NCERT Solutions for Class 11 Science Physics Chapter 8 Gravitation are provided here with simple step-by-step explanations. These solutions for Gravitation are extremely popular among Class 11 Science students for Physics Gravitation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Physics Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

#### Page No 201:

#### Question 8.1:

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 influence of nearby matter by putting it inside a hollow sphere or by some other means?

**(b) **An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?

**(c) **If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (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:

**Answer:** (a) No
(b) Yes

**(a)** Gravitational influence 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 (g).

**(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 No 201:

#### Question 8.2:

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 –*G Mm*(1/*r*_{2}– 1/*r*_{1}) is more/less accurate than the formula *mg*(*r*_{2}– *r*_{1}) for the difference of potential energy between two points *r*_{2}and *r*_{1}distance away from the centre of the earth.

#### Answer:

**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:

Where,

= 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:

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:

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*_{2} and *r*_{1} distance away from the centre of the Earth is respectively given by:

Hence, this formula is more accurate than the formula *m*g(*r*_{2}– *r*_{1}).

#### Page No 201:

#### Question 8.3:

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:

**Answer:** Lesser
by a factor of 0.63

Time taken by the Earth to complete one revolution around the Sun,

*T*_{e} =
1 year

Orbital radius of the
Earth in its orbit, *R*_{e }= 1 AU

Time taken by the planet to complete one revolution around the Sun,

Orbital radius of the
planet = *R*_{p }

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

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

#### Page No 201:

#### Question 8.4:

Io, one of the satellites of Jupiter, has an orbital period of
1.769 days and the radius of the orbit is 4.22 × 10^{8}
m. Show that the mass of Jupiter is about one-thousandth that of the
sun.

#### Answer:

Orbital period of

Orbital radius of

Satelliteis revolving around the Jupiter

Mass of the latter is given by the relation:

Where,

= Mass of Jupiter

G = Universal gravitational constant

Orbital radius of the Earth,

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

#### Page No 201:

#### Question 8.5:

Let us assume that our galaxy consists of 2.5 × 10^{11} 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 10^{5} ly.

#### Answer:

Mass of our galaxy
Milky Way, *M* = 2.5 × 10^{11} solar mass

Solar mass = Mass of
Sun = 2.0 × 10^{36}
kg

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

Diameter of Milky Way,
*d* = 10^{5} ly

Radius of Milky Way, *r*
= 5 × 10^{4} ly

1 ly = 9.46 ×
10^{15} m

∴*r* = 5 ×
10^{4} × 9.46 ×
10^{15}

=
4.73 ×10^{20} m

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

#### Page No 201:

#### Question 8.6:

Choose the correct alternative:

**(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.

#### Answer:

**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 infinity, 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 infinity 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 influence of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy.

#### Page No 201:

#### Question 8.7:

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:

g = Acceleration due to gravity

*R* = Radius of
the Earth

It is clear from
equation (i) that escape velocity *v*_{esc }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 No 201:

#### Question 8.8:

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 No 201:

#### Question 8.9:

Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem?

#### Answer:

**Answer: **(b),
(c), and (d)

**(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 No 201:

#### Question 8.10:

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 8.12) (i) a, (ii) b, (iii) c, (iv) O.

#### Answer:

**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 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 figure), then the net gravitational force acting on a particle located at centre O will be in the downward direction.

Since gravitational
intensity at a point is defined 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 No 202:

#### Question 8.11:

Choose the correct answer from among the given ones:

For the problem 8.10, 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:

**Answer: **(ii)

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 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 figure), 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 defined 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 No 202:

#### Question 8.12:

A
rocket is fired 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^{30}
kg, mass of the earth = 6 × 10^{24}
kg. Neglect the effect of other planets etc. (orbital radius = 1.5
× 10^{11}
m).

#### Answer:

Mass of the Sun, *M*_{s}
= 2 × 10^{30} kg

Mass of the Earth, *M*_{e}
= 6 × 10 ^{24} kg

Orbital radius,* r*
= 1.5 × 10^{11} m

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 influence of the Sun and the Earth as:

#### Page No 202:

#### Question 8.13:

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^{8}
km.

#### Answer:

Orbital radius of the Earth around the Sun, *r* = 1.5 × 10^{11} m

Time taken by the Earth to complete one revolution around the Sun,

*T* = 1 year = 365.25 days

= 365.25 × 24 × 60 × 60 s

Universal gravitational constant, G = 6.67 × 10^{–11} Nm^{2} kg^{–2}

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

Hence, the mass of the Sun is 2 × 10^{30} kg.

#### Page No 202:

#### Question 8.14:

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^{8}
km away from the sun?

#### Answer:

Distance of the Earth
from the Sun, *r*_{e} = 1.5 ×
10^{8} km = 1.5 ×
10^{11} m

Time period of the
Earth = *T*_{e}

Time period of Saturn,
*T*_{s} = 29. 5 *T*_{e}

Distance of Saturn from
the Sun = *r*_{s }

From Kepler’s third law of planetary motion, we have

For Saturn and Sun, we can write

Hence, the distance between Saturn and the Sun is.

