TPS Physics Papers with Solution 111 Gravitation

2 Gravitation - Paper with Solution

4. Two planets have density in the ratio

Section A - (4 Marks) 2 : 3 and radii in the ratio 1 : 2. Then
find the ratio of acceleration due to
gravity of their surface. [1]
1. The bulging of earth at the equator
and flattening at the poles is due to… Soln.: Given :
a) centripetal force ρ1 : ρ2 = 2 : 3
b) centrifugal force R1 : R2 = 1 : 2
c) gravitation force
To find : g1 : g2 = ?
d) electrostatic force [1]
Soln.: b) centrifugal force Calculation : For different planets

2. According to Kepler’s law, the areal 4

g = π GR ρ [½]
velocity of a planet around the sun 3
always… [1] g1 R1 ρ1
a) increases ∴
g2
= 
R1 ρ2
b) decreases
g1 1 2
c) remains constant ∴ = × [½]
g2 2 3
d) first increases and then decreases
Soln.: c) remains constant g1 1
∴ =
3. State the dimensions of universal g2 3
gravitational constant. [1]
 g1 : g2 = 1 : 3 [½]
m1 m2
Soln.: F = G 2
r
Section B – (12 Marks)
2
Fr
G =
m1 m2
5. State and explain Newton’s law of

[F] [r ] gravitation. [2]

2

[G] = Soln.: Newton’s law of gravitation

[m ] [m ]
1 2
Every particle of matter attracts every
[M L T ] [L ]
1 1 -2 2 other particle of matter with a force which is
= directly proportional to product of their
[m ] [m ] 1 1
masses and inversely proportional to square
of distance between them. [1]
= [M L T ]
-1 3 -2
[1]
TPS Physics Papers with Solution 112 Gravitation

Let m1 & m2 be masses of two particles (G = 6.67  10

-11 2
N.m / kg ,
2

and r be the distance between the two 2

R = 6400 km, g = 9.8 m/s ) [2]
particles then
Soln.:
m1 m2
F ∝ 2 2 GM
r Ve = [½]
R
G m1 m2
F = [1] 2  6.67  10–11  6  1024
r2
Ve = 6 [½]
Where G = universal gravitational constant. 6.4  10
6. Explain why an astronaut in an 20.01 7
orbiting satellite has feeling of Ve =  10
1.6
weightlessness [2]
20.01 8
Soln.: Explanation of feeling of weight : Ve =  10
16
Weight of a body is the force with
4.473
which the body is attracted by the earth Ve =  10
4
4
towards its centre. When a man is standing
on a surface, he presses the surface Ve = 1.1825  10
4

downwards with his weight in action. The

∴ Ve = 1.12  104 m/s
surface exerts a reaction force on the person
equal and opposite to the weight of the Ve = 11.2 km/s [1]
person. Due to this normal reaction he feels
OR
his weight. [1]
Explanation of feeling of weight : 7. The radii of orbits of two satellites
When an astronaut of mass m in a revolving around the earth are in the
satellite is orbiting around the earth, then ratio 3 : 8 compare their critical
satellite and every object inside the satellite speeds. [2]
has an acceleration equal to the acceleration Soln.: r 1 : r2= 3 : 8
due to gravity directed towards the centre of
To find : Vc1 : Vc2
the earth. So astronaut is unable to exert
weight on the floor of the satellite, in turn GM
∵ Vc = [½]
satellite does not provide normal reaction on r
the astronaut. Therefore the astronaut feels
Vc1 r2
weightlessness. =
Vc2 r1
Weightlessness does not mean the
absence of gravity, it simply means a Vc1 8
= [½]
situation by which an astronaut feels that he Vc2 3
is not attracted by any gravitational force. [1] Vc1 2.66
∴ =
OR Vc2 1

7. Calculate the escape velocity of a Vc1 1.632

∴ = [1]
body from the surface of the earth. Vc2 1
TPS Physics Papers with Solution 113 Gravitation

8. State any four uses of geostationary 10. Derive an expression for binding
satellite [2] energy of a body at rest on the earth’s
Soln.: Uses of geostationary satellite surface. [2]
1) For military purposes Soln.:
2) For weather forecasting
3) For broadcasting telecommunication
4) For astronomical observations
9. What should be the duration of year if
the distance between sun and the
earth gets doubled the present
distance. [2] Fig. Q. 10 [½]
Soln.: Let r be the present distance between
Let M = mass of earth
sun and earth and T be the period
R = radius of earth
Let r1 be the new distance between sun
m = mass of satellite
and earth and T1 be the period.
Gravitational potential on the surface of earth
∴ r1 = 2r [½]
GM
= – [½]
T = 365 days R

