GRAVITATION

CONTENTS
Particular's Page No.

Summary 01 – 03
Board Level Exercise 04

Exercise - 1 05 – 10
Part - I : Subjective Question
Part - II : Single Choice Objective Question
Part - III : Match the column

Exercise - 2 10 – 16
Part - I : Only one option correct type
Part - II : Single and double value integer type
Part - III : One or more than one options correct type
Part - IV : Comprehension

Exercise - 3 16 – 19
Part - I : JEE(Advanced) / IIT-JEE Problems (Previous Years)
Part - II : JEE(Main) / AIEEE Problems (Previous Years)

Answer Key 20 – 22

Advanced Level Problems 23 – 24

Answer Key 25

ALP Solutions 26 – 31

JEE (ADVANCED) Syllabus 2015

Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites
in circular orbits only ; Escape velocity.

BOARD EXAM SYLLABUS 2015

Kepler’s laws of planetary motion. The universal law of gravitation. Acceleration due to gravity and its variation
with altitude and depth.
Gravitational potential energy; gravitational potential. Escape velocity, orbital velocity of a satellite. Geostationary
satellites.

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Gravitation

GRAVITATION

SUMMARY
Newton's law of gravitation :
Fg Fg
Gravitational attraction force between two point masses m1 m2
G m1 m 2 r
Fg = and its direction will be attractive.
r2

Gravitational force on (1) due to (2) in vector form

G m1 m 2
F12 = – r̂12
r2

Gravitational Field :- Gravitational force acting on unit mass.

F
g=
m
Gravitational Potential : - Gravitation potential energy of unit mass

B
U dVg
Vg = g=– and VB – VA = – g·d r
m dr
A

(i) For point mass :

M
r <g
GM GM
g= 2
, V=–
r r
(ii) For circular ring
GMx
g= 2
(R x 2 )3 / 2
GM
V= –
R2 x2

(iii) For thin circular disc

2 GM 1
g= 1–
2
R2 R
1
x

– 2GM
V= 2
R2 x2 – x
R

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Gravitation
(iv) Uniform thin spherical shell : -

GM GM
gout = 2
gsurface =
r R2

gin = 0

Potential :

GM GM GM
Vout = – Vsurface = - Vin = -
r R R

(v) Uniform solid sphere :- (Most Important)

GM GM
gout = 2 gsurface =
r R2
GM
gin = r gcentre = 0
R3
GM GM
Potential : Vout = – Vin = – (3R 2 – r 2 )
r 2 R3
GM 3 GM
Vsurface = – Vcentre = –
R 2 R
Self energy :
1 GM2
Self energy of hollow sphere = Uself= –
2 R

3 GM 2
Gravitational Self energy of a Uniform Sphere = Uself= –
5 R
Escape speed from earth's surface

2GMe
Ve = = 11.2 km/sec.
R

If a satellite is moving around the earth is circular orbit, then its orbital speed is

GMe
V0 =
r
where r is distance of satellite from the centre of earth.

GMe m
PE of the satellite = –
r

1 GMe m
KE of the satellite = mv 20 =
2 2r

GMem
TE of the satellite = –
2r

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Gravitation
Time period of Geo - stationary satellite = 24 hours

Kapler's laws : -
(i) Law of Orbit :- If a planet is revolving around a sun, its path is either elliptical (or circular)
(ii) Law of Area : -
View (i) If a planet is revolving around a sun, the angular momentum of the planet about the sun
remains conserved
View (ii) The radius vector from the sum to the planet sweeps area at constant rate

dA L
Areal velocity = = = constant
dt 2m
(iii) For all the planets of a sun

4 2
T2 R3 T2 = GM R3
s

GMe 2
h 2h
1. Effect of Altitude : gh = R h 2 =g 1 ~ g 1 when h << R.
e Re Re

d
2. Effect of depth : gd = g 1
Re

3. Effect of the surface of Earth

The equatorial radius is about 21 km longer than its polar radius.

GM e
We know, g = Hence gpole > gequator.
R 2e

4. Effect of rotation of the Earth

2
Consider a particle of mass m at latitude . g = g – Re cos2
At pole = 90°
gpole = g , At equator = 0
gequator = g - 2Re . Hence gpole > gequator

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Gravitation

1. What is gravitational force ? [1 mark]

2. Which is greater, the attraction of the earth for 1 kg of iron or attraction of 1 kg iron for the earth ? Why ?
[1 mark]

3. Why is G called universal constant of gravitation ? [1 mark]

4. According to Newton’s law of gravitation , the apple and the earth experience equal and opposite forces due
to gravitation. But it is the apple that falls towards the earth and not vice-versa. Why ? [2 mark]

5. Choose the correct alternative: [1 mark]

(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its
kinetic/potential energy.
(b) The energy required to launch 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.

6. Does the escape speed of a body from the earth depend on : [1 mark]
(a) the mass of the body
(b) the location from where it is projected
(c) the direction of projection
(d) the height of the location from where the body is launched?

7. A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular
speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit?
Neglect any mass loss of the comet when it comes very close to the Sun. [1 mark]

8. Show that Kepler’s second law follow from the law of conservation of angular momentum. [2 mark]

9. Let us assume that our galaxy consists of 2.5 ×1011 stars each of one solar mass. How long will a star at a
distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the
Milky Way to be 105 ly. [2 mark]

10. Suppose that the gravitational force varies inversely as the nth power of distance .In that situation, what will
be the time period of planet in circular orbit of radius R around the sun ? [4 mark]

11. The largest and the shortest distance of the earth from the sun are r1 and r2 .What will be its distance from
the sun, when it is at perpendicular to the major-axis of the orbit drawn from the sun ? [4 mark]

12. Choose the correct alternative : [1 mark]

(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere
of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula G Mm(1/r1 – 1/r2) is more/less accurate than the formula mg (r2 – r1) for the difference of
potential energy between two points r2 and r1 distance away from the centre of the earth.

13. Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face,
(c) headache, (d) orientational problem. [1 mark]

14. Explain, why one can jump high on the surface of moon that on the earth. [1 mark]

15. Imagine a spacecraft going from the earth to the moon. How does its weight vary as it goes from the earth
to the moon ? [2 mark]
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Gravitation

PART - I : SUBJECTIVE QUESTIONS

A-1. The typical adult human brain has a mass of about 1.4 kg. What force does a full moon exert on such a brain
when it is directly above with its centre 378000 km away ? (Mass of the moon = 7.34 × 1022 kg)

A-2. Two uniform solid spheres of same material and same radius ‘r’ are touching each other. If the density is ‘ ’
then find out gravitational force between them.

A-3. Two uniform spheres, each of mass 0.260 kg are fixed at points ‘A’ and
‘B’ as shown in the figure. Find the magnitude and direction of the initial
acceleration of a sphere with mass 0.010 kg if it is released from rest at
point ‘P’ and acted only by forces of gravitational attraction of sphere at
‘A’ and ‘B’( give your answer in terms of G)

B-1. The gravitational potential in a region is given by V = (20x + 40y) J/kg. Find out the gravitational field (in
newton / kg) at a point having co-ordinates (2, 4). Also find out the magnitude of the gravitational force on a
particle of 0.250 kg placed at the point (2, 4).

B-2. Radius of the earth is 6.4 × 106 m and the mean density is 5.5 × 103 kg/m3. Find out the gravitational
potential at the earth’s surface.

C-1. A body which is initially at rest at a height R above the surface of the earth of radius R, falls fr eely
towards the earth. Find out its velocity on reaching the surface of earth. (Take g = acceleration due to
gravity on the surface of the Earth).

C-2. Two planets A and B are fixed at a distance d from each other as shown in the figure. If the mass of A is MA
and that of B is MB, then find out the minimum velocity of a satellite of mass MS projected from the mid point
of two planets to infinity.

D-1. A satellite is established in a circular orbit of radius r and another in a circular orbit of radius 1.01 r. How much
nearly percentage the time period of second-satellite will be larger than the first satellite .

