to other two bodies is
3Gm AGm
Gm2 2Gm2 ic) id)
ia) ib) r2 3r2
r2 r2

ANSWERS

84. (a.c) 85. (b,d) 86. (d) 87. (c) 88.(c) 89. (d)
GRAVITATION 8/125

91. Time period of revolution of satellite around the

n A rocket is fired vertically earth is

upwards with a speed of n (=5 km s"^) from (fl) 3550 s (b) 7100 s
the surface of earth. It goes up to a height h
(c) 5330 s {d) 8880 s
before returning to earth. At height h a body
is thrown from the rocket with speed Vq in 92. The energy to be spent in taking the satellite out
such a way so that the body becomes a of the gravitational field of the earth is (mass of
satellite of earth. Let the mass of the earth, the satellite is 2(X) kg)
Af s 6 X 10^ kg ; mean radius of the earth,
(a) 5-Ox 10^J (b) 10-0 X 10^ J
J? = 6*4 X 10*^ m ; G = 6-67 x ir^^ Nm^ kg-^ ;
g = 9‘8 ms“^. (c) 2-5 X lO'O J (d) 5-Ox 10’°J
Answer the following questions : 93. If this satellite is to be taken at double of the

present height from the surface of the earth, dien

ww
90. The value of h is
the new time period of revolution is
(a) 1-5 X 10^ m (b) 3-2x lO^m
(a) 9330 s ib) 20080 s

FF loo
(c) 3-2 X 10^ m (d) l-6x lO^m
(c) 11000 s (d) 29400 s

ree
09 Matching Type Questions

F rFee
oor r
DIRECTIONS. In each of the following questions, match column I and column II and select the correct
rur
match out of the four given choices.
s ff
94. A satellite of mass m is orbiting around the earth of mass M and radius R.
k
YYoou
ookos

Column I Column II
BBo

CMm
(A) Gravitational potential energy of satellite orbiting close to earth is (P)
re

2R
ouur
ad

GMm
(B) Kinetic energy of orbiting satellite close to earth is (q)
Yo

2R

CMm
dY

if)
Re

(C) Kinetic energy of orbiting satellite at height R from the surface of earth
idn

4/?
FFin

CMm
(D) Total mechanical energy of satellite orbiting close to earth is)
R

ia) A-p ; h-q ; C-r; D-j (b) A-q ; B-r; C-s ; D-p
(c) A-r; B-p ; Cs; D-g id) As; B-p ; C-r; D-^ “J ^

95. A body of mass m is to be escaped from the different positions w.r.t. earth as given below, M is the mass of
earth, R is the radius of earth and G is the universal gravitational constant.

ANSWERS

90.{d) 91.(b) 92.(a) 93.(u) 94. (d)

8/126
4 Fundamental Physics (XI) W«1BII

Column I Column II

%GM
(A) Escape speed of a body from the surface of earth. (P)
V 5/?
GM
(B) Escape speed of a body from the atmosphere at height R (?)
from the surface of earth. IR

(C) GM
Escape speed of a body from the satellite orbiting at a height R (r)
from the surface of earth. R

(D) 2GM
Escape speed of a body from the atmosphere at height R/4 from (■5)

ww
the surface of earth. R

(a) A-q ; B-p ; C-r ; D-s (b) A-p ; B-q ; C-s ; D-r (c) A-r ; B-s ; C-p ; D~q (d) As ; B-r; C-q ; D-p

Floo
96. A planet of mass M, has two natural satellites with masses mj and m2. The radii of their circular orbits are /?j

ree
and /?2 respectively. Ignore the gravitational force between the satellites. Define, L,. K^ and Tj to be,
respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite

rFee
1 and V2, L2, K2 and T2 to be the corresponding quantities of satellite 2. Given m^lm2 = 2 and /?i//?2 = 1/4,

F
Match the ratios in List-I to the numbers in List-II

oor r
rur
List'I
s ff
List'll

(A) U,/V2 (p) 1/8

osk
YYoou

(B) Lj/L2 (?) I

oook

(C) K^!K2 (.r) 2

eBB

(D) T’,/7'2 (s) 8

(a) As ; B-g ; C-p; D-r (b) A-r; B-q; Cs ; D-p (c) A-q ; B-r; C-p ; D-^ (d) A-q ; B-r; C-^; D-p
uur r
ad

(JEE Advanced 2018)

Yo

ca Matrix-Match Type Questions

p q
11
r
1 (
s

A © 0 o©
dY
Re

DIRECTIONS. Each of the following questions contains statements given in

idn

two columns, which have to be matched. The answers to th^e questions have B E o©
FFin

to be appropriately bubbled. If the correct matches are A-r ; As ; B-p ; B-^ ;

C-p and D-^ ; D-s, then the correctly bubbled matrix will look like the one c 0 0 Oi©
shown here.
D
® © 01©
97. Match the following columns :
Column 1 Column II

(A) Parking orbit (P) Polar satellite

(B) Polar orbit (?) Geostationary satellite
(C) Circular orbit (r) Communication satellite
(D) Elliptical orbit (●y) Remote sensing satellite

ANSWERS

95. id) 96, (b) 97. A-p, q. ,r s ; B-p, s ; C-p, q, r, s : D-r, s

 8/127
GRAVITATION

98. Match the following columns :

Column I Column II

(A) Every planet revolves around the sun in elliptical (p) Law of conservation of angular momentum
orbit and the sun is situated at one foci of ellipse.

(B) The position vector of the planet with respect to (q) Inverse square law
the centre of sun has a constant areal velocity.

(C) Geocentric theory of planetary motion (r) Copernicus

(D) Heliocentric theory of planetary motion (s) Ptolemy

VI. Integer Type Questions A 6 C D

w
DIRECTIONS. The answer to each of the following questions is a single digit
integer, ranging from 0 to 9. If the correct answers to the question numbers A, B, oooo

Flo
C and D (say) are 4, 0, 9 and 2 respectively, then the correct darkening of bubbles @@©@

reee
should be as shown on the side :
@@@@
@ 0®@

FFr
99. Gravitational acceleration on the surface of a planet is g, where g is the gravitational
11
® ©0®
uurr
2
for
acceleration on the surface of the earth. The average mass density of the planet is
® © ©©
— times that of the earth. If the escape speed on the surface of the earth is taken to be © ©©©
kkss
3

11 kms"^, the escape speed on the surface of the planet in kms“^ will be ;
® © © ®:
Yo
oooo

(irr 2010)
9H9U9)(9
eB

100. The earth takes 24 hours to rotate once about its axis. How much time (in min) does the sun take to shift by
1° when viewed from the earth ?
rr

101. A man can jump 1-5 m high on earth. He can jump on a planet to a height of 3 x at metre. The density of
ou
ad

planet is one quarter that of the earth and whose radius is one third of the earth. What is the value of a: ?
YY

102. A body of mass 100 kg falls on the earth from infinity. Its total energy on reaching the earth is 6-27 x 10" J.
6400 km and g = 9-8 m/s^. Air friction is neglected.
nndd

What is the value of n ? Given, radius of earth is

Re

103. Two satellites and S2 revolve around a planet in coplanar circular orbit in the same sense. Their periods
Fi

of revolutions are 1 hour and 8 hours respectively. The radius of orbit of 5 j is 10^ km. When S2 is closest to
5j, the speed of ^2 relative to Sj is it x 10" km/h. What is the value of n ?
104. The ratio of the radius of the earth to that of moon is 10. The ratio of acceleration due to gravity on the earth
and on the moon is 6. What is the ratio (in integral value) of the escape velocity from the earth’s surface to
that from the moon ?

105. A particle is projected vertically upwards from the surface of the earth of radius R, with a kinetic energy
equal to half of the minimum value needed for it to escape. The maximum height to which it irses above the
surface of earth is n R. What is the value of « ?