∴ T ∝r
2 3
[½] This potential represents P.E. of unit
2 3
mass
T r
∴ = GMm
T1
2
r1
3
∴ P.E. of satellite of mass m = –
R

T
2
r
3
GMm T
2
r
3

= ∴ P.E. = – [½]2 =
R
(2r)
2 3 3
T1 T1 8r
∵ satellite is rest on the surface of earth
2
T 1 ∴ K.E = 0
∴ 2 = 8 [½]
T1
T.E. = K.E. + P.E.
2
T 2
GMm
∴ = 8T ∴ T.E. = 0 –
T
2 R
1
B.E. = – T.E.
∴ T1 = 2 2T [½]
GMm
∴ T1 = 2 × 1.414 × 365 ∴ B.E = –
R
∴ T1 = 730 × 1.414 Negative value of T. E. shows that
T1 = 1032.22 satellite is bound to earth. In general T. E. of
object in bound condition is negative.
∴ T1 = 1032 days [½]
GMm
∴ B.E. = [½]
R
TPS Physics Papers with Solution 114 Gravitation

6.67  10– 11  6  1024  800

Section C – (9 Marks) = [½]
2 (6400  103 + 1800  103)

11. State Kepler’s laws of planetary 6.67  10– 11  6  8  1024  102

=
motion. [3] 2 (8200  103)
Soln.: (i) Kepler’s first law (law of orbit): 6.67  48  1026  10– 11
=
Every planet revolves in an elliptical 2  82  105
orbit round the sun, with the sun situated at
6.67  24  1015
one of the focii of the ellipse. This law is =
82  105
know as the law of orbit. [1]
(ii) Kepler’s second law (law of equal areas): T.E. = – 1.952  1010 J [1]

The radius vector drawn from the sun log 6.64 + log 24 – log 82
to any planet sweeps out equal areas in equal 0.8241
intervals of time. This law is known as the + 1.3802
law of areas i.e. the areal velocity of the
2.2043
radius vector is constant. [1]
–
(iii) Kepler’s third law (law of period): 1.9138

The square of the period of revolution 0.2905

of the planet round the sun is directly Antilog (0.2905) = 1.952
proportional to the cube of the semi-major
B.E. = – (T.E.) = – (– 1.952  1010) [½]
axis of the elliptical orbit. This law is know as
harmonic law (law of period). [1] = 1.952  10 J 10
[½]

12. Find the total energy and binding 13. What is the effect of altitude on
energy of an artificial satellite of mass acceleration due to gravity. [3]
800 kg orbiting at height of 1800 km Soln.: Let M be the mass of the earth and R
above the earth’s surface. be the radius of the earth. Consider body of
(G = 6.67  10
-11 2
N.m / kg ,
2 mass m is situated on the earth’s surface. Let
24
g be the acceleration due to gravity on the
R = 6400 km, M = 6  10 kg) [3] earth’s surface. [½]
Soln.:
We have
Given : Mass of satellite (m) = 800 kg
GM
Height (h) = 1800 km = 1800  103 m g = [½]
R2
G = 6.67  10– 11 Nm2/kg GM
gh = [½]
M = 6  1024 kg (R + h)2
R = 6400 km = 6400  103 m gh R2 gR2
= or gh =
To find : T.E. and B.E. = ? g (R + h) 2
(R + h)2
– GMm R2  h – 2
T.E. = [½] =1+
R 
2 (R + h) = [½]
 h 2

R2  1 +
 R 
TPS Physics Papers with Solution 115 Gravitation

2h mV2c GMm
= 1– +… [½]  = [½]
R r r2
h
Higher power of are neglected (here GM
R  V2c =
r
h
we assume that << 1) we get GM
R
 Vc = [½]
r
 2h 
gh = g  1 –
R 
 [½]

Section D – (10 Marks)
OR

13. Define critical velocity of a satellite 14. A) Obtain an expression for period of
and obtain an expression for it. [3] a satellite in a circular orbit round the
Soln.: Critical velocity – The minimum earth [3]
Horizontal velocity of projection of a satellite B) The mass of body on surface of
for which the satellite revolves in circular earth is 100 kg. What will be its
orbit round the earth is called critical mass and weight at an altitude of
velocity. [1] 1000 km ? [2]
Consider a satellite of mass m Soln.: A) Consider a satellite of mass m
revolving round the earth at a height ‘h’ projected in a circular orbit round the from a
above the surface of the earth. height h above the earth’s surface with radius
of orbit r = R + h
Let, in one revolution, the distance
covered by the satellite is equal to the
circumference of its circular orbit and the
time taken is the periodic time T. [½]
Critical speed (Vc) =
Fig. Q. 13 Circumference of the orbit
Period time
Diagram - ½ Mark
2r
Vc = [½]
Let M be the mass and R be the radius T
of the earth. The satellite is moving in a GM
circular orbit with critical velocity Vc and Vc = [½]
r
radius of circular orbit is r = R + h.
GM 2r
The centripetal force necessary for the  = [½]
r T
circular motion of the satellite is provided by
Squaring both sides
the gravitational force of attraction exerted
on the satellite by the earth. [½] GM 42 r2
=
Centripetal force = Gravitational force r T2
TPS Physics Papers with Solution 116 Gravitation