D-2. Two identical stars of mass M, orbit around their centre of mass. Each orbit is circular and has radius
R, so that the two stars are always on opposite sides on a diameter.
(a) Find the gravitational force of one star on the other.
(b) Find the orbital speed of each star and the period of the orbit.
(c) Find their common angular speed.
(d) Find the minimum energy that would be required to separate the two stars to infinity.
(e) If a meteorite passes through this centre of mass perpendicular to the orbital plane of the stars.
What value must its speed exceed at that point if it escapes to infinity from the star system.

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Gravitation
D-3. Two earth satellites A and B each of equal mass are to be launched into circular orbits about earth’s centre.
Satellite ‘A’ is to orbit at an altiude of 6400 km and B at 19200 km. The radius of the earth is 6400 km.
Determine-
(a) the ratio of the potential energy
(b) the ratio of kinetic energy
(c) which one has the greater total energy

D-4. The Saturn is about six times farther from the Sun than The Mars. Which planet has :
(a) the greater period of revolution ? (b) the greater orbital speed and
(c) the greater angular speed ?

E-1. The acceleration due to gravity at a height (1/20)th the radius of the earth above earth’s surface i s
9 m/s2. Find out its approximate value at a point at an equal distance below the surface of the earth.

E-2. If a pendulum has a period of exactly 1.00 sec. at the equator, what would be its period at the sout h
pole ? Assume the earth to be spherical and rotational effect of the Earth is to be taken.

PART - II : ONLY ONE OPTION CORRECT TYPE

A-1 Four similar particles of mass m are orbiting in a circle of radius r in

the same direction and same speed because of their mutual gravitational
attractive force as shown in the figure. Speed of a particle is given by
1
2
Gm 1 2 2 Gm
(A) (B) 3
r 4 r

Gm
(C) 1 2 2 (D) zero
r

A-2. Two blocks of masses m each are hung from a balance as shown in
the figure. The scale pan A is at height H 1 whereas scale pan B is at
height H2. Net torque of weights acting on the system about point 'C',
will be (length of the rod is and H1 & H2 are << R) (H1 > H2)
1 2 H1 mg 2mg H2 H1
(A) mg (B) ( H1 H2 ) (C) (H1 H 2 ) (D) 2mg H H
R R R 1 2

A-3. Three particles P, Q and R are placed as per given figure. Masses
of P, Q and R are 3 m, 3 m and m respectively. The gravitational
force on a fourth particle ‘S’ of mass m is equal to
3 GM 2
(A) in ST direction only
2d 2

3 Gm 2 3 Gm 2
(B) 2 in SQ direction and in SU direction
2d 2d 2

3 Gm 2
(C) in SQ direction only
2d 2

3 Gm 2 3 Gm 2
(D) in SQ direction and in ST direction
2d 2 2d 2
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Gravitation
A-4. Three identical stars of mass M are located at the vertices of an equilateral triangle with side L. The
speed at which they will move if they all revolve under the influence of one another’s gravitational force
in a circular orbit circumscribing the triangle while still preserving the equilateral triangle :

2 GM GM GM
(A) (B) (C) 2 (D) not possible at all
L L L

B-1. Let gravitation field in a space be given as E = – (k/r). If the reference point is at distance d i where
potential is Vi then relation for potential is :
1 r r r V
(A) V = k n V + 0 (B) V = k n d + Vi (C) V = n d + kVi (D) V = n d + i
i i i i k

B-2. Gravitational field at the centre of a semicircle formed by a thin wire AB

of mass m and length as shown in the figure. is :

Gm Gm
(A) 2 along +x axis (B) 2 along +y axis

2 Gm 2 Gm
(C) 2 along + x axis (D) 2 along + y axis

B-3. A very large number of particles of same mass m are kept at horizontal distances of 1m, 2m, 4m, 8m
and so on from (0,0) point. The total gravitational potential at this point (0, 0) is :
(A) – 8G m (B) – 3G m (C) – 4G m (D) – 2G m

B-4. Two concentric shells of uniform density of mass M 1 and M2

are situated as shown in the figure. The forces experienced by
a particle of mass m when placed at positions A, B and C
respectively are (given OA = p, OB = q and OC = r).
M1m (M1 m2 )m
(A) zero , G and G
q2 p2

(M1 M2 )m (M1 M2 )m M1m

(B) G 2
,G 2
and G
p q r2
M1m M1m
(C) G 2 , G and zero
q q2

G(M1 M2 )m M1m
(D) ,G and zero
p 2
q2

B-5. Figure show a hemispherical shell having uniform mass density. The
direction of gravitational field intensity at point P will be along:
(A) a (B) b
(C) c (D) d

B-6. Mass M is uniformly distributed only on curved surface of a thin

hemispherical shell. A, B and C are three points on the circular base of
hemisphere, such that A is the centre. Let the gravitational potential at
points A, B and C be VA, VB, VC respectively. Then
(A) VA> VB>VC (B) VC> VB>VA
(C) VB>VA and VB> VC (D) VA= VB=VC

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Gravitation

C-1. A body starts from rest at a point, distance R 0 from the centre of the earth of mass M, radius R. The
velocity acquired by the body when it reaches the surface of the earth will be

1 1 1 1 1 1 1 1
(A) GM R (B) 2 GM R (C) 2 GM (D) 2GM
R0 R0 R R0 R R0

C-2. Three equal masses each of mass ‘m’ are placed at the three-corners of an equilateral triangle of side ‘a’.
(a) If a fourth particle of equal mass is placed at the centre of triangle, then net force acting on it, is
equal to :

Gm 2 4G m 2 3G m 2
(A) (B) (C) (D) zero
a2 3a 2 a2

(b) In above problem, if fourth particle is at the mid-point of a side, then net force acting on it, is equal
to:

Gm 2 4G m 2 3G m 2
(A) (B) (C) (D) zero
a2 3a 2 a2

(c) If above given three particles system of equilateral triangle side a is to be changed to side of 2a,
then work done on the system is equal to :

3 G m2 3G m2 4 G m2 Gm 2
(A) (B) (C) (D)
a 2a 3a a

(d) In the above given three particle system, if two particles are kept fixed and third particle is released.
Then speed of the particle when it reaches to the mid-point of the side connecting other two masses:

2 Gm Gm Gm Gm
(A) (B) 2 (C) (D) 2a
a a a

D-1. Periodic-time of satellite revolving around the earth is - ( is density of earth)

1 1
(A) Proportional to (B) Proportional to (C) Proportional (D) does not depend on .

D-2. An artificial satellite of the earth releases a package. If air resistance is neglected the point where the
package will hit (with respect to the position at the time of release) will be
(A) ahead (B) exactly below
(C) behind (D) it will never reach the earth

D-3. The figure shows the variation of energy with the orbit radius of a
body in circular planetary motion. Find the correct statement about
the curves A, B and C
(A) A shows the kinetic energy, B the total energy and C the potential
energy of the system
(B) C shows the total energy, B the kinetic energy and A the potential
energy of the system
(C) C and A are kinetic and potential energies respectively and B is the
total energy of the system
(D) A and B are the kinetic and potential energies respectively and C is
the total energy of the system.

