NCERT Solutions for Class 11 Physics Chapter 7 Gravitation

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Physics
Chapter : 8
Chapter Name : Gravitation

Q8.1 Answer the following :
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can
you shield a body from the gravitational in uence of nearby matter by putting it inside a
hollow sphere 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 nd that the Sun’s pull is greater than the moon’s pull. (you can check this yourself
using the data available in the succeeding exercises). However, the tidal effect of the moon’s
pull is greater than the tidal effect of sun. Why ?

Answer. (a) No (b) Yes
(a) Gravitational in uence of matter on nearby objects cannot be screened by any
means. This is because gravitational force unlike electrical forces is independent of the
nature of the material medium. Also, it is independent of the status of other objects.
(b) If the size of the space station is large enough, then the astronaut will detect the
change in Earth's gravity
(c) Tidal effect depends inversely upon the cube Of the distance while, gravitational
force depends inversely on the square of the distance. Since the distance between the
Moon and the Earth is smaller than the distance between the Sun and the Earth, the
tidal effect of the Moon's pull is greater than the tidal effect of the Sun's pull.

Page : 200 , Block Name : Exercise

Q8.2 Choose the correct alternative :
(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 −GM m (1/r − 1/r ) is more/less accurate than the formula (r − r ) for
2 1 2 t

the difference of potential energy between two points r and r distance away from the centre
2 1

of the earth.

Answer. (a) Decreases
(b) Decreases

(c) Mass Of the body
(d) More
Explanation
(a) Acceleration due to gravity at depth h is given by the relation.
2h
g = (1 − )g
n Re

Where,
R= Radius of the Earth
g = Acceleration due to gravity on the surface of the Earth
It is clear from the given relation that acceleration due to gravity decreases with an
increase in height.
(b) Acceleration due to gravity at depth d is given by the relation:
d
g = (1 − )g
d Re

It is clear from the given relation that acceleration due to gravity decreases with an
increase in depth.
(c) Acceleration due to gravity of body of mass m is given by the relation:
GM
g = 2
R

Where,
G = Universal gravitational constant
M = Mass of the Earth
R= Radius of the Earth
Hence, it can be inferred that acceleration due to gravity is independent of the mass of
the body.
(d) Gravitational potential energy of two points r and r distance away from the centre of the
2 1

Earth is respectively given by:
GmM
V (r1 ) = −
r1

Gin M
V (r2 ) = −
r2

Difference in potential energy, V = V (r ) − V (r ) = − Gim M (
1 1
2 1 − )
r2 r1

Hence, this formula is more accurate than the formula mg (r 2 − r1 )

Page : 201 , Block Name : Exercise

Q8.3 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?

Answer. Lesser by a factor of 0.63
Time taken by the Earth to complete one revolution around the Sun,
T = 1 year
e

Orbital radius of the Earth in its orbit, R = 1AU
e

Time taken by the planet to complete one revolution around the Sun,T
1 1
P = Te =
2 2

Orbital radius of the planet = R p

From Kepler's third law of planetary motion, we can write:

3 2
Rr Tr
( ) = ( )
Re Te

2

Rp TP 3

= ( )
Rv Te

3 2
2
= ( ) = (0.5) 3 = 0.63
1

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

Page : 201 , Block Name : Exercise

Q8.4 In, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of
the orbit is 4.22 × 10 m. Show that the mass of Jupiter is about one-thousandth that of the
8

sun

Answer. Orbital period of I , T = 1.769 days = 1.769 × 24 × 60 × 60s 0 lo

Orbital radius of I , R = 4.22 × 10 m 0 lo
8

Satellite I is revolving around the jupiter
0

Mass of the latter is given by the relation:
2 3
4π R
±
Mj =
2
GT
li0

Where
MJ = mass of jupiter
G= Universal gravitational constant
Orbital period of the Earth,
Te = 365.25 days = 365.25 × 24 × 60 × 60s

