# CBSE Class 11 Physics Chapter 8 Gravitation Study Materials

### Class 11 Physics Chapter 8 Gravitation

Topics and Subtopics in  Class 11 Physics Chapter 8 Gravitation:

 Section Name Topic Name 8 Gravitation 8.1 Introduction 8.2 Kepler’s laws 8.3 Universal law of gravitation 8.4 The gravitational constant 8.5 Acceleration due to gravity of the earth 8.6 Acceleration due to gravity below and above the surface of earth 8.7 Gravitational potential energy 8.8 Escape speed 8.9 Earth satellite 8.10 Energy of an orbiting satellite 8.11 Geostationary and polar satellites 8.12 Weightlessness

### Gravitation Class 11 Notes Physics Chapter 8

• Kepler’s Laws of Planetary Motion
Johannes Kepler formulated three laws which describe planetary motion. They are as follows:
(i) Law of orbits. Each planet revolves around the sun in an elliptical orbit with the sun at one of the foci of the ellipse.
(ii) Law of areas. The speed of planet varies in such a way that the radius, vector drawn from the sun to planet sweeps out equal areas in equal times.

• Newton’s Law of Gravitation
Newton’s law of gravitation states that every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The direction of the force is along the line joining the particles.

• Universal constant of gravitation G is numerically equal to the force of attraction between two particles of unit mass each separated by unit distance.
• Important Characteristics of Gravitational Force
(i) Gravitational force between two bodies is a central force i.e., it acts along the line joining the centres of the two interacting bodies.
(ii) Gravitational force between two bodies is independent of the nature of the intervening medium.
(iii) Gravitational force between two bodies does not depend upon the presence of other bodies.
(iv) It is valid for point objects and spherically symmetrical objects.
(v) Magnitude of force is extremely small.
• Principle of Superposition of Gravitation

• Acceleration Due to Gravity
The acceleration produced in a body on account of the force of gravity is known as acceleration due to gravity. It is usually denoted by ‘g’. It is always towards the centre of Earth.
If a body of mass ‘m’ lying on the surface of the earth, the gravitational force acting on the body is given by

• Mass and Mean Density of Earth
Mass and Mean density of Earth is given in the following manner.

• Variation of Acceleration Due to Gravity
The value of acceleration due to gravity changes with height (i.e., altitude), depth, shape of the earth and rotation of earth about its own axis.
(a) Effect of Altitude. As one goes above the surface of Earth, value of acceleration due to gravity gradually goes on decreasing. If gh be the value of acceleration due to gravity at a height h from the surface of Earth, then

• Gravitational Field
The space around a body within which its gravitational force of attraction is experienced by other bodies is called gravitational field.
• Intensity of Gravitational Field
The intensity of the gravitational field of a body at a point in the field is defined as the force experienced by a body of unit mass placed at that point provided the presence of unit mass does not disturb the original gravitational field.

• Gravitational Potential
The gravitational potential at a point in the gravitational field of a body is defined as the amount of work done in bringing a body of unit mass from infinity to that point.
Gravitational potential at a point situated at a distance r from a body or particle of mass M is given by

• Gravitational Potential Energy
The work done in carrying a mass ‘m’ from infinity to a point at distance r is called gravitational potential energy.
The gravitational potential energy of the system is given by

i.e., Gravitational potential energy = gravitational potential x mass of the body.
It is a scalar quantity and measured in joule.
• Escape Velocity
The minimum velocity required to project a body vertically upward from the surface of earth so that it comes out of the gravitational field of earth is called escape velocity.

• Satellite
A satellite is a body which is revolving continuously in an orbit around a comparatively much larger body.
The orbit may be either circular or elliptical. A man-made object revolving in an orbit around a planet is called an artificial satellite.
• Orbital Velocity
Orbital velocity of a satellite is the minimum velocity required to put the satellite into a given orbit around earth.

• Geostationary Satellite
The satellite having the same time period of revolution as that of the earth is called geostationary satellite. Such satellites should rotate in the equatorial plane from west to east.
The orbit of a geostationary satellite is called ‘parking orbit’. These satellites are used for communication purposes.
A geostationary satellite revolves around the earth in a circular orbit at a height of about 36,000 km from the surface of earth.
• IMPORTANT TABLES

### CBSE Class 11 Physics Chapter-8 Important Questions

1 Marks Questions

1.Why is gravitational potential energy always negative?