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Gravitation
D-4. A planet of mass m revolves around the sun of mass M in an elliptical orbit. The minimum and maximum
distance of the planet from the sun are r1 & r2 respectively. If the minimum velocity of the planet is

2GMr1
(r1 r2 )r2 then it's maximum velocity will be :

2GMr2 2GMr1 2Gmr2 2GM

(A) (r1 r2 )r1 (B) g (r r )r (C) (r1 r2 )r1 (D) r1 r2
1 2 2

D-5. The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. If the body
is projected at an angle of 45º with the vertical, the escape velocity will be : [AIEEE-2003,4/300]

(A) 11 2 km/s (B) 22 km/s (C) 11 km/s (D) 11/ 2 m/s

E-1. If acceleration due to gravity on the surface of earth is 10 ms–2 and let acceleration due to gravitational
acceleration at surface of another planet of our solar system be 5 ms–2. An astronaut weighing 50 kg on
earth goes to this planet in a spaceship with a constant velocity. The weight of the astronaut with time
of flight is roughly given by

(A) (B) (C) (D)

PART - III : MATCH THE COLUMN

1. A particle is taken to a distance r (> R) from centre of the earth. R is radius of the earth. It is given
velocity V which is perpendicular to r . With the given values of V in column I you have to match the
values of total energy of particle in column II and the resultant path of particle in column III. Here 'G' is
the universal gravitational constant and 'M' is the mass of the earth.
Column I (Velocity) Column II (Total energy) Column III (Path)
(A) V = GM / r (p) Negative (t) Elliptical

(B) V = 2GM / r (q) Positive (u) Parabolic

(C) V > 2GM / r (r) Zero (v) Hyperbolic

(D) GM / r < V < 2GM / r (s) Infinite (w) Circular

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Gravitation
2. Let V and E denote the gravitational potential and gravitational field respectively at a point due t o
certain uniform mass distribution described in four different situations of column-I. Assume the
gravitational potential at infinity to be zero.The value of E and V are given in column- II. Match the
statement in column-I with results in column-II.
Column-I Column-II
(A) At centre of thin spherical shell (p) E = 0
(B) At centre of solid sphere (q) E 0
(C)A solid sphere has a non-concentric spherical cavity.
At the centre of the spherical cavity (r) V 0
(D) At centre of line joining two point masses of equal magnitude (s) V = 0

PART - I : ONLY ONE OPTION CORRECT TYPE

1. A spherical hollow cavity is made in a lead sphere of radius R, such

that its surface touches the outside surface of the lead sphere and
passes through its centre. The mass of the sphere before hollowing d
was M. With what gravitational force will the hollowed-out lead sphere
attract a small sphere of mass ‘ m ‘, which lies at a distance d from the
centre of the lead sphere on the straight line connecting the centres of R m
the spheres and that of the hollow, if d = 2R :
7 GMm 7 GMm 7 GMm 7 GMm
(A) (B) (C) (D)
18 R2 36 R2 9R2 72 R2

2. A straight rod of length extends from x = to x = + . as shown in the

figure. If the mass per unit length is (a + bx 2). The gravitational force it exerts
on a point mass m placed at x = 0 is given by

1 1 G m (a bx 2 )
(A) G m a b (B) 2

1 1 1 1
(C) G m a a
b (D) G m a b

3. A uniform ring of mass M is lying at a distance 3 R from the centre of a

uniform sphere of mass m just below the sphere as shown in the figure where
R is the radius of the ring as well as that of the sphere. Then gravitational force
exerted by the ring on the sphere is :
GMm GMm
(A) 2 (B)
8R 3R 2

GMm GMm
(C) 3 (D) 3
R 2
8R 2

4. The gravitational potential of two homogeneous spherical shells A and B (separated by large distance) of
same surface mass density at their respective centres are in the ratio 3 : 4. If the two shells coalesce into
single one such that surface mass density remains same, then the ratio of potential at an internal point of the
new shell to shell A is equal to :
(A) 3 : 2 (B) 4 : 3 (C) 5 : 3 (D) 3 : 5

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Gravitation
5. A projectile is fired from the surface of earth of radius R with a speed k e in radially outward direction
(where e is the escape velocity and k < 1). Neglecting air resistance, the maximum hight from centre
of earth is
R R
(A) 2 (B) k2 R (C) (D) kR
k 1 1 k2

6. Two small balls of mass m each are suspended side by side by two equal
threads of length L as shown in the figure. If the distance between the
upper ends of the threads be a, the angle that the threads will make with
the vertical due to attraction between the balls is

(a – x)g mG
(A) tan–1 (B) tan–1
mG (a – x)2 g

(a – x ) 2 g (a 2 – x 2 )g
(C) tan–1 (D) tan–1
mG mG

7. A block of mass m is lying at a distance r from a spherical shell of

mass m and radius r as shown in the figure. Then
(A) only gravitational field inside the shell is zero
(B) gravitational field and gravitational potential both are zero inside the shell
(C) gravitational potential as well as gravitational field inside the shell are not zero
(D) can’t be ascertained.
8. In a spherical region, the density varies inversely with the distance from the centre. Gravitational field at a
distance r from the centre is :
1
(A) proportional to r (B) proportional to (C) proportional to r2 (D) same everywhere
r

9. In above problem, the gravitational potential is -

1
(A) linearly dependent on r (B) proportional to
r
(C) proportional to r2 (D) same every where.

10. A point P lies on the axis of a fixed ring of mass M and radius R, at a distance 2R from its centre O. A
small particle starts from P and reaches O under gravitational attraction only. Its speed at O will be

2 GM 2 GM 2 GM 1
(A) zero (B) (C) ( 5 1) (D) (1 )
R R R 5

11. A body of mass m is lifted up from the surface of earth to a height three times the radius of the earth.
The change in potential energy of the body is (g = gravity field at the surface of the earth)
3 1 2
(A) mgR (B) mgR (C) mgR (D) mgR
4 3 3

12. Assuming that the moon is a sphere of the same mean density as that of the earth and one quarter of
its radius, the length of a seconds pendulum on the moon (its length on the earth’s surface is 99.2 cm)
is
99.2
(A) 24.8 cm (B) 49.6 cm (C) 99.2 (D) cm
2

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Gravitation
13. A satellite can be in a geostationary orbit around a planet at a distance r from the centre of the planet. If the
angular velocity of the planet about its axis doubles, a satellite can now be in a geostationary orbit around the
planet if its distance from the centre of the planet is
r r r r
(A) (B) (C) (D)
2 2 2 (4)1/ 3 (2)1/ 3

14. A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period
of a spy satellite orbiting a few hundred kilometers above the earth’s surface (REarth = 6400 km) will approximately
be : [JEE(Scr) - 2002, 3/84]
(A) 1/2 hr (B) 1 hr (C) 2 hr (D) 4 hr

15. A satellite of mass m revolves around earth of radius R at a height x from its surface. If g is the acceleration
due to gravity on the surface of the earth, the orbital speed of the satellite is : [AIEEE-2004, 4/300]

1/ 2
gR gR2 gR 2
(A) gx (B) (C) (D)
R x R x R x

PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE

1. A projectile is fired vertically up from the bottom of a crater (big hole) on the moon. The depth of the crater is
R/100, where R is the radius of the moon. If the initial velocity of the projectile is the same as the escape
velocity from the moon surface. The maximum aproximate height attained by the projectile above the lunar
(moon) surface is xR. Find the greatest integer of x. [JEE 2003(Main), 4/60]

2. The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is
increased to 4 times the previous value, the new time period becomes (in hrs). [AIEEE-2003, 4/300]

3. If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of
NmgR
mass m raised from the surface of the earth to a height equal to the radius R of the earth is . Find the
2
value of N : [AIEEE-2004, 4/300]

4. The gravitational field in a region is given by E (3 î 4 ĵ ) N/kg. Find out the work done (in joule) in displacing
a particle of mass 1 kg by 1 m along the line 4y = 3x + 9.