42 r3  42  3 0.9912 – 0.8692 – 08692

T2 = =  r
GM  GM  = 2.6036 – 1.7384
2
4 log gh = 0.8652
As is constant
GM
gh = antilog (0.8652)
 T2  r3
2
gh = 7.331 m/s [½]
Thus the square of the period of
revolution of the satellite is directly ∴ Weight of body = mgh
proportional to the cube of the radius of its
= 100 × 7.331
orbit.
= 733.1 N [½]
2
42 r3
T = OR
GM

r3 14. A) Derive an expression for escape

 T = 2 [½] velocity of an object from the surface
GM
of the earth. [3]
This is expression for the period of the
satellite. Period of the satellite depends on B) Show that escape velocity of a body
mass of the earth. [½] from the surface of earth is 2 vc
B) Mass of the body is independent of when it is orbiting very close to
acceleration due to gravity. earth’s surface. [2]

∴ Mass of body at altitude 1000 km is Soln.: A)

100 kg
gh R
2

= [½]
g (R + h )
2

6
R = 6.4 ×10 m
6
h = 1000 km = 1 × 10 m
Fig. Q. 14 [½]
2
gh  6.4 × 10
6

=   Let M = mass of earth
9.8  (6.4 × 10 + 1 × 10 )
6 6
  R = Radius of earth
gh 6.4 × 6.4 × 10
12 m = mass of satellite
= For a satellite rest on the surface of earth
9.8 7.4 × 7.4 × 10
12

Binding energy =
6.4 × 6.4 × 9.8
gh = [½] Gravitational potential energy
7.4 × 7.4
GMm
log gh = log 6.4 + log 6.4 + ∴ B.E. = [½]
R
log 9.8 – log 7.4 – log 7.4 When satellite is projected with escape
log gh = 0.8062 + 0.8062 + velocity.
TPS Physics Papers with Solution 117 Gravitation

K.E. of projection = Binding energy Soln.: Let M be the mass of the earth and R
1 GMm be its radius. The acceleration due to gravity
2
mve = [½] at the surface of the earth is given by
2 R
GM
2 GMm 2 g = …(1)
ve = × R2
R m
2 GM where G is gravitational constant.
2
∴ ve =
R

2 GM
∴ ve =
R
Escape velocity in terms of acceleration
due to gravity.
2
∵ GM = gR
2
2 gR
∴ ve = Fig. Q. 15
R

∴ ve = 2 gR [Diagram with labeling and explanation - 1 Mark]

2 GM
B) ve = …(1) Consider the earth to be a homogenous
R
sphere having mean density  then
GM
Vc = Mass = Volume  Density
r
4
GM = R2 
3
∴ Vc =
R+h
4 R3 
∴ Satellite is revolving close to the  g = G [½]
3 R2
surface of earth
4
R+h = R = RG …(2)
3
GM Consider a point P at a depth d below
∴ Vc = …(2)
R the surface of the earth where the
From equation (1) & (2) acceleration due to gravity be gd. The point P
Ve = 2 vc [½] at this depth will experience gravitational
force only due to remaining portion (inner
15. A) What is the effect of depth on the sphere) of the earth. If M be the mass of the
acceleration due to gravity. [3] inner sphere then
B) Obtain the relation between
4
universal gravitational constant and M =  (R – d)3 
3
the gravitational acceleration on the
surface of the earth. [2] GM
gd =
(R – d)2
TPS Physics Papers with Solution 118 Gravitation

4  (R – d)3
= G  [½]
3 (R – d)2
4
gd =  (R – d) G …(3)
3
Dividing Equation (3) by (2)
4
gd  (R – d) G
3
=
g 4
RG
3
gd R–d
= [½]
g R
gd d
= 1–
g R

 d
 gd = g  1 –
R 
[½]

B) Let M = mass of earth
R = Radius of earth
m = mass of satellite (1)
When satellite is rest on the surface of
earth
Weight of satellite = Gravitational force [1]
GMm
mg = 2
R
GM
∴ g = 2 …(1) [1]
R