Orbital radius of the earth ,
11
Re = 1AU = 1.496 × 10 m

Mass of Sun is given as:
2 3
4π Rr
Ms =
2
GTc

3 2 3 2
Mi 4π Re
2 GT Rc T
b lo
∴ = × = ×
MJ 2 3 3 2
GTe 4π 2 R R Tc
lu b

2 11
3
1.769×24×60×60 1.496×10
= ( ) × ( 8
)
365.25×24×60×60 4.22×10

= 1045.04
Ms
∵ ∼ 1000
MJ

Ms ∼ 1000 × MJ

Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.

Page : 201 , Block Name : Exercise

Q8.5 Let us assume that our galaxy consists of 2.5 × 10 stars each of one solar mass. How 11

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 10 ly . 5

Answer. Mass of our galaxy Milky Way, M = 2.5 × 10 11

Solar mass = Mass of Sun = 2.0 × 10 kg 36

Mass of our galaxy, M = 2.5 × 10 × 2 × 10 = 5 × 10 11 36 41
kg

Diameter Of Milky Way, d = 10 ly 5

Radius of Milky Way, r = 5 × 10 ly 4

15
1ly = 9.46 × 10 m

4 15
∴ r = 5 × 10 × 9.46 × 10

20
= 4.73 × 10 m

Since a star revolves around the galactic centre of the Milky Way, its time period is given by
the relation:
1

2 3 2
4π r
T = ( )
GM
1
1
2 3 60 2 30 2
4 × (3.14) × (4.73) × 10 39.48 × 105.82 × 10
= ( ) = ( )
−11 41 33.35
6.67 × 10 × 5 × 10

1

30 2 16
= (125.27 × 10 ) = 1.12 × 10 s

1 year = 365 × 324 × 60 × 60s

1
1s = years
365×24×60×60
16
1.12 × 10
16
∴ 1.12 × 10 s =
365 × 24 × 60 × 60
8
= 3.55 × 10 years

Page : 201 , Block Name : Exercise

Q8.6 Choose the correct alternative:
(a) If the zero of potential energy is at in nity, 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 in uence is
more/less than the energy required to project a stationary object at the same height (as the
satellite) out of earth’s in uence

Answer. (a) Kinetic energy
(b) Less
(a) Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and
potential energy (may be negative). At in nity, the gravitational potential energy of the
satellite is zero. As the Earth-satellite system is a bound system, the total energy of the
satellite is negative. Thus, the total energy of an orbiting satellite at in nity is equal to the
negative of its kinetic energy.
(b) An orbiting satellite acquires a certain amount of energy that enables it to revolve around
the Earth. This energy is provided by its orbit. It requires relatively lesser energy to move out
of the in uence of the Earth's gravitational eld than a stationary object on the Earth's surface
that initially contains no energy.

Page : 201 , Block Name : Exercise

Q8.7 Does the escape speed of a body from the earth depend on
(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?

Answer.
(a) No
(b) No
(c) No
(d) Yes
Escape velocity of a body from the Earth is given by the relation:
vesc = √2gR...(i)

g = Acceleration due to gravity
R = Radius of the Earth
It is clear from equation (i) that escape velocity vesc is independent of the mass of the body
and the direction of its projection. However, it depends on gravitational potential at the point
from where the body is launched. Since this potential marginally depends on the height of the
point, escape velocity also marginally depends on these factors.

Page : 201 , Block Name : Exercise

Q8.8 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.

Answer.
(a) No
(b) No
(c) Yes
(d) No
(e) No
(f) Yes
Angular momentum and total energy at all points of the orbit of a comet moving in a highly
elliptical orbit around the Sun are constant. Its linear speed, angular speed, kinetic, and
potential energy varies from point to point in the orbit.