ANSVtfERS

98. A-q ; B-p ; C-s ; D-q, r 99. (3) 100. (4) 101. (6) 102. (9)
103.(4) 104.(8) 105.(1)
 8/128
'4. Fundamental Physics (XI) VOL.II

VII.
Assertion-Reason Type Questions
FOR MEDICAL STUDENTS
Reason. Every body in this universe attracts every
DIRECTIONS. Read the following questions and other body with a force which is inversely
choose any one of the following four responses. proportional to the square of distance between
them.
(A) If both Assertion and Reason are true and the
Reason is the correct explanation of the (a) A (b) B (c) C (d)D
Assertion. 112. Assertion. The space rockets are usually launched
(B) If both Assertion and Reason are true but Reason from west to east in the equatorial line.
is not a correct explanation of the Assertion. Reason. It is easy to do so.
(C) If Assertion is true but the Reason is false. (a) A (b) B (c) C (£/) D
(D) If both Assertion and Reason are false. 113. Assertion. If the distance between the earth and

ww
106. Assertion. Angular speed, linear speed and KE the sun were half its present value, the number of
change with time but angular momentum remains days in a year would be 129.
constant for a planet orbiting the sun. Reason. According to Kepler’s law of period ;

FF loo
T^ocr^,
Reason. Angular momentum is constant as no
torque acts on the planet, (a) A ib) B (c) C (d)r>

ree
(a) A (b)B (c)C (d)D FOR ENGINEERING STUDENTS
107. Assertion. An astronaut in an orbiting space

rFee
station above the earth experience weightlessness. DIRECTIONS. Read the following two statements.
Of the four choices given, choose the one that best
Reason. An object moving around the earth under
oor rF
rur
the influence of earth’s gravitational force is in a describes the two statements.
s ff
stale of ‘free fall’. (A) Statement-1 is true ; Statement-2 is true ; State
(a) A {b) B (c) C (d) D ment-2 is a correct explanation of Statement-1.
k

(B) Statement-1 is true ; Statement-2 is true ;

YYoou

(AIIMS 2015)
ookos

108. Assertion. If ice cap of the pole melts, the day Statement-2 is not a correct explanation of
Statement-1.
BBo

length will shorten.

Reason. Ice will flow towards the equator and (C) Statement-1 is true ; Statement-2 is false.
re

decrease the moment of inertia of the earth. This (D) Statement-1 is false ; Statement-2 is true.
increase the frequency of rotation of the earth.
ouur
ad

114. Statement-1. A body released from a height equal

(a) A (b) B (c) C (d) D
Yo

to the radius (R) of the earth. The velocity of the

109. Assertion. Earth is continuously pulling moon body when it strikes the surface of the earth will
towards its centre, but moon does not fall to earth.
Yd

be ^2gR .
Re

Reason. Attraction of sun on moon is greater than

idn

that of earth on moon. Statement-2. As -i- 2 as.

FFin

(a) A (b) B (c) C (iO D (a) A (b) B (c) C (d)D

110. Assertion. The time period of revolution of a 115. Statement-1. An artificial satellite moving in a
satellite close to surface of earth is smaller than circular orbit around the earth has a total energy
that revolving away from surface of earth. (i.e., sum of potential energy and kinetic energy)
Reason. The square of lime period of revolution Eq. Its potential energy is - Eq.
of a satellite is directly proportional to cube of its Statement-2. Potential energy of the body at a
orbital radius.
GM m
{a) A ib) B (c) C {d)D point in a gravitational field of earth is -
R
111. Assertion. We can not move even a finger without
disturbing all the starts. (a) A ib) B (c) C (iO D
ANSWERS
106. (a) 107.(a) 108. id) 109. (c) 110. ia) 111. ia) 112. (c)
113. (a) 114. id) 115. id)
 8/129
GRAVITATION

116. Statement-1. Two solid spheres of radius r and Statement-2. Orbital velocity of satellite is
2 r, made of same material, are t in contact. The V -
GM

mutual gravitational force of attraction between y r

them is proportional to \!r^. (a) A (b) B (c) C ((/) D
Statement-2. Gravitational attraction between 119. Statement-1: The escape velocity from the earth
two point mass bodies varies inversely as the e' The escape velocity from a planet whose
is V

square of the distance between them. radius is twice that of the earth and mean density
(c) C (rf) D.
is same as that of the earth is 2 Vg.
(a) A (b) B
117. Statement-1. If both the mass and radius of the Statement-2,
earth decrease by 1%, the value of acceleration (a) A (b) B (c) C (d) D
due to gravity will increase by 2%. 120. Statement-1: A body weighs W newton on the
Statement-2, g = GMIR^ surface of the earth. Its weight at a height equal

ww
(a) A (b) B (c) C (d) D to half the radius of the earth will be 2 W/5.

118. Statement-1. Two satellites of mass 3 M and M

orbit the earth in circular orbits of radii r and 3 r
Statement-2, g' = g
(R + h)^

Flo
respectively. The ratio of their speeds is -J3:\. (a) A (b) B (c) C id) D.

e
rere
r FF
For Difficult Questions
uurr
for
Multiple Choice Questions (with one correct Answ^)|
kss
Yoo
ooook

1. Let r^ and r2 be the shortest and longest distances 0/2 n3/2

eBB

A-01/? 3/2
of the earth w.r.t. sun which will be when earth is = (1-01)
at A and B as shown in Fig. 8(CF).17. If R is the
Tl R
1 I R
uurr

distance of the earth from sun in perpendicular 0/2

3
ad

1
positions, then from the property of the ellipse 1 + = 1 +
Yo

100 2x100
1 1 1 1
1-m
dY

—I— = —I—
or
R R r, r2 R
n'2 . . (T'o-T'y xlOO
%change in time period = -= i-
Re
innd

or R =
l
Fi

(T
_ iz -1 xl00 = —XlOO = 1-5%
FIGURE S(CF).17 T.I 200

3. Here, T, = 7 hour, /?, = /?, 72 = ?,/?= 3 R

0/2 0/2

Bl A As, T2 = T]
f^2 -7f—
R

=1 = 1\^ =21 X 1-732 = 36-37

* 36 hours

ANSWERS

116.id) 117. id) 118. ia) 119.(c) 120. id)

8/130
’4. Fundamental Physics (XI) VOL.II

For Difficult Questions

The force on 1 kg at P due to body of mass 600
kg at B is
4. As T'^ocr’ or T'^=Kr^ ...(0 „ Gx600xl 80,000^ , no
Diiferentialing it, we have ^ : G, along PB
" {0-15)2 3
2TAT=3Kr^Ar ...(«) Since (0-25)2 = (0-2)2 + (0-15)2
Dividing (n) by (/), we have AB2=Ap2 + pg2
2TAT _ 3Kr^Ar or AT = -T
3_ Ar hence ZAPS = 90°
t2 KP 2 r Resultant force on I kg body at P is,
5. When two binary stars are rotating about their
common centre of mass, they will continue to
rotate on the same orbital path all the time, nl/2
otherwise their centre of mass will be changing. = (20,000G)2 +
80,000 G

ww
3
FIGURE 8(CF).18

-1I/2

Floo
fflA '
= 10‘^G 4 + ^

&
9

ree
CM 10
A\ /
/ B = 10‘*x6-67xI0-‘* X —

rFee
3
s

= 2-22 X N

F
oor r
rur
9. Gravitational force of attraction on mass m at P
Due to it, the angular velocities of two binary stars due to solid sphere is
s ff
must be the same. As (o = 2 n/T, therefore, their
time periods of revolution must be the same, i.e., ^ _ GM m __ GM m GMm
osk

or
= 4F ...(1)
~(2P)2 ~ 4p2
YYoou
Ta = Tb‘ p2
oook

6. According to Kepler’s law T^ocr^

Mass of the spherical portion removed from
eBB

1 r 1 sphere
and as
^ r 2
M

x^Ti{Rl2f =
m
M
M' =
uur r

x3/2
8
/j 0/2
ad

T r ~kR^
= 2“2/2
s s
3
Yo

T r 2
m m
Gravitational force of attraction on mass m at P if
-3/2 T
r, = 2 mass of the spherical portion removed is present
dY