5. In a solid sphere of radius ‘R’ and density ‘ ’ there is a spherical

cavity of radius R/4 as shown in figure. A particle of mass ‘m’ is
released from rest from point ‘B’ (inside the cavity). Find out Velocity
(in mm/sec.) of the particle at the instant when it strikes the cavity
10 20
(R = 3m, = 10 3 kg/m3, G = 10 11
Nm2kg–2)
3

6. A ring of radius R = 8m is made of a highly dense-material. Mass of the ring is mR = 2.7 × 109 kg distributed
uniformly over its circumference. A particle of mass (dense) mp = 3 × 108 kg is placed on the axis of the ring
at a distance x0 = 6m from the centre. Neglect all other forces except gravitational interaction. Determine
speed (in cm/sec.) of the particle at the instant when it passes through centre of ring. :

7. Our sun, with mass 2 × 1030 kg revolves on the edge of our milky way galaxy, which can be assumed
to be spherical, having radius 10 20m. Also assume that many stars, identical to our sun are uniformly
distributed in the spherical milky way galaxy. If the time period of the sun is 10 15 second and number
20
of stars in the galaxy are nearly 3 × 10 (a), find value of ‘a’ (take 2 = 10, G × 10–111 in MKS)
3

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Gravitation
8. Assume earth to be a sphere of uniform mass density. The energy needed to completely disassemble
the planet earth against the gravitational pull amongst its constituent particles is x × 10 31 J. Find the
value of x. given the product of mass of earth and radius of earth to be 2.5 x 1031 kg-m and g = 10 m/s2.
[JEE '92, 10 Part (b)]
9. The two stars in a certain binary star system move in circular orbits. The first star, moves in an orbit
of radius 1.00 109 km. The other star, moves in an orbit of radius 5.00 108 km. What is the ratio
of masses of star to the star ?

10. If the radius of earth is R and height of a satellite above earth's surface is R then find the minimum co-latitude
(in degree) which can directly receive a signal from satellite.

PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

1.* For a satellite to appear stationary to an observer on earth
(A) It must be rotating about the earth’s axis.
(B) It must be rotating in the equatorial plane.
(C) Its angular velocity must be from west to east.
(D) Its time period must be 24 hours.

2.* Inside an isolated uniform spherical shell :

(A) The gravitation potential is not zero (B) The gravitational field is not zero
(C) The gravitational potential is same everywhere (D) The gravitational field is same everywhere.

3.* Which of the following statements are correct about a planet rotating around the sun in an elliptic orbit:
(A) its mechanical energy is constant
(B) its angular momentum about the sun is constant
(C) its areal velocity about the sun is constant
(D) its time period is proportional to r 3

4.* A tunnel is dug along a chord of the earth at a perpendicular distance R/2 from the earth’s centre. The
wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel.
The pressing force by the particle on the wall and the acceleration of the particle varies with x (distance
of the particle from the centre) according to :

(A) (B)

(C) (D)

5.* A planet revolving around sun in an elliptical orbit has a constant

(A) kinetic energy
(B) angular momentum about the sun
(C) potential energy
(D) Total energy

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Gravitation
6.* A satellite close to the earth is in orbit above the equator with a period of revolution of 1.5 hours. If it is
above a point P on the equator at some time, it will be above P again after time
(A) 1.5 hours
(B) 1.6 hours if it is rotating from west to east
(C) 24/17 hours if it is rotating from east to west
(D) 24/17 hours if it is rotating from west to east

7.* An object is weighed at the equator by a beam balance and a spring balance, giving readings W b and
W s respectively. It is again weighed in the same manner at the north pole, giving readings of Wb' and Ws'
respectively. Assume that intensity of earth gravitational field is the same every where on the earth’s
surface and that the balances are quite sensitive.
(A) W b = W b' (B) W b = W s (C) W b' = W s' (D) W s' > W s

8.* If a body is projected with speed lesser than escape velocity :

(A) the body can reach a certain height and may fall down following a straight line path
(B) the body can reach a certain height and may fall down following a parabolic path
(C) the body may orbit the earth in a circular orbit
(D) the body may orbit the earth in an elliptic orbit

9.* A double star is a system of two stars of masses m and 2m, rotating about their centre of mass only
under their mutual gravitational attraction. If r is the separation between these two stars then their time
period of rotation about their centre of mass will be proportional to
(A) r3/2 (B) r (C) m1/2 (D) m–1/2

10.* An orbiting satellite will escape if :

(A) its speed is increased by ( 2 1)100 %
(B) its speed in the orbit is made 1.5 times of its initial value
(C) its KE is doubled
(D) it stops moving in the orbit

11*. In case of an orbiting satellite if the radius of orbit is decreased :

(A) its Kinetic Energy decreases (B) its Potential Energy decreases
(C) its Mechanical Energy decreases (D) its speed decreases

12*. In case of earth :

(A) gravitational field is zero, both at centre and infinity
(B) gravitational potential is zero, both at centre and infinity
(C) gravitational potential is same, both at centre and infinity but not zero
(D) gravitational potential is minimum at the centre

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Gravitation

PART - IV : COMPREHENSION
Comprehension 1
Many planets are revolving around the fixed sun, in circular orbits of different radius (R) and different
time period (T). To estimate the mass of the sun, the orbital radius (R) and time period (T) of planets
were noted. Then log10 T v/s log10 R curve was plotted.
The curve was found to be approximately straight line (as shown in figure) having y intercept = 6.0
20 11
(Neglect the gravitational interaction among the planets [Take G = 10 in MKS, 2
= 10]
3

1. The slope of the line should be :

3 2 19
(A) 1 (B) (C) (D)
2 3 4
2. Estimate the mass of the sun :
(A) 6 × 1029 kg (B) 5 × 1020 kg (C) 8 × 1025 kg (D) 3 × 1035 kg

3. Two planets A and B, having orbital radius R and 4R are initially at the
closest position and rotating in the same direction. If angular velocity
of planet B is 0, then after how much time will both the planets be
again in the closest position ? (Neglect the interaction between planets).
2 2 2 2
(A) 7 (B) 9 (C) (D) 5
0 0 0 0

COMPREHENSION 2
An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of
escape velocity from the surface of earth. R is the radius of earth and g is acceleration due to gravity at the
surface of earth. (R = 6400 km)
4. Then the distance of satellite from the surface of earth is
(A) 3200 km (B) 6400 km (C) 12800 km (D) 4800 km

5. The time period of revolution of satellite in the given orbit is

2R 4R 8R 6R
(A) 2 g (B) 2 g (C) 2 g (D) 2 g

6. If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, the speed with which
it hits the surface of the earth.

gR gR
(A) gR (B) 1. 5 g R (C) (D)
2 2

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Gravitation
COMPREHENSION 3
A pair of stars rotates about their center of mass. One of the stars has a mass M and the other has
mass m such that M = 2m. The distance between the centres of the stars is d (d being large compared
to the size of either star).
7. The period of rotation of the stars about their common centre of mass (in terms of d, m, G.) is

4 2 3 8 2 3 2 2 3 4 2 3
(A) d (B) d (C) d (D) d
Gm Gm 3Gm 3Gm

8. The ratio of the angular momentum of the two stars about their common centre of mass ( L m/ LM) is
(A) 1 (B) 2 (C) 4 (D) 9

9. The ratio of kinetic energies of the two stars ( K m/KM.) is

(A) 1 (B) 2 (C) 4 (D) 9

PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)

* Marked Questions may have more than one correct option.
1. A double star system consists of two stars A and B which have time period TA and TB. Radius RA and RB and
mass MA and MB. Choose the correct option. [JEE 2006, –3/184,–1]
(A) If TA > TB then RA > RB (B) If TA > TB then MA > MB
2 3
TA RA
(C) (D) TA = TB
TB RB

2. Column describes some situations in which a small object moves. Column describes some characteristics
of these motions. Match the situations in Column with the characteristics in Column
[IIT-JEE 2007, 6/162]
Column Column
(A) The object moves on the x–axis under a conservative (p) The object executes a simple
force in such a way that its "speed" and "position" harmonic motion.

satisfy v = c 1 c 2 x 2 , where c1 and c2

are positive constants.
(B) The object moves on the x–axis in such a way that (q) The object does not change its
its velocity and its displacement from the origin satisfy direction,
v = –kx, where k is a positive constant.
(C) The object is attached to one end of a massless spring (r) The kinetic energy of the
of a given spring constant. The other end of the spring is object keeps on decreasing.
attached to the ceiling of an elevator. Initially everything is
at rest. The elevator starts going upwards with a constant
acceleration a. The motion of the object is observed from
the elevator during the period it maintains this acceleration.
(D) The object is projected from the earth's surface vertically (s) The object can change its
upwards with a speed 2 GMe / R e , where Me is the mass direction only once.
of the earth and Re is the radius of the earth. Neglect forces
from objects other than the earth.