Page : 201 , Block Name : Exercise

Q8.9 Which of the following symptoms is likely to af ict an astronaut in space
(a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem

Answer. (a) Legs hold the entire mass of a body in standing position due to gravitational pull.
In
space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen
feet of an astronaut do not affect him/her in space.
(b) A swollen face is caused generally because of apparent weightlessness in space. Sense
organs such as eyes, ears nose, and mouth constitute a person's face. This symptom can affect
an astronaut in space.
(c) Headaches are caused because of mental strain. It can affect the working of an astronaut in
space.
(d) Space has different orientations. Therefore, orientational problem can affect an astronaut
in space.

Page : 201 , Block Name : Exercise

Q8.10 In the following two exercises, choose the correct answer from among the given ones:
The gravitational intensity at the centre of a hemispherical shell of uniform mass density has
the direction indicated by the arrow (see Fig ) (i) a, (ii) b, (iii) c, (iv) 0.

Answer. (iii)
Gravitational potential (V) is constant at all points in a spherical shell. Hence, the
gravitational potential gradient ( is zero everywhere inside the spherical shell. The
dV
)
dr

gravitational potential gradient is equal to the negative of gravitational intensity. Hence,
intensity is also zero at all points inside the spherical shell. This indicates that gravitational
forces acting at a point in a spherical shell are symmetric.
If the upper half of a spherical shell is cut out (as shown in the given gure), then the net
gravitational force acting a particle located at centre O will be in the downward direction.
Since gravitational intensity at a point is de ned as the gravitational force per unit mass at
that point, it will also act In the downward direction. Thus, the gravitational intensity at
centre O of the given hemispherical shell has the direction as indicated by arrow c.

Page : 201 , Block Name : Exercise

Q8.11 For the above problem, the direction of the gravitational intensity at an arbitrary point
P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Answer. Gravitational potential (V) is constant at all points in a spherical shell. Hence, the
gravitational potential gradient is zero everywhere inside the spherical shell. The gravitational
potential gradient ( is equal to the negative of gravitational intensity. Hence, intensity is
dV
)
dr

also zero at all points inside the spherical shell. This indicates that gravitational forces acting
at a point in a spherical shell are symmetric.
If the upper half of a spherical shell is cut out (as shown in the given gure), then the net
gravitational force acting on a particle at an arbitrary point P will be in the downward

direction.
Since gravitational intensity at a point is de ned as the gravitational force per unit mass at
that point, it will also act in the downward direction. Thus, the gravitational intensity at an
arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.

Page : 201 , Block Name : Exercise

Q8.12 A rocket is red from the earth towards the sun. At what distance from the earth’s
centre is the gravitational force on the rocket zero ? Mass of the sun = 2 × 10 kg, mass of 30

the earth = 6 × 10 kg , Neglect the effect of other planets etc. (orbital radius =1.5 ×10 m)
24 11

)

Answer. Mass of the Sun, M = 2 × 10 kg
30
s

Mass of the Earth, M = 6 × 10 kg e
24

Orbital radius, r = 1.5 × 10 m 11

Mass of the rocket =m
Let x be the distance from the centre of the Earth where the gravitational force acting on
satellite P becomes zero. From Newton's law Of gravitation, we can equate gravitational forces
acting on satellite p under the in uence of the Sun and the Earth as:
GmMs Mc

2
= Gint
(r−x) x2

2
r−x Mx
( ) =
x Me

1
30
r−x 2×10 2

= ( ) = 577.35
x 24
60×10

11 11
1.5 × 10 − 1.5 × 10
11
1.5×10 8
x = = 2.59 × 10 m
578.35

Page : 201 , Block Name : Exercise

Q8.13 How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the
earth around the sun is 1.5 × 10 km 8

Answer. Orbital radius of the Earth around the Sun, r = 1.5 × 10 m 11

Time taken by the Earth to complete one revolution around the Sun,
T = 1 year = 365.25 days
T = 1 year days
= 365.25 × 24 × 60 × 60s