^m
Re

7. If gravitational constant becomes ten times, i.e., there is

idn

G' = 10 G ; then the acceleration due to gravity

FFin

increases, [because g G], therefore option {d)

OC _ G{M /8)m _ GM m 1 4
is wrong. All other options are true. (3P/2)2 p2 ^8^9
8. Refer to Fig. 8(CF). 19, the force on 1 kg at P due .4C 1 4 2P
to body of mass 800 kg at A is =4Px-x--—
8 9 9
Gravitational force of attraction on mass m at
P due to remaining part of the sphere is
F"=F-F'=F- 2F _ IF
9 " 9
10. Let M be the mass of solid sphere of radius R.
The force on mass m at P due to solid sphere is
GxSOOxl
^A = = 20,000 G, along PA F = GM m _ GM m
1 ...O')
(0-2)2 (3P)2 “ 9p2
 8/131
GRAVITATION

Total force experienced by point mass due to

For Difficult Questions
whole rod is

Mass of cavity created. 2L

GM^ dx GM^
F =
xijc(/?/2)3
M M
M' =
L
L r~ 11}
8
-kR^
3 14. Net force on the mass m at the left end of the rod
The force on mass m due to mass of cavity is
_ GMm Gir}
G(M/S)m GM rn
F' =
OR-RIlf 50 R^ Acceleration of the mass'm' at the left end of the
The force on mass m due to remaining part of rod
sphere
1 GMm Gm~

w
GM m GM m _ 4\ GM m a
1 "
F^ = F^-F' = 9/?2 in (31} }
50/?2 450 r}
Similarly net force on the mass m at right end of
. ^2

Flo
41/450 41 the rod
F.1 1/9 “50
GMm Grr}

ee
11. Gravitational acceleration due to solid sphere at
(4/)
2^

Fr
GM
a point on the surface of sphere, g ~ Acceleration of mass m at right end of the rod
r}
_ 1 GMm Gn}
Gravitational acceleration at a point on the surface for
ur
2 “m ^(41}'^ }
a

of sphere of radius (Rll) which is to be evacuated

Since tension in the rod is zero. So, acceleration
G(M/8) ^ GM
ks
g' = in the two masses should be same i.e. = ^2
(Rll} ~ 2R}
IS
Yo
oo

Gravitational acceleration at a point just outside 1 iGMm Gm^ _ _1_ GMm ^ Gm'^
eB

the sphere when cavity is created m (31} } ~ m (4/)^ }

, GM GM ^ GM IGm}
r

= g-8 =
R^ 2/?2~2R2 GMm J__J_
ou
ad

or
} [9 16
12. Weight of a passenger = mg, which changes with
Y

change in g. While going from Earth to Mars, there GMm 1 _ IGn}

must be a point where the value of g becomes or
} ^9x16 }
nd
Re

zero and hence weight becomes zero. Therefore,

the curve C is true.
Fi

m 7 7
or
13. Consider an element of the rod of length <ix at a M 9x16x2 288
distance x from point mass as shown in Fig.
8{CF).20. kM
or m = —M = (Given)
288 288
FIGURE 8(CF).2Q
M- b -H so it = 7
dx M
15. Refer to Fig. 8(CF).21, Resultant force on
particle 1,
M- ■H
X ■7
Gm X m Gm x m mv~

Mass of element = (MIL) dx

F; = V2F + F' = V2 (^flr} 4r^ r

Force experienced by point mass due to this

iGm' 2-Jl +1
GMx(MIL)dx GM^ dx or u =
element is dF =
x^ L x^
r I, 4
8/132
“Px<t<Ue^'4. Fundamental Physics (XI) voL.n

Por Difficult Questions

w
Gravitational force on particle at A due to particle

roow
e
at Sis

re
„ GMM ,
16. Let the two spheres collide at point P, after time Fj = a
— along AB

reF
t, when the distance covered by smaller sphere is

uFFll
Xi and of bigger sphere is X2- Gravitational force on particle at A due to particle

e
at D is.
Gravitational force on each sphere
GMM ,

sFr
^_GMx5M ^2 = -— along AD.
a

foro
(12F)2
uor
FIGURE B(CF).22
fk Gravitational force on particle at A due to particle
at C is:
okso
2R
^1
F _G^ along AC
Y
Yo

R P ^ 2a^
oo

y
BB

9R
Resultant force on particle at A along AC
Acceleration of smaller sphere, - Fj cos 45° + F2 cos 45° + F3
rYree
ouu

a
F G5M GA/2 1 GM2 1 GA/2
1 ~
M (12F)2
ad
Ydo

Acceleration of bigger sphere,

GM2 r 2 1] GM
nidn

a
F GM
a ^ LV2"'2.- a

^ 5A/ (12F)2
Re

This force will act as a centripetal force for

F

Using equation of motion,

Fi

circular motion of the particle of radius r =

1 9 ^ 1‘ 2 Therefore,
●^ = 2^^ and
Mv^ GA/2 [2^2+1’
2
= ^ orj:i=5j:2 ..(0 r a 2

Here, Xj + ^2 = 9 R ...(«●) or Mv^ _ GM'^ [2^2+1’

From (0 and (i7), we get Xj = 7*5 R al-Jl cP- 2

17. Refer to Fig. 8{CF).23,

or
V
2 GMr2-^+ll GM xI-35
AC = -yja^ +a^ = -<Jl a a 2-Jl a

a
IGM
AO = OC-r =V2fl/2 = 0= 116
V2 a
 8/133
GRAVITATION

d^ (R-d r
For Difficult Questions
g' = S I-T =8 ...(0
R R

18. Weight of body on earth = mg = 72 N, where, R - d = r= distance of location from the

GM
centre of the earth. When r=0, g' = 0.
where 8 -
R^ From (i) g'« r, till r= R, for which g' = g
Weight of body at height h (= R/2) is For r> R, g' -
gR^ _§R^ or g' oc
1

r2 r2
mGM __ 4 GM
mg' =
(R + RJ2)^ ”9"" R^ Here /? + /t = r

Therefore, the variation of g with distance r from

4 4 centre of earth will be as shown in figure (4). Thus,
= -mg = -x72 = 32N

ooww
9 9 option (a) is correct.
23. Here, mg = 200 N,
1
19. Given, or
4 = 5“. weight of body at depth d = RI2 is
4
8p R/2 mg
mg'= mg 1-— =mg 1- 2

e
R
M

ree
or 200

rFl
GM^IRl 4
M
^ J
4 = 100N

Fre
2

24. Let Rp be the radius of the planet. Then

rrF
R
P f _ ^ 9 ^R
p
1 R
p _ 1
or 9x
4
or
R 4
or
R 2 R
R^ 6x10^ = 6 X 10^ m
R
ouur
k ®) ^)
sffoo e
p 10

GM _ G 4 -] 4
10

R As 8 = 2^ — 71R^ p = — TiGpR
or
R? ~ R 3 ^3
okks
P 2 2
Yo
So g« R
20. Refer to Fig. 8(CF).24(o)
oo
Y
BB

FIGURE 6(Q).24 . If. Rp I

= — or gp - — = 11 ms -2^
M 2m/3 4m/3 8e R, 10 ^ 10 10
rre

M
Acceleration due to gravity at depth .v on a planet is
© © © ©
ouu
YY
ad

!♦ r
♦I r

(a) (b) Let X be the mass per unit length of wire.