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Gravitation
3. A spherically symmetric gravitational system of particles has a mass density [JEE 2008, 3/82,–1]
0 for r R
0 for r R
where 0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field
of particles. Its speed V as a function of distance r (0 < r < ) from the centre of the system is represented
by
V V V
V

(A) (B) (C) (D)

r r R r R r
R R

4. STATEMENT -1 [JEE 2008, 3/82,–1]

An astronaut in an orbiting space station above the Earth experiences weightlessness.
and
STATEMENT -2
An object moving around the Earth under the influence of Earth's gravitational force is in a state of 'free-fall.
(A) STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT -2 is a correct explanation
for STATEMENT -1
(B) STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT -2 is NOT a correct explanation for
STATEMENT -1
(C) STATEMENT -1 is True, STATEMENT -2 is False
(D) STATEMENT -1 is False, STATEMENT -2 is True.

5. A thin uniform annular disc (see figure) of mass M has outer radius 4R and P
inner radius 3R. The work required to take a unit mass from point P on its axis
to infinity is : [JEE 2010, 3/163, –1] 4R

2GM 2GM
(A) 4 2 5 (B) 4 2 5 3R
7R 7R 4R

GM 2GM
(C) (D) 2 1
4R 5R

6. A binary star consists of two stars A (mass 2.2 MS) and B ( mass 11 MS) where Ms is the mass of the sun.
They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio
of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass
is : [JEE 2010, 3/163]

6
7. Gravitational acceleration on the surface of a planet is g, where g is the gravitational acceleration on the
11
2
surface of the earth. The average mass density of the planet is times that of the earth. If the escape speed
3
on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of the planet in kms–1
will be : [JEE 2010, 3/163]
8. A satellite is moving with a constant speed 'V’ in a circular orbit about the earth. An object of mass ‘m’ is
ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its
ejection, the kinetic energy of the object is [JEE 2011, 3/160, –1]
1 3
(A) mV 2 (B) mV2 (C) mV
2
(D) 2mV2
2 2
9.* Two spherical planets P and Q have the same unfirom density , masses MP and MQ, with surface areas A
and 4A, respectively. A spherical planet R also has unfirom density and its mass is (MP + MQ) . The escape
velocities from the planets P, Q and R, are VP, VQ and V respectivley. Then [IIT-JEE-2012, Paper-2; 4/66]
1
(A) VQ > VR > VP (B) VR > VQ > VP (C) VR/VP = 3 (D) VP /VQ =
2
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Gravitation
10.* Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the
midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct
statement(s) is (are) : [JEE (Advanced) 2013, 3/60, –1]
GM
(A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4 .
L

GM
(B) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2 .
L

2GM
(C) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is .
L
(D) The energy of the mass m remains constant.

1
11. A planet of radius R = × (radius of Earth) has the same mass density as Earth. Scientists dig a well of
10
R
depth on it and lower a wire of the same length and of linear mass density 10–3 kgm–1 into it. If the wire is
5
not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the
radius of Earth = 6 × 106 m and the acceleration due to gravity on Earth is 10 ms–2)
[JEE (Advanced)-2014, 3/60, –1]
(A) 96 N (B) 108N (C) 120N (D) 150N

12. A Bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its
maximum height, its acceleration due to the planet's gravity is 1/4th of its value at the surface of the planet.
If the escape velocity from the planet is v esc = v n , then the value of N is (ignore energy loss due to
atmosphere) [JEE(Advanced) 2015 ; P-1, 4/88]

13. A large spherical mass M is fixed at one position and two identical
point masses m are kept on a line passing through the centre of M
(see figure). The point masses are connected by a rigid massless
rod of length and this assembly is free to move along the line
connecting them. All three masses interact only through their
mutual gravitational interaction. When the point mass nearer to M
is at a distance r = 3 from M, the tension in the rod is zero for
M
m=k . The value of k is [JEE(Advanced) 2015 ; P-2,4/88]
288

14. A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity.,
If P(r) is the pressure at r(r < R), then the correct option(s) is (are) [JEE(Advanced) 2015 ; P-2,4/88, –2]
P(r 3R / 4) 63 P(r 3R / 5) 16 P(r R / 2) 20
(A) P(r = 0) = 0 (B) (C) (D)
P(r 2R / 3) 80 P(r 2R / 3) 21 P(r R / 3) 27

PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)

1. The change in the value of ‘g’ at a height ‘h’ above the surface of the earth is the same as at a depth ‘d’ below
the surface of earth. When both ‘d’ and ‘h’ are much smaller than the radius of earth, then, which one of the
following is correct ? [AIEEE-2005, 4/300]
h 3h
(1) d = (2) d = (3) d = 2h (4) d = h
2 2

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Gravitation
2. A particle of mass 10 kg is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find
the work to be done against the gravitational force between them, to take the particle far away from the
sphere (you may take G = 6.67 × 10–11 Nm2/kg2 ); [AIEEE-2005, 4/300]
(1) 13.34 × 10–10 J (2) 3.33 × 10–10 J (3) 6.67 × 10–9 J (4) 6.67 × 10–7 J

3. If gE and gm are the accelerations due to gravity on the surfaces of the earth and the moon respectively and
if Millikan's oil drop expriment could be performed on the two surfaces, one will find the ratio

to be [AIEEE-2007, 3/120]

(1) 1 (2) 0 (3) gE/gM (4) gM/gE

4. A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller.
Given that the escape velocity from the earth is 11 km s–1, the escape velocity from the surface of the planet
would be [AIEEE-2008, 3/105]
(1) 11 km s–1 (2) 110 km s–1 (3) 0.11 km s–1 (4) 1.1 km s–1

g
5. The height at which the acceleration due to gravity becomes (where g = the acceleration due to gravity on
9
the surface of the earth) in terms of R, the radius of the earth, is [AIEEE-2009, 4/144]
R R
(1) (2) (3) 2 R (4) 2R
2 2

6. Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line
joining them where the gravitational field is zero is : [AIEEE - 2011, 4/120, –1]
4Gm 6Gm 9Gm
(1) zero (2) – (3) – (4) –
r r r
7. Two particles of equal mass ‘m’ go around a circle of radius R under the action of their mutual gravitational
attraction. The speed of each particle with respect to their centre of mass is:[AIEEE 2011, 11 May; 4/120,–1]

Gm Gm Gm Gm
(1) (2) (3) (4)
4R 3R 2R R
8. The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The
value of 'g' and 'R' (radius of earth) are 10 m/s2 and 6400 km respectively. The required energy for this work
will be : [AIEEE 2012 ; 4/120, –1]
11 8 9
(1) 6.4 × 10 Joules (2) 6.4 × 10 Joules (3) 6.4 × 10 Joules (4) 6.4 × 1010 Joules
9. What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M
and radius R in a circular orbit at an altitude of 2R? [JEE(Main) 2013, 4/120,–1]
5GmM 2GmM GmM GmM
(1) (2) (3) (4)
6R 3R 2R 3R
10. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the
action of their mutual gravitational attration. the speed of each particle is : [JEE-(Main) - 2014,4/120,–1]
GM GM GM 1 GM
(1) (2) 2 2 (3) 1 2 2 (4) 1 2 2
R R R 2 R

11. From a solid sphere of mass M and radius R, a spherical portion of

radius R/2 is removed , as shown in the figure. Taking gravitational
potential V = 0 at r = , the potential at the centre of the cavity thus
formed is : (G = gravitational constant) [JEE(Main)-2015; 4/120, –1]