Universal gravitational constant, G = 6.67 × 10 Nm kg −11 2 −2

Thus, mass of the Sun can be calculated using the relation,

2 3
4π r
M =
2
GT
3
2 11
4 × (3.14) × (1.5 × 10 )
=
−11 2
6.67 × 10 × (365.25 × 24 × 60 × 60)

133.24 × 10
30
= = 2.0 × 10 kg
4
6.64 × 10

Hence, the mass of the sun is 2 × 10 30
kg

Page : 201 , Block Name : Exercise

Q8.14 A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the
earth is 1.50 × 10 km away from the sun?
8

Answer. Distance of the Earth from the sun, r = 1.5 × 10 km = 1.5 × 10 e
8 11
m

Time period Of the Earth = T e

Time period of Saturn, T = 29.5T s e

Distance of Saturn from the Sun = r s

From Kepler's third law of planetary motion, we have
1

2 3 2
4π r
T = ( )
GM

For saturn and Sun , we can write , we have
3 2
rs Ts
=
3 2
re Te

2

Ts 3

rs = re ( )
Te

2

29.5Te 3
11
= 1.5 × 10 ( )
Te
2
11
= 1.5 × 10 (29.5) 3

11
= 1.5 × 10 × 9.55
11
=14.32 × 10 m

Hence, the distance between Saturn And the Sun is 1.43 × 10 12
m

Page : 201 , Block Name : Exercise

Q8.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due
to the earth at a height equal to half the radius of the earth?

Answer. Weight of the body, W = 63 N
Acceleration due to gravity at height h from the Earth's surface is given by the relation:
Re = Radius of Earth

For h =
Re

2

g g 4
′
g = 2
= 2
= g
Rc 1
9
(1+ ) (1+ )
2×Re 2

Weight of g =
g
′
2
1+h
( )
Re

Where
g = Acceleration due to gravity on the Earth's surface
Re = Radius of the Earth
Re
For h =
2

g g 4
′
g = 2
= 2
= g
Rc 1
9
(1+ ) (1+ )
2×Re 2

Weight of a body of mass m at height h is given as
′
W = mg
4 4
= m × g = × mg
9 9
4
= W
9

4
= × 63 = 28N
9

Page : 201 , Block Name : Exercise

Q8.16 Assuming the earth to be a sphere of uniform mass density, how much would a body
weigh half way down to the centre of the earth if it weighed 250 N on the surface?

Answer. Weight Of a body Of mass m at the Earth's surface, W = mg = 250 N
Body of mass m is located at depth, d = R 1

2
v

Where,
R = Radius of the Earth
e

Acceleration due to gravity at depth g (d) is given by the relation:
d
′
g = (1 − )g
Re

Re 1
= (1 − )g = g
2 × Re 2

Weight of the body at depth d,
′
W = mg

1 1 1
= m × g = mg = W
2 2 2
1
= × 250 = 125N
2

Page : 201 , Block Name : Exercise

Q8.17 A rocket is red vertically with a speed of 5kms from the earth's surface. How far −1

from the earth does the rocket go before returning to the earth? Mass of the earth 6×10 kg
24

mean radius of the earth = 6.4 × 10 m; G = 6.67 × 10 Nm kg 6 −11 2 −2

Answer. 8 × 10 m from the centre of the Earth
6

Velocity of the rocket v = 5km/s = 5 × 10 m/s 3

Mass of the earth ,M = 6.0 × 10 kg e
24

Radius of Earth R = 6.4 × 10 m e
6

Height reached by rocket mass, m = h
At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
1 2 −GMe m
= mv + ( )
2 Re

At highest point h,
v=0
And Potential energy = −
GMe m

Rc +h

Total energy of the rocket = 0 + (−
GMe m GMe m
) = −
Re +h Re +h

From the law of conservation of energy, we have
Total energy of the rocket at the Earth's surface = Total energy at height h
1 2 GMe m GMe m
mv + (− ) = −
2 Re Rc +h