Given, X, = 10“^ kg m“*.
dd

GMm GM'^ Force on a small segment of wire of length dx at

Re
iinn

F =
r2 R^ a depth x from surface is
F

In Fig. 8(CF).24(i»), ( x\
= XdxX 1--
dF = (Xdx)g^ RJ
8p
F' =
G (2m/3) X (4m/3) 8 CM—^ 8„
= — f
r2 9 9 x\
= A 1--
R
gpdx
GM Ag ^AR Total force on wire is
21. g = _ ; ___2 —
8 R
R/5 / 2
%change in the value of g F
= /A 1- 8p
^xl00 = -2—xlOO 0 ' -*0
8 R
9R
R__^
= -2(l%)=-2% = A
5 50
8p = 10"3 50
8p
Negative sign shows the decrease in g.
22. The acceleration due to gravity at a depth d below = 10"3x —xbxIO^xl =108N
the surface of earth is 50
8/134
“Pn^uCee^ ^ Fundamental Physics (XI) VOL.II

For Difficult Questions

-7r/?^p h mG—TiR^ p h
or mG
3 e i 3 P^p p
25. At the surface, mg= yt x I e
R^
P
At a certain height, mg' = kxx or
Pe Pp
R^
- =£ R
or h ^-X^Xh = 2 4
-x-xl-5 = 0‘8m
1 8 (R + hf P R 0 ^ 3 5
p ^p

6400 f 16 29. Observed weight of body on equator due to

rotation of earth = m g - m r (i-?.
6400 + 1600 5 25
Observed weight of body due to motion of train

oww
16 16
x = —xl = — cm
from west to east with velocity v (as centrifugal
25 25 forces will be acting ) will be
GM
W = {mg - m r co^) - mv^lr
26. As, 8 = = mg-mu(0-mu(0 =mg-2mvfo
R^

ee
= mg{\ - 2v (o/g) =W{l^2v co/g)
Ag 2AR

FFrlo
30. Let g be the acceleration due to gravity on the

r
8 R
surface of earth. Acceleration due to gravity at

rF
ee
% change in the value of g depth d from the surface of the earth is
Ag xl00 -2AR

rF
= xlOO
ouru
8 R 8
R

-2(-21km)
XlOO =0-67% ffosor
os k
(6300 km) Given, g' = -8 or g 1-- =-
27. Here R = 6-4x 10^ km = 64 x 10^ m.
n R n
ook

g = 10 ms~^
Yo
Y

d 1 n-1
At the pole, weight is same as the true weight or 1-- or
Bo

R n R n n
As weight = mg
reeB

100N = mx lOm/s^ or m = 10 kg rn-n R

or d=
At the equator, latitude, X = 0, apparent weight is
ooY

n
uur

given by
31. g at height h is ;
ad

mg' = mg- mRa? cos- X~mg- mRiip- cos^ 0

dY

mg' = mg - mR(£i^ sR^ 4

Si =
Also angular speed of an equatorial point on (R + h)- (i? + /?/2)2 9^
nind

earth’s surface is
Re

d
2k
g at depth d is Si =S 1--
F
Fi

03 =
= 7-27 X 10'5 rad/s R
24x60x60
Now mg' = 100 - 10 X 64 x 10^ (7-27 x 10^5^2 As per question, — g
4
= 100-338-25 X lO'^x lO"**^ R
Si = 82.
= 100-338-25 X 10-3
= 100-0-33825 or 1=1-^ or £^1_4^5
9 R R 9 "9
« 9966 N

R
32. At equator, acceleration due to gravity is,
28. Here, -4 I ● gj = g - R(X)^ cos^ 9 = g - Roy^
= 1-5 m ; /ip = ?
R
p 3 ’ Pp At height h from the pole, the acceleration due to
K.E. of person = RE. at the maximum height to gravity is
which he can jump. 2h
GM GM 82=8 1- R
rng^h^ = mgphpQT m- = m

Rl ^ R^
P
P
Given, gj = g^
GRAVITATION 8/135

35. Since, / = — dVIdr or dV = - I dr

For Difficult Questions

So, V = ~Idr = -Kr-^dr

2h 2gh
g-R(sP- =g 1 - = g~ 0 0
R R

= -K
^-3+1 y K

RaP- = 2gh -3 + 1 2x^

R Jo

RW 36. Point P is outside the spherical shell A and inside

or h = the spherical shell B. So potential at point P is
2g given by
33. Here, mass of a person, m = 48 kg, Vp = potential due to shell A
Radius of earth, R = 6400 km = 6-4 x lO'^ m. + potential due to shell B
Weight of the person at pole is = = 48 x 10 N Gx4Af GM ■4 1

ww
=-GM -+
Weight of the person at equator - mg-m R(sP- r 2R r 2R

= (48x 10 - 48 X 0-034) N 37. M is the mass of sphere, let p be the density of

FF loo
'48x10 48x0-034 4
kg wt sphere. So M = V x p = - kR^q
10 10 3

ree
Similarly if hf is the mass of cavity
= 48-0 1632 = 47-837 kg wt
4 R
= 47-84 kg wt

reFe
M' = -n
34. When r < /?, the force on the test mass m at the 3

oroFr
rur
surface of the sphere = mg. Force on the test mass 1 (A
M' = ~ d3 1
- nR-’p = —
W
s ff
at distance r from the centre of sphere is
8 [3
or
8
FIGURE 8(CF).25
Gravitational potential at a point lying inside the
k
YYouo

sphere at distance r from centre is

okso

S
/
GM
BBoo

t \
/ V V = - (3r2_;.2)
2R^
I
r ee

t 1
I
\ I
\ I
\ /
\ / GM R
^P = 2R^ 3R2- il
ad
ouur

●●● 2
Yo

r
F = mg GM [llR^ WGM
d
Re
idnY

R
2R^ 4 BR
2
mv
FFin

Therefore, mg — = or V =
Gravitational potential due to cavity portion of
R r R
or V r for r<R
sphere at its centre is

When r > /?, the test mass is outside the sphere. 3GA/' 3 G{M/B) 3GM
2
v = 2r 2 R/2 8/?
GM m mv
Then,
r~
●7
r Net gravitational pot. at P
\
WGM 3GM GM
8
or V = ^R SR SR R
\r
38. Consider element of the rod of width dx as shown
1
or V oc in Fig. 8(CF).26.
rT m
Mass of element, dm = — dx
Therefore, the graph shown in option (c) is /
correct.
8/136 ’4. Fundamental Physics (XI)

~GMm
~ —mv^ +
1 2 (-~GM m
For Difficult Questions
a 2 a

2 _ 2GM 2 GM 2GM 1
or II 1-
a V2 a a

2GM 1

/"Vi.
or V =
a

■►x
40. Total gravitational potential at point 0 due to each
of mass m located at positions shown in Fig.
8(CF).27 is

ooww
FIGURE 8{CF).27

-8 -4 -2 -1 x=0 1 2 4 8
Gravitational field intensity at P due to this
element is Gxl Gxl Gxl Gxl
V = 2 -

e
Gdm G m , 1 2 4 8
dl = = —;; ~x—dx along PC

ere
+d'^
= -2G l+i + ± + ±+...

rFl
In AOPC, X = tan 9 and dx = d sec^ 0 dQ L 2 2^ 2^

Fre
x^ + d^ = d^ (tan^ 0 + 1) = sec^ 8

rrF
1
= -2G = -4G
Gm dsQC^BdQ Gm 1
dl = dQ 1--
I sec^ 0 Id
sffoo
ouur
2
-»
Magnitude of V s: 4 G
Resolving d / into two rectangular components
kosk
we have dl cos 0 acts along PO and dl sin 0 acts
GM m f GMm
41. Gain in PE, ^1 = ”
Yo

perpendicular to PO. iR + h) ^ )
oo
Y

If we find the gravitational field due to other 1 1 GM mh

BB

elements of the rod at P and resolve them into = GM m

R R + hj R(R + h)
two rectangular components, we note that the
rre

/
components of gravitational field intensity 1 2 _
1 GM
perpendicular to OP will cancel out and Gain in KE. ^2 = 2^^ 0
ouu