Ans. Gravitational potential energy is always negative because gravitational force is always attractive in nature.

2.At what height above the surface of the earth value of acceleration due to gravity is reduced to one fourth of its value on the surface of the earth?

Ans.g h = g/4 = g

R = h

2R – R = h

3.Name two factors which determine whether a planet has atmosphere or not?

Ans.(1) Acceleration due to gravity at the surface of planet

(2) Surface temperature of the planet.

4.The gravitational force between two blocks is F what would happen if a mass of both the blocks as well as distance between them is doubled?

Ans. We know F=

Here m1 = m2 (2m)

r1 = r = 2r

i.e. force will remains the same.

5.A body is weightless at the centre of earth. Why?

Ans.At the centre of the earth g = o

6.Where will a body weigh more at Delhi or at Shimla? Why?

Ans.A body will weigh more at Delhi because at higher altitudes the value of g decreases.

7.On which fundamental law of physics is keplers second law is based?

Ans.Low of conservation of angular momentum.

8.Which is greater the attraction of the earth for 1 kg of aluminum or aluminum or attraction of 1kg of aluminum for the earth?

Ans.In accordance with the universal law of gravitation both the forces are equal and opposite.

9.Distance between two bodies is increased to three times its original value. What is the effect on the gravitational force between them?

Ans.Since

r1  3r

Force will be decreased to 1/9 times

10. 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.

Ans. (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.

Very Short Answer Questions 1 Marks

1. The mass of moon is nearly 10% of the mass of the earth.  What will be the gravitational force of the earth on the moon, in comparison to the gravitational force of the moon on the earth?
Ans. Both forces will be equal in magnitude as gravitational force is a mutual force between the two bodies.
2. Why does one feel giddy while moving on a merry round?
Ans. When mowing in a merry go round, our weight appears to decrease when We move down and increases when we move up, this change in Weight makes us feel giddy.
3. Name two factors which determine whether a planet would have atmosphere or not.
Ans. (i) Value of acceleration due to gravity (ii) surface temperature of planet.
4. The force of gravity due to earth on a body is proportional to its mass, then why does a heavy body not fall faster than a lighter body?
Ans.  but  and does not depend on ‘m’ hence they bodies fall with same ‘g’.
5. The force of attraction due to a hollow spherical shell of uniform density on a point mass situated inside is zero, so can a body be shielded from gravitational influence?
Ans. No, the gravitational force is independent of intervening medium.
6. The gravitational force between two bodies in 1 N if the distance between them is doubled, What will be the force between them?
Ans. F = 1
7. A body of mass 5 kg is taken to the centre of the earth. What will be its (i) mass (ii) weight there.
Ans. Mass does not change.
8. Why is gravitational potential energy negative?
Ans. Because it arises due to attractive force of gravitation.
9. A satellite revolves close to the surface of a planet. How is its orbital Velocity related with escape velocity of that planet.
Ans.   When r = R
10. Does the escape Velocity of a body from the earth depend on (i) mass of the body (ii) direction of projection.
Ans. No,
11. Identify the position of sun in the following diagram if the linear speed of the planet is greater at C than at D.

Ans. Sun should be at B as speed of planet is greater when it is closer to sun.
12. A satellite does not require any fuel to orbit the earth. Why?
Ans. The gravitational force between satellite and earth provides the necessary centripetal force for the satellite to orbit the earth.
13. A satellite of small mass burns during its desent and not during ascent. Why?
Ans. The speed of satellite during descent is much larger than during ascent, and so heat produced is large.
14. ls it possible to place an artificial satellite in an orbit so that it is always Visible Over New Delhi.
Ans. No, A satellite will be always visible only if it revolves in the equatorial plane, but New Delhi does not lie in the region of equitorial plane.
15. If the density of a planet is doubled without any change in its radius, how does ‘g’ change on the planet.
Ans. ‘g’ gets doubled as g  (density)
16. Mark the direction of gravitational intensity at (I) centre of a hemispherical shell of uniform mass density (ii) any arbitrary point on the upper surface of hemisphere.
Ans.