1 2 1 1
v = GMe ( − )
2 Re Re +h

Re +h−Rc
= GMe ( )
Re (Re +h)

1 2 GMe h Re
v = ×
2 Rc (Rc +h) Rc

gRe h
1 2
× v =
2 Rc +h

Where g = GM

2
= 9.8m/s
2
(Acceleration due to gravity on the Earth's surface)
Rc

2
∴ v (Re + h) = 2gRv h

2 2
v Re = h (2gRe − v )

2
Re v
h = 2
2gRe −v
2
6 3
6.4×10 ×(5×10 )
= 2
6 3
2×9.8×6.4×10 −(5×10 )

12
6.4×25×10 6
h = 6
= 1.6 × 10 m
100.44×10

Height achieved by the rocket with respect to the centre of the Earth
= Re + h

6 6
= 6.4 × 10 + 1.6 × 10

6
= 8.0 × 10 m

Page : 202 , Block Name : Exercise

Q8.18 The escape speed of a projectile on the earth's surface is 11.2kms . A body projected −1

out with thrice this speed. What is the speed of the body far away from the earth? Ignore the
presence of the sun and other planets.

Answer. Escape velocity of a projectile from the Earth, V esc = 11.2km/s

Projection velocity of the projectile, v = 3v p esc

Mass of the projectile = m
Velocity of the projectile far away from the Earth = V e

Total energy of the projectile on the Earth = mv 1

2
2

f

Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy Of the projectile far away from the Earth = mv 1

2
2

f

From the law of conservation of energy, we have
1 2 1 2 1 2
mvp − mvcc = mv
2 2 2 f

2 2
vr = √vr − vrse

2 2
= √(3vsec ) − (vesc )

= √8vcx

= √8 × 11.2 = 31.68km/s

Page : 202 , Block Name : Exercise

Q8.19 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 earth's gravitational in uence? Mass of the
satellite = 200 kg; mass of the earth = 6.0 × 10 kg; radius of the earth 6.4 24

6 −11 2 −2
×10 m; G = 6.67 × 10 Nm kg

Answer. Mass of the Earth, M = 6.0 × 10 kg 24

Mass of the satellite, m = 200 kg
Radius of the Earth, R = 6.4 × 10 m e
6

Universal gravitational constant, G = 6.67 × 10 Nm kg −11 2 −2

Height of the satellite, h = 400km = 4 × 10 m = 0.4 × 10 m 5 6

Total energy of the satellite at height h = 1 2 −GMe m
mv + ( )
2 Re +h

Orbital velocity Of the satellite, v = √
GMe

Re +h

Total energy of height, h= 1 GMe GMe m 1 GMe m
m( ) − = − ( )
2 Rc +h Re +h 2 Re +h

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy
of the satellite.
Energy required to send the satellite out of its orbit =— (Bound energy)
1 GMe m
=
2 (Re +h)
−11 24
1 6.67×10 ×6.0×10 ×200
= ×
6 6
2 (6.4×10 +0.4×10 )

1 6.67×6×2×10 9
= × = 5.9 × 10 J
6
2 6.8×10

Page : 202 , Block Name : Exercise

Q8.20 Two stars each of one solar mass (= 2 × 10 30
kg) are approaching each other for a head
on collision. When they are a distance 109 krn, their speeds are negligible. What is the speed
with which they collide? The radius of each star is 104 km. Assume the stars to
remain undistorted until they collide. (use the known value of G)

Answer. Mass of each star, M = 2 × 10 kg 30

Radius of each star, R = 10 km = 10 m 4 7

Distance between the stars, r = 10 km = 10 m 9 12

For negligible speeds, v = 0 total energy of two stars separated at distance r
−GM M 1 2
= + mv
r 2

−GM M
= + 0
r

Now, consider the case when the stars are about to collide:
Velocity of the stars = v
Distance between the centers of the stars = 2R
Total kinetic energy of both stars = M v + M v = M v 1