— m
2
Y

R +h
ad

components of gravitational intensity along PO

will be added up. Therefore, total gravitational GM m
dY

intensity at P due to entire rod is 2(R + h)

innd

+a +a
GM mh _ GM m
Re

■ Gm
/ = dl cos 0 = cosO dQ Given, Ej = E2, so.
Id R(R + h) " 2{R + h)
Fi
F

-a -a

Gm IGm 1/2 A =i
[2sina] = R~ 2
Id Id
^(l/2f + d^ R 6-4x10^
or h = — = 3-2x10^ km
2Gm 2 2
1 1 GM^
d^l^+4d^ 42. Initial KE of the object, =^mv^ =~m
2 r

39. Gravitational potential at P, GM m

2
or e
= mv ...(0
-GM -GM r

a Let K be the kinetic energy given to object to

-GM
escape, then using law of conservation of
Gravitational potential at C, - mechanical energy, we have
a
-GM _m GM m 2
According to law of conservation of energy.
e
S—^K^QotK = = m II
r r
 GRAVITATION 8/137

47. Total energy of satellite in circular orbit of radius

For Difficult Questions
2R
-l2
43. Refer to Fig. 8(CF).28, GMm 1 'GiW
£.1 =PE+KE = - + —m
FIGURE 8(CF).28 2R 2 2R

GM rn

4R

Total energy of satellite in circular orbit of radius

4R
i2
GM m 1 GM GM m
+ — w
Gravitational potential at P 4R 2 4R 8£
Vp = Gravitational potential due to spherical shell
GM m GM m

ww
on surface
Energy spent = £2 - - Jr 4R
+ Gravitational potential due to mass Af at O
GM ( GM
_GMm _gR?-m _mgR

FF loo
3GM
SR SR 8
a a/2 a

400x9-80x(6-4xl0^)

ree
3GM 8
Magnitude of Vp = a = 3-13xl0’j
1
48. Let i' be the velocity of projection of the particle

reFe
44. Work done = RE. + K.E.; - 3 = PE+ — xlx2^
or RE. = -3-2 = -5 J ^ from the cavity at a depth r {= £/100) in the moon.
oroFr
rur
or Gravitational potential where
v = ^2GM/R
s ff
= RE./mass = - 5/1 = - 5 J/kg We know that if point lies inside the sphere of
45. Workdone = final PE - initial PE
k
radius R, mass M at distance r from the centre of
YYouo
okso

sphere, where r<R, then gravitational potential

GM m GM m __GM m at that point is
{R + R) ~R 2^
BBoo

FIGURE 8(CF}.30
r ee

mgR
2
[v GM = gR^]
ad
ouur

46. Refer to Fig. 8(CF).29, AC = BD =l>pi

Yo

FIGURE B(CF).29
d
Re

Ai
idnY

B
\
71 2
V
FFin

I V V =-
GMOR-?-r^)
2/?3
✓ S

✓
4 S 3 In the given question, r-{R- R/lOO)
D C
/ = (99 R/lOO)

Total energy of the system of four masses placed From law of conservation of total energy
at the vertices of square will be (PE + K£) at the cavity = PE at height h
= W12 + (Wj3 + W23) + + W24 + W34)
3/?2- ("99/? f
GMm 1 2 _ GMm
+ —mv
Gm^ , Gm^ GnP' Gm^ Gm^ Gm^ 2/?3 _ 100 2 R-¥h
I I I
GMm 1 (2GM\ GMm
or - [3/?2_o.98/?2] + i m
IGm^ ■ 1 2/?3 2 I ^ R^h
2+
I V2 On solving, we get, = 99 R
8/138
‘Pxtxdee^ Fundamental Physics (XI)
52. Let K be the KE given to the satellite on the
For Difficult Questions
surface of earth to take it to the desired location.
Following the law of conservation of total energy,
49. Initial velocity of body is zero. Total energy of we have
GM m mgR-?
the body = - GmM GmM 1
/
GM ^
r r + K = - + — m
R (2R + R) 2 3R
Let i> be the velocity acquired by body on reaching
the surface of earth then total energy of the body GMm GMm GMm
on the surface of earth - +
3R 6R 6R
1 2
= —mv = — - mg R GMm GMm SGMm
or K =

oww
2 R 2
R 6R 6R
According to law of conservation of energy
53. Consider an element of the disc of radius x and
1 2 mgR^ thickness dx. Mass of the element is
— - mgR = -
r

2gR^ 1 1

e
or =2gR-

FFrlo
re
r R r

ree
F
or V

rF
50. Total energy of rocket on the surface of earth
1 2 GM m 1
= —mu = —mu^ - mgR
fsoor
ouur
2 R
total energy of rocket at the height h
skf
GM m _ mg R^
ooko
= 0-
(R + h)~~(R + h)
Yo
M x2nxdx 2M
dM = xdx
Y

According to law of conservation of energy [7l(4/?)2-7l(3/?)2] 7/?2

Bo

mg /?2
reB

1
— mu ^-mgR = - Gravitational potential at P due to element of the
2 (R + h)
2M 1
1 2 o2 r i 1 mg Rh disc dV = —G xdx X
uur
oY

— mu^=mgR^ — IR?-
2 L/? + (/? + /i) ●jx^+(4R)^
ad

R + h _ 2gR Workdone in taking a unit mass from point P to

dY

or
h u
2 infinity is
R
2gR R^2gR ^
innd

AR
Re

or - + 1 = or 2M 1
h 2 xdx
u h IV =-V = + G
IRp-
^x^+(4Rf
Fi
F

R 3/?
or /i =
2gR u 2 R 4R
2GM xdx
U‘
7/?2 J [X^ +(4/?)2]>/2
3R
51. As/v is the escape velocity of the body from the
platform, then using the law of conservation of
total energy
Let
^x^+(4R)^ =z
GM m l , . .2 r\ Differentiting it, we have
or
- + -m(f vy = 0
{R + R) 2
|u2+(4R)2]
-1/2
x2xdx = dz
gR^
or fv =
R
= V^- xdx
or = dz
The value of escape velocity from the surface of
= v!'j2
■ylx^+{4Rf
earth v = -^2gR or

When, x=3R\ z = R)^ + (4Rf =5R

fv = v/->l2 or / = l/-s/2
GRAVITATION 8/139

For Difficult Questions FIGURE 8(CF).32

I X

When, x = AR ; z = >/(4^)^T(4^ = 4V2/?

✓
✓ I \
/ s
/ N

/
t ^^0 \
\

A^R I
o
\

IGM IGM
w =
IR} I “ ir}
[^] 5R
a-Hr
\
\
R R ^^2 /
5R \ /
\ t
\ /
X

2GM 2GM
[4^R~5R] [4V2-5]
IR^ 7R
Using law of conservation of energy, we have
54. Here, v = k v^, where k<lsov<Vg. Let M, m be
the masses of earth and particle respectively, h be 1‘ 7 ( 2GMm^

ww
-mvt + - = 0
the height reached by the particle and R be the 2 ° R
radius of the earth.
ACM
Using law of conservation of mechanical energy, or V
0 “
R

Flo
we have

'4x(6-67xIQ-^^)x3xlQ^^

ee
1 2 GMm _ G Mm
— mv
10’*

eer
2 R (/? + /»)
= 2-83 X 105 -I
= 2-8 X 105 ms-*

FFr
ms

_GM GM GMh 56. Gravitational field is maximum at the surface of

oorr
uur r
or
a sphere. Therefore,
2 R (R + h) R(R + h)
s ff
GM GM
1 I
= 2 or = 2
i./:V = GMh
R?1 l2
...(/)
sk
or
YYoo
2 ^ R(R + h)
ooko

and
GM^ = 3 GM^ = 3 ...(»■)
eBB

or
1 '2GM\ GMh
or - i^2 X
2 R
/?! 22
R(R + h)
Dividing (/) by (it), we get
uurr
ad