In both cases it will be downward
17. Why an astronaut in an orbiting space craft is not in zero gravity although weight less?
Ans. The astronaut is in the gravitational field of the earth and experiences gravity. However, the gravity is used in providing necessary centripetal force, so is in a state of free fall towards the earth.
18. Write one important use of (i) geostationary satellite (ii) polar satellite.
Ans. Geostationary satellite are used for tele communication and polar satellite for remote rensing.
19. A binary star system consists of two stars A and B which have time periods TA and TB, radius RA and RB and masses ma and me which of the three quantities are same for the stars. Justify.
Ans. Angular velocity of binary stars are same is wA = wB,
20. The time period of the satellite of the earth is 5 hr. If the separation between earth and satellite is increased to 4 times the previous value, then what will be the new time period of satellite.
Ans.
21. The distance of Pluto from the Sun is 40 times the distance of earth if the masses of earth and Pluto he equal, what will be ratio of gravitational forces of sun on these planets.
Ans.
22. If suddenly the gravitational force of attraction between earth and satellite become zero, what would happen to the satellite?
Ans. The satellite will move tangentially to the original orbit with a velocity with which it was revolving.

2 Marks Questions

1.What is kepler’s law of periods? Show it mathematically?

Ans.It states that the square of the period of revolution of a planet around the sun is proportional of a planet to the cube of the semi-major axis of the elliptical orbit.

i.e. T R3

T2 = KR3

where T is time period of evolution

R is the length of semi major axis

K is constant for all planets

2.With two characteristics of gravitational force?

Ans.(1) It is a central force

(2) It is a conservation force

(3) It obeys inverse square law.

(4) It is a universal force and is always attractive in nature.

3.Assuming earth to be a uniform sphere finds an expression for density of earth in terms of g  nd G?

Ans.Since g =

If earth is uniform sphere of mean density P

g =

g =

P =

4.If radius of earth is 6400km, what will be the weight of 1 quintal body if taken to the height of 1600 km above the sea level?

Ans.R = 6400km = 6400 x 103m

h = 1600km

w = mg = 1 quintal = 100 kg = 100×9.8 N

weight (w) = mgh

w = mg

w = 100×9.8

w = 64×9.8N = 64kg

5.The distance of the planet Jupiter from the sun is 5.2 times that of the earth. Find the period of the Jupiter’s revolution around the sun?

Ans. Te = 1 year    RJ = 5.2 Re

TJ  =(5.2) 3/2 x 1year

TJ  =  11.86 year

6.Show that for a two particle system

Ans.  = (1)

(2)

& (1) and (2) cam be written as

Since

Hence proved.

7.State two essential requisites of geostationary satellite?

Ans.(1)The period of revolution of a satellite around the earth should be same as that of earth about its own axis (T=24hrs)

(2)The sense of rotation of satellite should be same as that of the earth about its own axis i.e. from west to east in anti-clockwise direction

8.Show that an artificial satellite circling round the earth in an orbit of radius obeys keeper’s third low?

Ans.Orbital velocity of a satellite is

=

Where M is the crass of earth

Time period of satellite T =

T =

T =

T =

Thus

Hence proved.

9.A 400kg satellite in a circular orbit of radius 2 Re about the earth calculate the kinetic energy potential energy and total energy of the satellite?

RE = 6.4×106m

M = 6×1024kg

Ans.M = 6×1024 kg M = 400 kg

RE = 6.4×106m

Hence r = 2RE = 12.8×106m

G = 6.67×10-4 N m2/kg2

KE =

P. E. =

PE = -2 x 6.25 x 10 = -12.5× 109 Joules

T. E. = K. E + P. E

T. E. = 6.25 ×109 – 12.50 × 109

T. E. = 6.25 × 109 Joules

(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?

Ans. (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.

11. 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?

Ans.

(a) No

(b) No

(c) No

(d) Yes

Escape velocity of a body from the Earth is given by the relation:

………………(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.

12. 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.

Ans.

(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.

13. 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 =  kg, radius = 6400 km.