GM GM 2GM 2GM
(1) (2) (3) (4)
2R R 3R R
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Gravitation

the sun and the planet form one system and it is

BOARD LEVEL EXERCISE : HINT & SOLUTIONS not acted upon by any external torque; the angular
momentum of the system must remain unchanged.
1. The force between two bodies by virtue of their As the sun is at rest, the angular momentum of
masses is called gravitational force. the sun is zero, and therefore angular momentum
2. In accordance with Newton’s law of gravitation, the of the planet is the angular momentum of the
force is equal in the two cases. It is because , system
when the two bodies interact due to their masses L = m v R = m R2 = constant
(gravitational interaction), they exert equal forces Suppose that the planet moves from the point A to
other but in opposite directions. Bin small time dt. Then,
AB = v dt = R dt
3. It is because the value of G is independent of the The area swept by the planet in time dt
two object taking part in the gravitational interaction.
1 1
4. Let M and m be the masses of the earth and the = R × R dt = R2 dt
2 2
apple respectively. If r is the distance between the
earth and the apple, Hence, the area swept per unit time

Mm 1 2 mR2 L
F G = R = = = constant
r2 2 2m 2m
W hat this force a acts on the apple, the Hence , Kepler’s second law of follows from the
acceleration produced in the motion of the apple, law of conservation of angular momentum .
F GM 9. M = mass of 2.5 × 1011 stars
a apple
m r2 = 2.5 × 1011 solar masses
On the other hand, when this force acts on the = 2.5 × 1011 × 2 × 1030 kg = 5 × 1041 kg
earth, the acceleration produced in the motion of radius of the orbit of a star,
earth, r = 50,000 ly = 50,000 × 9.46 × 1015
= 4.73 × 1020 m
F Gm There period of revolution of a star is given by
aearth
M r2
Since m << M, aapple >> aearth . r3
T 2
For this reason, it is apple, which falls towards the GM
earth.
(4.73 1020 )3
5. (a) It is negative of the kinetic energy. = 2 1.12 1016 s
(b) It is lesser. 6.67 10 11 5 1041
6. (a) The escape speed of a body of from the earth 1.12 1016
does not depend upon the mass of the body . = = 3.55 × 108 year
365 24 60 60
(b) No
(c) No 10. Let M be the mass of the sun and m, the mass of
(d) yes, It depends on the height of the location the planet. Further, let R be the radius of the earth.
from where the body is projected. It is because, The required centripetal force,
the value of the escape velocity depends upon the mv 2
value of g, where g decreases with height, though F
R
slowly. where v is orbital velocity of the planet. Since the
7. (a) The linear speed varies. gravitational force varies inversely as the nth power
(b) The angular speed varies. of the distance , gravitational force on the planet is
(c) Angular momentum remains constant given by
(d) The kinetic energy varies. GMm
(e) The potential energy varies F
Rn
(f) The total energy remains constant. From the question (i) and (ii), we have
8. Consider a planet of mass m revolving around the mv 2 GMm
sun along an elliptical path.Suppose that at any =
R Rn
instant, the planet is at the point A ,Such that its
distance from the sun is SA = R. Let be the GM
1/ 2
angular velocity of the planet at that instant. Since or v
Rn 1
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Gravitation
Hence , period of revolution of the planet,
Let l be the distance of the earth from the sun,
1/ 2 when it is at perpendicular to the major-axis of the
n 1
2 R R
T 2 R orbit drawn from the sun. It is called semi-latus
v GM rectum and is given by
1/ 2 b2 ( r1r2 )2 2r1 r2
Rn 1 l
= 2 a (r1 r2 ) / 2 r1 r2
GM
12. (a) Acceleration due to gravity decreases with
11. When the earth is at the points A and B of its orbit, increasing altitude.
it is at the largest and shortest distance r1 and r2 (b) Acceleration due to gravity decreases with
.Let l be the distance of the earth from sun, when it increasing depath.
is at perpendicular to the major-axis of the orbit (c) Acceleration due to gravity is independent of
drawn from the sun as shown in fig. the mass of the body.
(d) The formula G M (1/r1 –1/r2) is more accurate
than the formula m g (r2 – r1)

13. (a)On the earth, the weight of the whole body of a

person is carried by his legs, In the space, the
astronaut is in the state of weightlessness.
Likewise, the feet of an astronaut do not get swollen.
(b) In the gravity free space, the face of an astronaut
The distance l is called semi-latus rectum of the
elliptical orbit. Let a and b the semi-major axis gets swollen
and semi-minor axis of the elliptical orbit of the (c) In the gravity free space, an astronaut may suffer
earth and e be its eccentricity. from headache.
From the coordinate geometry, the maximum (d) In the gravity free space, an astronaut is bound
to have orientational problem.
distance of the earth from the sun (at focus),
r1 = a (1+ e)
and the minimum distance of the earth from the 14. Acceleration due to gravity on the moon is one
sun, sixth of that on the surface of earth. Therefore, one
r2 = a(1– e) can jump higher on the surface of the moon than
Adding the equations (i) and (ii) , we have on the earth.
r1 + r2 = a (1+e) +a (1–e) = 2a
15. As the spacecraft moves away from the surface of
r r the earth towards moon, there will be no change in
or a 1 2
2 the mass of the spacecraft. However, its weight
Dividing the equations (i) by (ii) , we have will keep on changing as described below.
1. Its weight will decrease in the beginning.
r1 a(1 e) (1 e) 2. It will become zero at some point, where the
r2 a(1 e) (1 e) force of attraction on the spacecraft due to the earth
Applying componendo and dividendo, we have and that due to the moon become just equal and
opposite.
r1 r2 (1 e) (1 e) 1 3. It will again start increasing as the spacecraft
r1 r2 = (1 e) (1 e) e further moves towards the moon.
r1 r2
or e
r1 r2 EXERCISE - 1
Also, for an elliptical orbit, we have
PART - I
b = a 1 e2

r1 r2 r –r
2
m1m 2 1.4 7.34 10 22
–11
1
= 1– 1 2 A-1. F=G = 6.67 × 10 ×
2 r1 r2 r2 (378 106 )2
= 4.8 × 10–5 N
r1 r2 (r1 r2 )2 (r1 r2 )2 1 4 2 2 4
= 2 4r1 r2 A-2. Gr
(r1 r2 )2 2 9
A-3. 31.2 G m/sec2 = 2.1 × 10–9 m/s2, towards mid
or b r1 r2 point
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Gravitation

B-1. 20 î 40 ĵ , | F | = 5 5 N, F 5 î 10 ĵ E-1. (A)

4 PART - III
B-2. G 5.5 × 103 × (6.4 × 106)2 J/Kg
3 1. I II III
= 6.3 × 107 J/Kg A p w
B r u
C q v
C-1. gR D p t
2. (A) p,r (B) p,r (C) q,r (D) p,r
G(M A MB )
C-2. 2
d
EXERCISE - 2
D-1. 1.5% PART - I
1. (B) 2. (A) 3. (D) 4. (C)
GM2 GM R3 5. (C) 6. (B) 7. (C) 8. (D)
D-2. (a) F = (b) ; T=4
4R 2 4R GM 9. (A) 10. (D) 11. (B) 12. (A)
13. (C) 14. (C) 15. (D)
GM GM2 4GM
(c)
4R 3 (d)
4R
(e)
R PART - II
1. 99 2. 40 3. 1 4. 0
5. 2 6. 9 7. 11 8. 15
UA 25600 9. 2 10. 30
D-3. (a) = =2
UB 12800

KA mA rB
PART - III
(b) = =2 1. (A,B,C,D) 2. (A,C,D) 3. (A,B,C)
KB mB rA
4. (B,C) 5. (B,D) 6. (B,C)
(c) B is having more energy. 7. (A,C,D) 8. (A,B,C,D) 9. (A,D)
10. (A.C) 11. (B,C)
D-4. (a) The Saturn (b) The Mars (c) The Mars
12. (A,D)