2GM AM M
' I
— =1
Yo

or
e
R M, 3x4 6
dY

12 = /* 57. Here, m = 60/1(X)0 kg, F = 3-0 N

or k^ Rk^ + h/^ = h
Re

or
innd

R +h p 30
Gravitational intensity = — =
FFi

Rk^ m 60/1000
or h{\~k^) = Rk^ or h =
= 50 N/kg
2tc 2t: 271 2ti
55. Here, Af, = A/2 = A/ = 3 x lO^* kg, 58. r, = or(0^ = ; TB or 0)
T,I ' ~ B -
2R = d = 2x 10“ (0
if '^2
/e=10“m(Fig. 8(CF),32) 1 = 1 [v7i = r2 = 71
or

Gravitational potential at O due to two stars 5j ■■ "2 27t/72 7;

and $2 59. As time period of revolution
GM 2GM
V =- x2 = - 2%r 2nr GM
R R r = V =
V -Jgm V r
Let Vq be the minimum speed of meteorites while
passing at O in order to escape. Let m be the mass
of the meteorites. squaring both sides we gel
8/140 Fundamental Physics (X1)EEZ9H

4te^
For Difficult Questions
64. Orbital velocity of a satellite, v =
j?
7-2 = = Kr^ (Given), X r = CM = constant
GM

4tc^ (3 V)2x4/? =y|x/?

K = or GMK = 4n^
GM V
B = -^36V^ =6V
GmM 47c2 4^2 ^n+1
60. ~ mR or = 65. Let u be the minimum initial velocity of the mass
T~ GM rn to escape. According to law of conservation of
2n total energy, we have
or T = ^(«+i)/2 or roci?(«+l)/2
.Jgm 2GMm 1
+ - mv^ = 0
L 2
61. Given, ll-2kms“* - 2GM^
2 _ 4GM

w
or t; or u = 2.1
L L
Here, = 3 x 10^ ; Let r be the distance of
rocket from sun. Then r = 2*5 x 10^ Rg. Let m be GM

Flo
the mass of the rocket. 66. Orbital speed of satellite, V =
r

According to law of conservation of mechanical

e
energy when rocket is to leave the sun earth system Escape speed, Vg = V2 x Orbital speed

rree
GM
(-GMm'\ (-GM^m = 0

r FF
~ mvl + = V2
\ r
2 ^
I r
J KE of the object to escape from satellite
uurr
or Uj =
2GM^L+ 2GM^L
r
for
= -J mV^
x,2 = -1 mx 2GM GMm
= mV2
2 ^ 2
kss
r r

2GMg 2Gx3xlO^M^
ooook

GM GM
Yo

2-5x10^/?^ 67. Orbital velocity, Vq = y—

2GM( 3x1()5 2GM
eB

(as h « R)
V 1 1+ ^xI3
s
2-5x10*^ R
e
Velocity required to escape.
l2G~M _ I 2GM
urr

= {\\-2)^ =42kms
-1

yR+h~i R
ad
Yo

62. When a dust comes out with speed v w.r.i.

satellite, then thrust on the satellite, 2GM 'GM
dY

dM Increaseinvelocity = Ug- ^ R
F = -u = - V (p u) = -13
Re
innd

dt
=^j2jR-^ =(.S-l)4g»
F
Fi

Acceleration of satellite = —- 68. According to law of conservation of mechanical

M M energy.

J¥ GM m GM m
63. Orbital speed of satellite, v = ...(0
im(A:Ug)^+
2 V
- R (/? + /!)
If r decreases, v increases differentiating (i) we get
^[gm GM m GM m _ GM mh
or =
dv = -—-jGM r~^'^ dr =~ 3/2
dr
(R + h)~ R(R + h)
2 2r

1 -7^ -mk^x2gR = mg Rh _ tngR(r-R)

dv 1 dr

V 2 r^/2
dr/^(GM/r) = 2 r
or
2 R + h r

dv [A&h = r-R]
%increase in velocity = xlOO
V r-R R R
or - = 1 or r =
1 dr 1 1
xl00 = — X XlOO = 0*5% r r \-k^
2 r 2 100
GRAVITATION 8/141

For Difficult Questions On comparing it, with v-Jn , we get A/ = 2

71. Escape velocity from Earth’s surface
69. The earth will become black hole if the escape
velocity on earth is equal to the velocity of light, [2GM _ I 2G 4
-KR^p ISnG p
^ i R VR
V X
= R
2GM 2 CM y 3
i.e., V = c or R =
^ "V R From Planet, escape velocity is
■y
C~

2 X (6-67 X 10~‘J) X (5-98 x 10^^) 8tcGp V

e
/. /? = Vp = (4/?) 3 4R~ 4
(3 X 10^)2 V
p

oww
= 8-86 X 10-3 m « 10-2 m.
or Vp =4Vg = 4v (V v^ = v)
1 GM 1 GM 72. We know escape velocity is given by,
70. Here
Sp 4 Sp or
(.R + x)'^ ~4 /?2 2GM
V
or (R + x)2 = 4R^ or R+x=2R
e
R

e
or x==R
^GM/R

re
V
A _

FFrlo
GMm 1 GMm = 1
Also -
R
+ — mv 2 _
2
V
B V2G(Af/2)/(/?/2)

rF
R-¥R

ee
V
1 This,
2 _ GMm GM or n = 4
or —mv or V = V 4
ouru
2 2R R
B

rF
73. Angular velocity to = ^'
1
As escape velocity l.e., (Ooe —

{
2GM GM
ffosor T
os k
or = V tor- 8
R R i = -2
“2 1
1
ook

m
Yo
Y

Multiple Choice Questions (with One or More than One Correct Ahsi^j|
Bo
reeB

74. Gravitational potential at a point inside the earth 75. Inside the earth
ouY

at a distance r from the centre of earth is,

R-d^
ur

- 1 “ gr
V =
-GM(3/?2-r2) 8 = 8 1-- = 8
ad

R R R
Yo

2R}
where r is the distance from the centre of earth.
d

At the centre of earth, r = 0,

GM
g increases when r increases provided r<R
nidn

^GM
V = - x(3/?2-0) = - (Minimum) At the centre of earth r = 0, g' = g x QIR = 0
Re

2/?3 2R
2h
F

For a point outside the earth at distance r from Outside the earth g' - g 1-
Fi

as h increases
R
GM g' decreases.
centre of earth, V = ~
r 76. The speed and time period of revolution of a
GM satellite are independent of mass of satellite but
When r = oo, V =- = 0 energy and angular momentum of satellite
OO

depends upon mass of satellite.

Gravitational intensity at a point distance r from

i
GM 2GM
the centre of earth is / = 77. As =
, Vg is independent of mass
r2 ’ R

When r = oo, 7 = 0 of the object and the direction of projection of

When point is inside the earth, then the object but depends upon mass (M) and radius
(R) of the planet.
7 = G 4 T
-:rX~nrp or
47cGpr
7 =
r2 3 ^ 3 78. Work done W = F .ds ; for a displacement over
When r = 0, 7 = 0 —^ ● ,

any small part of the orbit F and ds are

8/142 ‘P’uuU^ 4 Fundamental Physics (XI)
The reduced mass of the two objects,
For Difficult Questions
mx4m _ 4m
m + 4m 5
perpendicular to each other, hence work done is
zero. Also for one complete revolution, If Vf is the relative velocity of approach and r is
displacement is zero, hence work done is zero. the distance between the two objects, then
79. Refer to Fig. 8(CF).33, O is the centre of mass of total KE of objects = loss in PE of objects
Gmx4m 4Gm^

w
the system of 3 particles present at A, B and C
respectively. Here AD = AB sin 60° = r -73 / 2 r r