2
2 1

2
2 2

Total potential energy of both stars =
−GM M

2R

Total energy of the two stars = M v 2 GM M
−
2R

Using the law of conservation of energy, we can write:
2 GMM −GM M
Mv − =
2R r

2 −GM GM 1 1
v = + = GiM(− + )
r 2R r 2R

−11 30 1 1
= 6.67 × 10 × 2 × 10 [− + ]
12 7
10 2×10

19 −12 −8
= 13.34 × 10 [−10 + 5 × 10 ]
10 −3
−13.34 × 10 × 5 × 10

12
∼ 6.67 × 10

√ 12 6
v = 6.67 × 10 = 2.58 × 10 m/s

Page : 202 , Block Name : Exercise

Q8.21 Two 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 force and potential at the mid point of the line
joining the centers Of the spheres? Is an object placed at that point in equilibrium? If so, is the
equilibrium stable or unstable?

Answer.
0

−8
−2.7 × 10 J/kg

Yes;

Unstable

Explanation
The situation is represented in the given gure:

Mass of each sphere , M = 100 kg
Separation between the spheres, r = 1m
X is the mid point between the spheres. Gravitational force at point X will be zero. This is
because gravitational force exerted by each sphere will act in opposite directions.
Gravitational potential at point X:
−GM GM GM
= − = −4
r r r
( ) ( )
2 2

−11
4 × 6.67 × 10 × 100
=
1
−8
= −2.67 × 10 J/kg

Any object placed at point X will be in equilibrium state, but the equilibrium is unstable. This
is because any change in the position of the object will change the effective force in that
direction.

Page : 202 , Block Name : Exercise

Q8.22 As you have learnt in the text, a geostationary satellite orbits the earth at a height of
nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at
the site of this satellite ? (Take the potential energy at in nity to be zero). Mass of the earth
kg , radius = 6400 km.
24
= 6.0 × 10

Answer. Mass of the Earth, M = 6.0 × 10 kg 24

Radius of the earth, R = 6400 km R = 6400km = 6.4 × 10 m 6

Height of a geostationary satellite from the surface of the Earth,
7
h = 36000km = 3.6 × 10 m

Gravitational potential energy due to Earth's gravity at height h,
−GM
=
(R+h)

−11 24
6.67 × 10 × 6.0 × 10
= −
7 7
3.6 × 10 + 0.64 × 10

6.67 × 6
13−7
= − × 10
4.24
6
= −9.4 × 10 J/kg

Page : 202 , Block Name : Additional Exercise

Q8.23 A star 2.5 times the mass of the sun collapsed to a size of 12 km rotates with a speed of
1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain
stellar objects called pulsars belong to this category). Will an object placed on its equator
remain stuck to its surface due to gravity? f = .
9
GM m

R2

Answer. Yes
A body gets stuck to the surface of a star if the inward gravitational force is greater than the

outward centrifugal force caused by the rotation of the star.
Gravitational force, f = 9
GM m

2
R

Where,
M = Mass of the star = 2.5 × 2 × 10 = 5 × 10 30 30
kg

Mass of the body
R = Radius of the star = 12km = 1.2 × 10 m 4

−11 10
6.67×10 ×5×10 ×m 11
∴ fg = = 2.31 × 10 mN
2
4
(1.2×10 )

Centrifugal force, f = mrω c
2

ω = Angular speed = 2πv

v = Angular frequency = 1.2 rev s −1

2
fc = mR(2πv)
4 2 2 5
= m × (1.2 × 10 ) × 4 × (3.14) × (1.2) = 1.7 × 10 mN

Since f g > fc , the body will remain stuck to the surface of the star.