PART - IV
19 1. (C) 2. (A) 3. (A) 4. (B)
E-1. m/s2 = 9.5 m/s2
2 5. (C) 6. (A) 7. (D) 8. (B)
9. (B)
1 4 2 103
E-2. T = 1– 6400 = 0.998 s
2 (86400 )2 9.8
EXERCISE - 3
PART - II
PART - I
A-1 (A) A-2. (B) A-3. (C) A-4. (B) 1. (D)
2. (A) (p) ; (B) (q, r);(C) (p) ; (D) (q, r)
3. (C) 4. (A) 5. (A) 6. 6
7. 3 8. (B) 9. (B,D) 10. (B,D)
B-1. (B) B-2. (D) B-3. (D) B-4. (D)
11. (B) 12. 2 13. 7 14. (B,C)
B-5. (C) B-6. (D)
PART - II
C-1. (C) 1. (3) 2. (4) 3. (1) 4. (2)
C-2. (a) (D)1 (b) (B) (c) (B) (d) (B) 5. (4) 6. (4) 7. (1) 8. (4)
9. (1) 10. (4) 11. (2)

D-1. (B) D-2. (D) D-3. (D) D-4. (A)

D-5. (C)

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Gravitation

SUBJECTIVE QUESTIONS
1.# Let a star be much brighter than our sun but its mass is same as that of sun. If our earth has average
life span of a man as 70 years. In the earth like planet of this star system having double the distance
from our star find the average life span of a man on this planet in terms of our year.

2. Consider a spacecraft in an elliptical orbit around the earth. At the lowest point or perigee, of its orbit
it is 300 km above the earth’s surface at the highest point or apogee, it is 3000 km above the earth’s
surface.
(a) What is the period of the spacecraft’s orbit ?
(b) Using conservation of angular momentum, find the ratio of the spacecraft’s speed at perigee
to its speed at apogee.
(c) Using conservation of energy, find the speed at perigee and the speed at apogee.
(d) It is derised to have the spacecraft escape from the earth completely. If the spacecraft‘s rockets are
fired at perigee, by how much would the speed have to be increased to achieve this ? What if the
rockets were fired at apogee ? Which point in the orbit is the most efficient to use ?
[ k = 1 10 13 sec2/m3 ]

3. A planet A moves along an elliptical orbit around the Sun. At the moment when it was at the distance r0 from
the Sun its velocity was equal to v0 and the angle between the radius vector r0 and the velocity vector v0 was
equal to . Find the maximum and minimum distance that will separate this planet from the Sun during its
orbital motion. (Mass of Sun = MS)

4. A satellite is put into a circular orbit with the intention that it hover over a certain spot on the earth’s
surface. However, the satellite’s orbital radius is erroneously made 1.0 km too large for this to happen.
At what rate and in what direction does the point directly below the satellite move across the earth’s
surface ?
R = Radius of earth = 6400 km
r = radius of orbit of geostationary satellite = 42000 km
T = Time period of earth about its axis = 24 hr.

5. What are : (a) the speed and (b) the period of a 220 kg satellite in an approximately circular orbit
640 km above the surface of the earth ? Suppose the satellite loses mechanical energy at the average
rate of 1.4 105 J per orbital revolution. Adopting the reasonable approximation that due to atmospheric
resistance force, the trajectory is a “circle of slowly diminishing radius”. Determine the satellite’s
(c) altitude (d) speed & (e) period at the end of its 1500 th revolution.(f) Is angular momentum
around the earth’s centre conserved for the satellite or the satellite-earth system.

6. A planet of mass m moves along an ellipse around the Sun so that its maximum and minimum distance from
the Sun are equal to r1 and r2 respectively. Find the angular momentum J of this planet relative to the centre
of the Sun. (Mass of Sun = MS)

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Gravitation
7.# A solid sphere of mass m and radius r is placed inside a hollow spherical y
shell of mass 4 m and radius 4r find gravitational field intensity at :
(a) r < y < 2r (b) 2r < y < 8r (c) y > 8r
here y coordinate is measured from the point of contact of the sphere and the shell.

x
8.# A sphere of density and radius a has a concentric cavity of radius b, as shown in the figure.

a
b r
m

(a) Sketch the gravitational force F exerted by the sphere on the particle of mass m, located at a
distance r from the centre of the sphere as a function of r in the range 0 r .
(b) Sketch the corresponding curve for the potential energy u (r) of the system.

9. (a) What is the escape speed for an object in the same orbit as that of Earth around sun (Take
orbital radius R) but far from the earth ? (mass of the sun = Ms)
(b) If an object already has a speed equal to the earth’s orbital speed, what minimum additional
speed must it be given to escape as in (a) ?

10.# A cosmic body A moves towards the Sun with velocity v0 (when far from the Sun) and aiming parameter , the
direction of the vector v0 relative to the centre of the Sun as shown in the figure. Find the minimum distance
by which this body will get to the Sun. (Mass of Sun = MS)

11. Two stars of mass M1 & M2 are in circular orbits around their centre of mass. The star of mass M 1 has
an orbit of radius R1, the star of mass M2 has an orbit of radius R2. (assume that their centre of mass
is not accelerating and distance between stars is fixed)

(a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their
masses, that is R1/R2 = M2/M1.

(b) Explain why the two stars have the same orbital period and show that the period,

(R 1 R 2 )3 / 2
T=2 .
(
G M1 M 2 )

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Gravitation

1. 25 years
4 v1 94
2. (a) T = a3 / 2 = 7.16 103 sec. (b) = = 1.4
R g v2 67

94
(c) Vp = 896×102 m/sec. = 8.35 103 m/s ,
67 161

67
Va = 896×102 m/sec= 5.95 103 m/s
94 161

(d) V = 14× 102 67 – VP = 3.09 103 m/s, perigee

r0
3. rm =
2
[1 ± 1 (2 ) sin 2 ],

where = r0v 02/GMS.

3 rR
4. Vrel = = m/sec 1.66 cm/sec., to the east along equator
rT 189

448 220
5. (a) km/s = 7.527 km/s (b) 3520 sec. 1.63 hour
3520 7

22 14 64 2 7040 448
(c) 2
6400 km 411.92 km (d) km/sec. 7.67 km/s
22 14 64 7040 6 3406

1703
(e) 3406 sec. 1.55 hour (f) No
56

6. J = m 2GMS r1r2 /(r1 r2 )

Gm ( y r ) Gm 4 Gm Gm
7. (a) ( ĵ ) (b) 2
( ĵ ) (c) 2
( ĵ )
r 3
(y r) ( y 4r ) ( y r )2

8. (a) (b)

2GM s GMs
9. (a) (b) 2 1
R R

10. rmin = (GMS / v 02) [ 1 ( v 02 / GM S ) 2 – 1]

2
4 [1.5 1012 ]3
11. M = = 3.376 1029 kg , M = 2M = 6.75 1029 kg
3G[44.5 365 86400]2

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Gravitation

1. The mass of the sun is same for both the cases, so we can apply
2 3 2 3
Tplanet Rplanet Tplanet 2r
= =
Tearth Rearth 1 year r
Tplanet = 23/2 years
70
The life span of the man is 70 years . During 70 years , the revolution completed by that planet = 25
23/2
revolutions. So he will see 25 summers, 25 winters, 25 springs ... so according to that planet his age will be
25 years.
2. Total distance from apogee to perigee
(a) 300 + 2(6400) + 3000 = 2a
a = 8050 km
Time period of the spacecarft

4 2 3
T2 = a
GMe

2
T= a3 / 2
GMe

4
= a3 / 2
GMe 2
R
R2

4 3/2 4
T= R ga = 3
(8050 10 3 )3 / 2
6400 10 9.8
(b) Apply angular momentum conservation about the centre of earth, between perigee and Apogee.
mv 1 rmin = m v 2 rmax .............(i)
(v 1) (300 + 6400) = v 2(3000 + 6400)
v1 94
v2 =
67
(c) Also apply energy conservation between perigee an Apogee