As total KE of objects = loss in PE of objects, so

OA = -AD = — X = OB = OC 1 4Gm^
3 3 2 "73

e
r

e
1 4m 7 4Gm^ lOGm

orw
or —X
or r
=

r
2 5 r

81. The mass of the sphere with cavities is

F
symmetrically situated w.r.t. the origin. The circle

ullo
^ + = S6 has a radius 6 and all the points on it

FF
are at a distance 6 from the centre of sphere where

e
the whole mass of the sphere can be supposed to

sre
r
be concentrated. The circle is outside the sphere.

oF
The similar explanation is valid for + ^ = 4. The

k
Gravitational force on the particle at B due to gravitational pull at the centre of sphere is zero.
oofr
uor
Gmxm
particle at C is FBC along BC 82. Gravitational potential at the centre of the irng
sf
r2 even if mass is non-uniformly distributed on irng
ko
Y
Gravitational force on the particle at B due to GM
y
rBB
Yo
IS
GmXm c
R
along BA
oo

particle at A is =
eY

r2 Gravitational potential at infinity is, V^= 0

The resultant force on particle at B will act along Workdone = increase in potential energy
re

BO is
GM GM m
u

G'^m^ G'^m' 2Gm^ Grr? = m (Vy- V^) = m 0- R R

d

F = cos 60°
ou

X
o
ad

r" r2 r2 2nr Inr 2ttr3/2

83. Time period, T =
nY

This force is providing the centripetal force, so V VGMTt (GM) 1/2

1/2
SGm^ 4jt2 i.e.. Toe l/(G)
nid

mr
Re

If G is decreasing with time, then Tincreases, i.e.,

73 r2
F

r2
earth takes longer time for one revolution. Due
Fi

47c2r3 to it, the length of day, on earth, will decrease.

or 7^ = 4 or Toe But the length of year will increase w.r.t. the
y 3Gm m
original year of earth. The centripetal force on
80. Initially both the objects are at rest at an infinite satellite
separation, hence their initial PE and KE are zero. mv^ mGM / r GMm
Therefore, following law of conservation of F =
r r r2
mechanical energy, total energy of the two
As G decreases with time, F also decreases with
objects at separation r will be zero.
time so the earth will move away and describes
There will be two equal and opposite forces acting
a spiral path of increasing radius.
on two objects. The net torque on two objects is
1 I GM
zero. So, the angular momentum will remain Kinetic energy, £ = — mij2 = — m
conserved. InitiaUy, both the objects were at rest, K~2 2 r

therefore, the angular momentum about any i.e., E G OC

point is zero. so as G decreases KE decreases.

 8/143
GRAVlTATfON

So total mechanical energy of the second is

For Difficult Questions greater than that of the first satellite,
84. Using Kepler’s third law, 85. Let rp, Vq are the radii of planets P and Q
/ n2/3
respectively. Then
T
^2 256

«2 = ^l f
= R = 4R
Anri = A, A%tq=AA ; rQ = 2rp

w
\ 32;
For planet R,
1 2 1 GM GM ni
K.E. of satellite E K - mvi =-m A .3
2 ^ 2 r 2r P

e
1 or =
As EK oc —

Escape velocity,

e
r

wr
A

oo
_1__R i 2G-7cr p

r
<E 87cGp
% 2GM
or
so
K 3
E 4R 4
V r ^
1 V =
K ^■2

F
1 r

FFllu
Total mechanical energy, £ = PE + KE i.e., D o« r
1/3
GMm GMm GM m = I :2:9
Vp'.VqWp - rp:2Tp.9
2r 2r 1/2
r
Vp>VQ>Vp and Vp/VQ =
Here -ve sign shows that as r increases £ Thus options (6) and (d) are true.

rese
uro
increases.

Fkr
o
Dn Multiple Choice Questions (Based on the given Passage/Compreh ension)
foo
fr When each body is describing a circular orbit with
kso
86. Gravitational force on body at C due to body at A
Y
centre at 0, the force F provides the required
Gm X m
centripetal force. If u is the speed of the body in
Y

along CA
B

IS F.1 =
Yo
circular orbit, then
oo

2
V3Gm2 Gm
eBr

mv
or V =

(r/^) r2 \ r
rue

88. When the three bodies are at the vertices of

oud

triangle A, B and C, then the gravitational potential

no

energy of the system is

ad

/ \ /
-Gm^ -Gm^ 3Gm^
Y

Gmxm
U1 +
\ r J \ r
ndi

r r

When one body at C is taken for way from the

Re
F

Gravitational force on body at C due to body at B other two bodies at A and B, there will be no
Fi

Gm X m interaction between bodies at far off distance with

is along CB.
r2 the bodies at A and B. Then the gravitational
potential energy of the system is
The angle between £j and F2 is 60® Gm^
So, the total gravitational force on body at C is
= [F^ ■¥ F^ A-2FyFy(M2f'^ = Work done in taking the body at C for away
from other bodies
[V F, = F^\
V3Gm2 Gm^ 3Gm^ 2Gm^
= t/2-t/l r r r
r2
87. Here, CO = lcD=-AC sin 60® 89. Here, AD = r sin 60® =rS/2-
3 3
The gravitational field at point D due to equal
= -rx
2 masses at B and C will be equal and opposite,
3 2 hence cancel out each other’seffect. Thus the total
8/144
Fundamental Physics (XI) VOL.II

For Difficult Questions

_ 2x(22/7)x(6-4xl0^ + 1-6x10^)
7-1x103
gravitational field at point D is due to mass at A = 7082 » 7100 s
only and is given by 92. Total energy of satellite = RE. + K.E.
Gm Gm 4GM GM m 1
E =
along DA + -mv^0
{ADf (r-^5/2)2 “ 3r2 (R + h) 2
90. According to law of conservation of total =
GMm
+ ~m
\ GM GM m
mechanical energy, {R + h) 2 {R + h) 2{R + h)

w
total energy of rocket at the surface of earth Energy spent to take the satellite out of the earth’s
= total energy of rocket at the highest point gravitational field is
1 -GMm -CMm^
or -mv^ + = 0+ = - (total energy of satellite)
2 R R+h J GM m

e
roow
or
GM GM gR^ gR^ 2 {R + h)

re
2 R {R + h) R {R + h)
(6-67xlQ~^^)x(6x!0^^)x200
R \ ( h
= gR 1- = gR 2 (6-4x10^+1-6x10^)

reF
R + hJ \R + h)

uFFll v'^{R + h) = 2gRh

= 5-0 X 10® J

e
or
93. Initial orbital radius,
or
Rv^ = 2gRh~v^h
/?i =6-4x 10^+ 1-6 X 10^= 8 X 10^ m

sFr
= {2gR-v^)h
Rv^ Time period of revolution Tj = 7100 s

foro
or h = ofk New orbital radius,
uor
(2gR-v2) /?2 = 8 X 10^ + l-6x 10^ = 9-6 X 10^ m
64x10^x(5x1Q3)2 New time period of revolution 72 = 2
kos
(2 x 9-8 X 6-4 X10^) - (5 x lO^)^
Y
\3/2 n3/2
Yo

= 1-6 X 10* m ^2 9-6x10*

reeBB
oo

72=7; = 7100
91. Time period of revolution of satellite, R
K “I J I 8x10*
2n{R + h) = 7100(1-2)3/2 = 9330 s
uurY

V
0
09 Matching Type Questions
ad
doo

94. (A) Gravitational potential energy of a satellite 95. (A) Escape speed of a body from the surface of
nY

orbiting close to earth = gravitational pot. on

2GAf
surface of earth earth =
nid

V R
Re

X mass of satellite
F

(B) If escape speed of a body from the atmosphere

Fi

GMm
R
at height R from the surface of earth is v, then
(B) Kinetic energy of orbiting satellite close to I -GMm
—mv^ + = 0
earth =-mv^ I (GM
= ~m —
1
^ GMm 2 2R
2 0 2 t R IR