Page : 202 , Block Name : Additional Exercise

Q8.24 A spaceship is stationed on Mars. How much energy must be expended on the spaceship

to launch it out of the solar system? Mass of the spaceship 1000 kg; mass of the Sun
kg ; mass of mars = 6.4 × 10 kg; radius of mars = 3395 km; radius of the orbit of
30 23
= 2 × 10

mars = 2.28 × 10 kg; G = 6.67 × 10 8 −11 2 −2
m kg

Answer. Mass of the spaceship, ms = 1000 kg
Mass of the Sun, M = 2 × 10 kg 30

Mass Of Mars, m = 6.4 × 10 kg m
23

Orbital radius of Mars, R = 2.28 × 10 kg = 2.28 × 10 8 11
m

Radius of Mars, r = 3395km = 3.395 × 10 m 6

Universal gravitational constant, G = 6.67 × 10 m kg −11 2 −2

Potential energy of the spaceship due to the gravitational attraction of the Sun
−GM ms
=
R

Potential energy of the spaceship due to the gravitational attraction of Mars =
−GMm ms

r

Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero.

Total energy of the spaceship =
−GM ms −GMm
−
R r

M mm
= − Gm( + )
R r

The negative sign indicates that the system is in bound state.
Energy required for launching the spaceship out of the solar system
= (Total energy of the spaceship)
M mm
= Gms ( + )
R r

30 33
−11 3 2×10 6.4×10
= 6.67 × 10 × 10 × ( 11
+ 6
)
2.28×10 3.395×10

−8 17 17
= 6.67 × 10 (87.72 × 10 + 1.88 × 10 )

−8 17
= 6.67 × 10 × 89.50 × 10
∘
= 596.97 × 10

11
= 6 × 10 J

Page : 202 , Block Name : Additional Exercise

Q8.25 A rocket is red 'vertically' from the surface of mars with a speed of 2kms − 1. If 20% of
its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from
the surface of mars before returning to it? Mass of mars = 6.4 × 1023kg; radius of mars =
3395 km; G = 6.67 × 10 Nm kg −11 2 −2

Answer. Initial velocity of the rocket, v = 2km/s = 2 × 10 m/s 3

Mass of Mars, M = 6.4 × 10 kg 23

Radius of Mars, M = 6.4 × 10 kg 23

Universal gravitational constant, G = 6.67 × 10 −11
Nm kg
2 −2

Mass Of the rocket = m
Initial kinetic energy of the rocket = mv 1

2
2

Initial potential energy of the rocket =
−GM m

R

Total initial energy = 1

2
mv
2
−
GM m

R

If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 %
of its kinetic energy helps in reaching a height.
Total initial energy available =
80 1 2 GM m 2 GM m
× mv − = 0.4mv −
100 2 R R

Maximum height reached by the rocket = h
At this height, the velocity and hence, the kinetic energy of the rocket will become zero.
Total energy of the rocket at height h = 80 1 2 GM m 2 GM m
× mv − = 0.4mv −
100 2 R R

Applying the law of conservation of energy for the rocket, we can write:

Total energy of the rocket at height h = − GM m

(R+h)

Applying the law of conservation of energy for the rocket, we can write:
2 GM m −GM m
0.4mv − =
R (R+h)

2 GM GM
0.4v = −
R R+h

1 1
= GM ( − )
R R+h

R+h−R
= GM ( )
R(R+h)

GM h
=
R(R+h)

R+h GM
=
h 0.4v2 R

R GM
+ 1 = 2
h 0.4v R

R GM
= 2
− 1
h 0.4v R

R
h =
GM
− 1
0.4v2 R

2 2
0.4R v
=
2
GM − 0.4v R
2 2
6 3
0.4 × (3.395 × 10 ) × (2 × 10 )
=
2
−11 23 3 6
6.67 × 10 × 6.4 × 10 − 0.4 × (2 × 10 ) × (3.395 × 10 )

18
18.442 × 10 18.442
6
= = × 10
12 12
42.688 × 10 − 5.432 × 10 37.256

3
= 495 × 10 m = 495km

Page : 202 , Block Name : Additional Exercise