#### Page No 202:

#### Question 8.15:

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:

Where,

g = Acceleration due to gravity on the Earth’s surface

*R*e = Radius of
the Earth

Weight of a body of
mass *m* at height *h* is given as:

#### Page No 202:

#### Question 8.16:

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,

Where,

= Radius of the Earth

Acceleration due to
gravity at depth *g *(*d*) is given by the relation:

Weight of the body at
depth *d*,

#### Page No 202:

#### Question 8.17:

A
rocket is fired vertically with a speed of 5 km s^{–1}
from the earth’s surface. How far from the earth does the
rocket go before returning to the earth? Mass of the earth = 6.0 ×
10^{24}
kg; mean radius of the earth = 6.4 ×
10^{6}
m; G=
6.67 ×
10^{–11}
N m^{2}
kg^{–}^{2.}

#### Answer:

**Answer:** 8 ×
10^{6} m from the centre of the Earth

Velocity of the rocket,
*v* = 5 km/s = 5 × 10^{3}
m/s

Mass of the Earth,

Radius of the Earth,

Height reached by
rocket mass, *m *= *h*

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

At highest point *h*,

Total energy of the rocket

From the law of conservation of energy, we have

Total energy of the
rocket at the Earth’s surface = Total energy at height *h*

Height achieved by the rocket with respect to the centre of the Earth

#### Page No 202:

#### Question 8.18:

The escape speed of a
projectile on the earth’s surface is 11.2 km s^{–1}.
A body is projected 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 }= 11.2 km/s

Projection velocity of
the projectile, *v*_{p} = 3v_{esc}

Mass of the projectile
= *m*

Velocity of the
projectile far away from the Earth = *v*_{f}

Total energy of the projectile on the Earth

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth =

From the law of conservation of energy, we have

#### Page No 202:

#### Question 8.19:

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 influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 ×10^{24} kg; radius of the earth = 6.4 ×10^{6} m; G = 6.67 × 10^{–11} N m^{2} kg^{–2}.

#### Answer:

Mass of the Earth, *M*
= 6.0 × 10^{24} kg

Mass of the satellite,
*m* = 200 kg

Radius of the Earth, *R*_{e}
= 6.4 × 10^{6} m

Universal gravitational
constant, G = 6.67 × 10^{–11} Nm^{2}kg^{–2}

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

Total energy of the
satellite at height *h*

Orbital velocity of the
satellite, *v*_{ }=

Total energy of height,
*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)

#### Page No 202:

#### Question 8.20:

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 km, 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^{30 }kg

Radius of each star, *R* = 10^{4} km = 10^{7} m

Distance between the stars, *r *= 10^{9} km = 10^{12}m

For negligible speeds,* v* = 0 total energy of two stars separated at distance *r*

Now, consider the case when the stars are about to collide:

Velocity of the stars = *v*

Distance between the centers of the stars = 2*R*

Total kinetic energy of both stars

Total potential energy of both stars

Total energy of the two stars =

Using the law of conservation of energy, we can write:

#### Page No 202:

#### Question 8.21:

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:

**Answer: **

0;

–2.7 × 10^{–8 }J /kg;

Yes;

Unstable

**Explanation:**

The situation is represented in the given figure:

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: *

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 No 202:

#### Question 8.22:

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 infinity to be zero). Mass of the earth = 6.0
× 10^{24}
kg, radius = 6400 km.

#### Answer:

Mass of the Earth, *M*
= 6.0 × 10^{24} kg

Radius of the Earth, *R*
= 6400 km = 6.4 × 10^{6} m

Height of a geostationary satellite from the surface of the Earth,

*h* = 36000 km =
3.6 × 10^{7} m

Gravitational potential
energy due to Earth’s gravity at height *h,*

#### Page No 202:

#### Question 8.23:

A star 2.5 times the
mass of the sun and 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? (Mass of the sun = 2 × 10^{30}
kg).

#### Answer:

**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*_{g}

Where,

*M *= Mass of the
star = 2.5 × 2 × 10^{30} = 5 × 10^{30 }kg

*m *= Mass of the
body

*R =* Radius of
the star = 12 km = 1.2 ×10^{4} m

Centrifugal force, *f*_{c}^{
}= *mr**ω*^{2}*ω*
= Angular speed = 2π*ν*

ν = Angular
frequency = 1.2 rev s^{–1}

*f*_{c} =
*mR* (2π*ν*)^{2}

= *m* ×
(1.2 ×10^{4}) × 4 × (3.14)^{2} ×
(1.2)^{2} = 1.7 ×10^{5}*m*^{ }N

Since *f*_{g}
> *f*_{c}, the body will remain stuck to the surface
of the star.

#### Page No 202:

#### Question 8.24:

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 space ship = 1000 kg; mass of the Sun = 2 × 10^{30} kg; mass of mars = 6.4 × 10^{23} kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10^{8}kg; G= 6.67 × 10^{–11} m^{2}kg^{–2}.

#### Answer:

Mass
of the spaceship, *m*_{s
}= 1000 kg

Mass
of the Sun, *M*
= 2 × 10^{30}
kg

Mass
of Mars, *m*_{m}
= 6.4 × 10 ^{23}
kg

Orbital
radius of Mars, *R*
= 2.28 × 10^{8 }kg
=2.28 × 10^{11}m

Radius
of Mars, *r *=
3395 km = 3.395 × 10^{6}
m

Universal
gravitational constant, G = 6.67 × 10^{–11}
m^{2}kg^{–2}

Potential energy of the spaceship due to the gravitational attraction of the Sun

Potential energy of the spaceship due to the gravitational attraction of Mars

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

Total energy of the spaceship

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)

#### Page No 202:

#### Question 8.25:

A rocket is fired
‘vertically’ from the surface of mars with a speed of 2
km s–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× 1023 kg;
radius of mars = 3395 km; G = 6.67× 10^{-11} N m^{2}
kg^{–2}.

#### Answer:

Initial velocity of the
rocket, *v* = 2 km/s = 2 × 10^{3} m/s

Mass of Mars, *M*
= 6.4 × 10^{23} kg

Radius of Mars, *R*
= 3395 km = 3.395 × 10^{6} m

Universal gravitational
constant, G = 6.67× 10^{–11} N m^{2} kg^{–2}

Mass of the rocket = *m*

Initial kinetic energy of the rocket =

Initial potential energy of the rocket

Total initial energy

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

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*

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

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