(C) Kinetic energy of orbiting satellite at height or V

GM
R from the surface of earth R
\
1 1 GM GMm
'2 =
_ —m
= -mv (C) If v' is the escape speed of a body from the
2 0 2 V
2R y
4R
’ satellite orbiting at a height R from the surface of
earth, then
(D) Total mechanical energy of satellite orbiting
close to earth
1 '2I +
-mv^ '' —
—GMm 1
+—m
GM
= 0
- GMm , GMm GMm 2 2R 2 2R
= PE + KE = '' / \
R 2R 2R
 8/145
GRAVITATION

For Difficult Questions

V
1 - h-1
1
^2
\GM
On solving, v' = (B) L = mv R
2R

(D) If v" is the escape speed of a body from the 2 2 1,

=-x-x-=l
atmosphere at height {R/4) from the surface of Lj m2 i>2 ^ 1 1 4
earth, then
GM m m
1 , ( -GMm = 0 (C) Kinetic Energy, K = or K OC —

~mv + rz^ R R
2
V
R + R/4) >

oww
KI m
1 ^2 = -x-
l-x^ 2 4 = 8_
I8GM
1 1
^2 m2 R^
//
or V
5R

^ R 1 ^12
s3/2
T 1 i
GM 1
iD)TccR^'^ -f =

e
96. (A) V = or V oe
^2 4j 8

FFrlo
\ R

re
ree
F
K9 Matrix-Match Type Questions

rF
97. Knowledge based question. 98. Knowledge based question.
ffsoor
ouur
VI. Integer Type Questions
kosk
101. Let p and p' be the density of earth and planet
Yo
oo
Y

99. g = respectively R and R' be their radii, then p = 4 p'

BB

and /? = 3 R'. Let h and h' be the height of jump

of a man on earth and planet respectively, g and
re

3«
R = g' be the acceleration due to gravity on earth and
AnGp
planet respectively. Here, /i = 1*5 m; /i' = 3 x a: m
uur
oY
ad

i.e. R
g
GM G f4 _3 1 4
dY

OC —

g =
P R^ R^ L3 n 3
innd
Re

gp
Hence, X

Pp
11 2
...(0 and g'~
/?'2
-nR'^ p' ^tiGR'p'
Fi
F

The gain in potential energy at the highest point

Escape velocity, v - -^2gR will be the same in both the cases. Hence
mg'h' = mgh or h'= ghig'
X

Se 11 11 2} (A
-tiGRo
or h' = .-
l3 4x/j = -^X-^X/l
V6x(3/2) R P
= — X11 = 3 km/s -nGR'p'
or Vp = 11
v,=
11 3 J
100. Time taken for the shift of 360“ = 24 h
= 24 X 60 min = H^x^xI-5 = 18 = 3 XX(given)
R' p'
24x60
Time taken for the shift of 1“ - = 4min
or X = 6
360
 8/146
Fundamental Physics (XI) voL.n

For Difffcult Questions 4x10^ 10^

= 2tc
102. A body projected upwards with escape speed will 8 1

go to infinity. Thus, the speed of the body falling = 7t X

10‘* km/h = 7C X 10" km/h (Given)
on earth from infinity will be equal to the escape n = 4
speed of the earth which is given by
104. V
V
^ =^|2gR = V2x9-8x6400xlCp
= ll-2x 10^ m/s and Vm
m
R
m

Here, 10 g, = 6 g
K.E. acquired by body = m

V R
e Se
- = V6xl0 = 8

w
1 X

= -xl00x(lI-2xl03)2=6-27xl09j V
m m
R
m

As per question, 6-27 x 10" = 6-27 x 10^ 105. According to law of conservation of total energy

Flo
so n = 9
(PE + KE) on the surface of earth = PE at height h

e
GM m \( —mv
\ 7

ree
103. Here, Tj = I h, Tj = 8 h ; rj = lO'* km ; T2 = ? + —
GM m
R 2 2 e

FFr
From Kepler’s law iR + h)

n3/2
GMm 1 (2GM
urr
T'l GM m
'2=n
M
= 10*^ -
V
I /
= 4x lO^km
for
or -

R
+ —m
4 [ R {R + h)
kkss
Relative speed of $2 w.r.t. = | U2 - Uj I or 1 GM m _ GM m
2 R ~ '{R + h)
Yo
ooo

_ 27cr2 I
= 2n ^--L
r~ ,r
or R + h = 2R or h = R = nR (as per question)
^2 .T
L^2 ^ij
eB

1
n = 1
ur
ad

VII.
YYo

Assertion-Reason Type Questions

d

FOR MEDICAL STUDENTS

Re

111. Here, both the Assertion and the Reason are

in

106. Here both assertion and reason are true and reason correct and the Reason is correct explanation
of Assertion.
F

is correct explanation of assertion.

112.
In case of planet orbiting the sun, r = 0 Here, the Assertion is correct but the Reason is
—i
wrong. Since, the earth also rotates from west to
east.
dL dL
as X = = 0 or L is constant 113.
dt dt Here, both Assertion and Reasons are correct
because
107. Here both Assertion and Reason are true and the
\3/2
Reason is the correct explanation of Assertion. 'i r/2
\^I2

108. Both the Assertion and the Reason are wrong. T'l-T'x = 365
\ r J
VI >
109. Here, Assertion is correct but Reason is wrong.
110. Both the Assertion and the Reason are correct and 365
Reason is the correct explanation of Assertion. = 129 days.
2V2
 8/147
GRAVITATION

116. Here, statement-1 is wrong but statement-2 is

For Difficult Questions correct. Let p be the density of the material of
each sphere.
FOR ENGINEERING STUDENTS
M ,=-K r p
4 3
Then,
114. Here, statement-1 is wrong and statement-2 is ' 3 ^
correct.

At height R, velocity of the body, u = 0. and =

(2r^) p
K.E. of body = 0 ;
Gm GM m Distance between their centres = r + 2r = 3r
RE. of body = —
(R + R) 2R
Now F =
GMj A^2
GM m Or?
Total energy = K.E. + P.E. = -

ww
2R
( 47t U4
On the surface of earth, let v be the velocity of G
V 3
p
/ \ ^TtSr^p
the body

Flo
9^2
1

K.E. ofbody =^mv^ :

e
This gives F »= r**

rere
117. Here, statement-1 is wrong but statement-2 is

r FF
GM m
correct.
P.E. of body = -
R
Given dm/m - - 1/100
uurr
Total energy =-mv
1 2 GM m and foordRIR = - 1/100
2 R As g = GM/R^
kss
Taking log of both the sides and differentiating
Yoo
According to law of conservation of energy
ooook

we get
1 2 GM m _ GM m
eBB

—m V
2 R 2R dg _dM IdR
g M R

\ 2 gm gm gR^
uurr

or — V = = =
dg
ad

2 R 2R 2R % change in g = — x 100
Yo

v = 4s^
dY

or
1
115. Here, statement-1 is wrong but statement-2 is = .^-2 X100 = 1%
Re
innd

100 100
correct. As for a satellite
Fi

GM m
Positive sign shows that there is increase in the
K.E.= value of g by 1%.
2R
118. Here, both statement-1 and statement-2 are
GM m correct.
and P.E. = -
R
GM 1
As V = i.e., V oc
Total energy, Eq = K.E. + P.E. y r rT '
GM m GM m
2R R V
so 1 _ r2
^2 ~y r
GM m PE

2R 2 119. Here, statement-1 is correct but statement-2 is

wrong.
PE = 2 Eq.

I
8/148
Fundamental Physics (XI) voCi?

● ■/
! :3>iv^^c?Yr<Lrr;(et^ 120. Here statement-1 is wrong and statement-2 Is
For Difficult Questions correct As

2G 4 . GM m
As = —x-nR^ p lV=m5 =
R 3 ^ /?2
so
v^«/? GM m
and W' = mg' =
V
p _ 2R
(/?+/i)2
= 2
V R
e
GM m 4
= - W
or
(R + R/2)2 9

w
Flo
reeee
FFr
for
ur
kkss
Yo
oo
eB
r
ou
ad
YY
ndd
Re
Fi