1 GmMe 1 GmMe
mv 12 + rmin = mv 22 + rmax .............(ii)
2 2
Where rmin = (300 + 6400)km and rmax = (3000 + 6400)km
From eqn. (i) & (ii) we get
v 1 = 8.35 × 103 m/sec
v 2 = 5.95 × 103 m/sec.
(d) To escape, velocity at r should be zero.
Applying energy conservation between perigee and r .
ki + U = kf + Uf

1 GM 2m
mv 12 + =0+0
2 ( 300 6400 )
v 1 = 11.44 × 103 m/sec.
Increase in speed = 11.44 × 103 – 8.35 × 103
= 3.09 × 103 m/sec.
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Gravitation

3.

at max and min. distance, Velocity will be perpendicular to the radius vector, Applying angular momentum
conservation about the sun, between initial position and the position of max or minimum distance.
MV0 r0 sin = MVr ..........(1)
applying energy conservation :

1 GMSm 1 GMS m
MVo2 + r0 = mv 2 +
2 2 r

from equ (1) and (2) get

ro
r= 1 1 (2 ) sin 2
2

ro v o
where
GMs
have sign will give rmax and sign will give rmin .

4. R = Radius of earth
r = radius of orbit of geostationary satellite
T = Time period of earth about its axis
T r3 / 2
r– 3 / 2
–3 r
= ×
2 r

–3 r
= × ×
2 r
Vrel = ( 1 – 2 ) R = – ×R
3 r
= × ×R×
2 r
3 r 2 3 rR
Vrel = × ×R× =
2 r T rT
earth
Alternately
earth
2 2
= T =
earth
earth 24 hours
s
If the satellite were geo–stationary its T would also be 24 hours.
But radius is slightly increased, so its T will also be increased.
4 2
T2 = R3 , taking log on both the sides
GM

4 2
2 log T = log GM 3log(R)

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Gravitation
Differentiating
dT 3dR
2 =0+
T R
3 dR
dT = T (here R = radius of geo–stationary satllite = 42000 km)
2 R
3 (1km) 3 24
dT = (24 hours) = hours
2 ( 42000 km) 2 42000
Now, angular velocity of satellite relative to the earth
2 2
= – =
s/earth s earth T dT T
1
2 dT
= T 1 1
rel T
Using binomial expansion
2 dT
= 1 1
rel T T

2
= dT
rel
T2
Velocity of the point directly below the satellite relative to earth's surface will be
v = ( rel)Rearth
2
v= dT (R )
T2 earth

2 3 24
v= hours (6400 km). = m/s = 1.66 cm/sec
( 24 hours) 2 2 42000 189

5. (a) Orbital speed

11
GMe 6 .6 10 ( 6 10 24 )
v= =
r ( 6400 640 )km
= 7.53 km/sec.
(b) Time period

4 2
T2 = GMe r3

2
4
2
T = 11 (6400 + 640 km)3
(6.6 10 6 1024 )
T = 1.63 hours
GMem
(c) Initial mechanical energy =
2r

( 6.6 10 11 )( 6 10 24 )(220 )
= J
2 ( 6400 640 )km

Total loss in mech. energy during 1500 rev.

= (1.4 × 105) × 1500
= 21 × 107 J.

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Gravitation
Final mechanical energy remaining after 1500 rev.
TEf = TEi – Eloss

GMem (6.6 10 11 6 10 23 220)

2rf – 2 (6400 640)km – 21 × 107 J

Solving get rg = 6812 km

Height from earth's surface = 6812 – 6400 = 412 km
(d) Final orbital velocity

GMe 6 .6 10 11 6 10 23
vf = rf =
(6812 km )
v f = 7.67 km/sec.
2 r 2 6812
(e) Time period  T = =
v 7 .67
= 1.55 hours
(f) Due to Air resistance, net torque about the earth is non–zero.
So, angular momentum about the earth will not remain conserved.

6. from angular momentum conservation about the sun,

J = m v 1 r1 = m v 2 r2 ..........(1)
from energy conservation

1 GMS m 1 GMSm
mv 12 + = mv 22 + ..........(2)
2 r1 2 r2
Solving eq (1) and (2) get

2GM s r1r2
J= m (r1 r2 )

7. (a) r < y < 2r y

Field due to outershell = 0
Distance from centre of solid spere = (y – r)
Gravitation field intensity
GM
= – × distance from centre
(radius)3
y
r
GM x
= – (y – r) in y - direction y-
r3
GM
= – (y – r) ĵ
r3
GM
= (y – r) (– ĵ ) Ans.
r3
(b) Field due to outshell = 0 y
Distance from centre of solid spere = (y – r)
GM
E = 0– ĵ y
( y – r )2
4r
GM
= (– ĵ ) Ans.
( y – r )2 r
x
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Gravitation
(c) y > 8r
For any point outside, the shells acts as point situated at centre.
Distance from centre of hollow shell = (y – 4r)
4GM
Field due to hollow shell = –
( y – 4r )
Distance from centre of solid spere = (y – r)
GM
Field due to solid spere = –
( y – r )2

4GM GM
Total field = (– ĵ )
y – 4r ( y – r )2

8. (a) Force will be due to the mass of the sphere upto the radius r
In case (i) 0 < r < b ; Mass M = 0, therefore F(r) = 0

4 4 b3
In case (ii) b < r < a ; Mass M = (r3 – b3), therefore F(r) = Gpm r
3 3 r2

4 4 a3 b3
(iii) a < r < ; Mass M = (a3 – b3), therefore F(r) = Gpm
3 3 r2

4 4 b3
In case (ii) b < r < a ; M = (r3 – b3), F(r) = Gpm r
3 3 r2

4 4 a3 b3
(iii) a < r < ;M= (a3 – b3), F(r) = Gpm
3 3 r2

r2

(b) Uf – Ui = – Fc .dr
r1

(i) 0 < r < b ; u(r) = - 2 G m(a2 — b2)

2 G m
(ii) b < r < a ; u(r) = (3ra2 - 2b3 - r3)
3r

4 G m 3
(iii) a < r < ; u (r) = (a b3 )
3r

1 GMsm 2G S
9. (a) mv 2 = or V =
2 R R
1 G m G
(b) mv e2 – = 0 or Ve =
2 2R R
1 GmMs
m (V + Ve)2 =
2 R
2GMs
or V + Ve =
R
GMs
V = 2 –1
R
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Gravitation
10. Applying angular momentum conservation :
mv 0 = mvd
v 0 = vd .......... (i)
1
Intial energy = mv 02 + 0
2
1 GMs
Final energy = mv 2 –
2 d
Applying energy conservation ,
1 1 GMsm
mv 02 = mv 2 –
2 2 d
2 GM s
v 02 = v 2 – .......... (ii)
d
From equation (i) and (ii) :
v02 2 2GMs
v 02 = 2
–
d d
2GMs
d2 + d– 2
= 0
v 02
Solving this quadratic
2
GMs GM s 2
d = – +
v02 v0
2

2
GMs v 20
1 –1
= GM Ans.
v 20
11. (a) Since centre of rotation is the centre of mass of M1 and M2 .
M1 R1 = M2 R2
M1 R2
or M1 R1 .................... (i)
(b) Since force on M1 and M2 must be towards CM their radial line
should be along same line therefore they must have same orbital period
2 R1
Now T = V1 .................... (ii)

G M1 M2 M1V12
and ................................. (iii)
(R1 R2 )2 R1
from (i), (ii) and (iii)
(R1 R 2 )3 / 2
T=2 G (M1 M2 )
(a) Solving equation
We get
M1R1 = M2R2
M1 R1
M2 = R 2 ....................... (iv)

2 R1
(b) T = V1
On solving equation
3
2 (R 1 R 2 ) 2 2 R2
T = = = T2
G (M1 M2 ) V2
T1 = T2
R1 R2
Because =
V1 V2
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