Ans. Mass of the Earth, M =kg

Radius of the Earth, R = 6400 km =  m

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

h = 36000 km =  m

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

=

SHORT ANSWER TYPE QUESTIONS (2 MARKS)

1. If the radius of the earth Were to decrease by 1%, keeping its mass same, how will the acceleration due to gravity change?
Ans.  If R decrease by 1% it becomes

g’ increases by 0.02  therefore increases by 2%.
2. If ‘g’ be the acceleration due to gravity on earth’s surface, Calculate the gain in potential energy of an object of mass m raised from the surface of earth to a height equal to the radius or earth in term of ‘g’.
Ans. Gain in
3. A satellite is moving round the earth with velocity v what should be the minimum percentage increase in its velocity so that the satellite escapes.
Ans. The maximum orbital velocity of a satellite orbiting near its surface is

For the satellite to escape gravitational pull the velocity must become v.
But  = 1.414 vo = (1 + 0.414) vo
This means that it has to increases 0.414 in 1 or 41.4%
∴ The minimum increase required, as the velocity of satellite is maximum When it is near the earth.
4. Two planets of radii r1, and r2 are made from the same material. Calculate the ratio of the acceleration due to gravity on the surface of the planets.
Ans.

5. If earth has a mass 9 times and radius 4 times than that of a planet “P”. Calculate the escape velocity at the planet ‘P’ if its value on earth is 11.2 kms -1.
Ans.

= 7.47 km/sec
6. At what height from the surface of the earth will the value of ‘g’ be reduced by 36% of its value at the surface of earth.
Ans.

7. At what depth is the value of ‘g’ same as at a height of 40 km from the surface of earth.
Ans. gd = gh
8. The mean orbital radius of the earth around the sun is 1.5 x 108 km. Calculate mass of the sun if G = 6.67 x 1011 Nm2/kg-2?
Ans. R = 1.5 × 108 Km = 1.5 × 1011 m
T = 365 days = 365 × 24 × 3600 s
Centripetal force = gravitational force

Ms = 2.01 × 1030 kg
9. Draw graphs showing the variation of acceleration due to gravity with
(i) height (ii) depth from the surface of earth.
Ans.

above surface of earth; r = R + h
below surface of earth; r = R – d
G is max for r = R on surface of earth.
10. Which planet of the solar system has the greatest gravitational field strength? What is the gravitational field strength of a planet where the weight of a 60 kg astronaut is 300 N.
Ans. Jupiter has maximum gravitational field strength gravitational field strength

= 5 N kg -1
11. Two satellites are at different heights from the surface of earth which would have greater velocity. Compare the speeds of two satellites of masses m and 4m and radii. 2R and R respectively.
Ans.
where M is mass of the planet,
v0 is independent of mass of the satellite.
12. What is (i) inertial mass, (ii) gravitational mass. Are the two different?
Ans. Inertial mass is the measure of inertia of the body
Gravitational mass of a body determine the gravitational pull between earth and the body.

Both inertial mass and gravitational mass are not different but are equivalent.
13. Why the space rockets are generally launched West to East?
Ans. Since the earth revolves from West to east, so when the rocket is launched from west to east the relative velocity of the rocket increases which helps it to rise without much consumption of fuel.
14. Explain why a tennis ball bounces higher on hills than in plane?
Ans. The value of ‘g’ on hills is less than at the plane, so the weight of tennis ball on the hills is lesser force than at planes that is why the earth attract the ball on hills with lesser force than at planes. Hence the ball bounces Higher.
15. The gravitational force on the earth due to the sun is greater than moon. However tidal effect due to the moon’s pull is greater than the tidal effect due to sun. Why?
Ans. The tidal effect depends inversely on the cube of the distance, while gravitational force depends on the square of the distance.
16. The mass of moon is (where M is mass of earth). Find the distance of the point where the gravitational field due to earth and moon cancel each other. Given distance of moon from earth is 60 R, where R is radius of earth.
Ans.

Gravitational field at C due to earth
= Gravitational field at C due to earth moon

81x2 = (60R – x)2
9x = 60R – x
X = 6 R
17. The figure shows elliptical orbit of a planet m about the sun S. The shaded area of SCD is twice the shaded area SAB. If t1 is the time for the planet to move from D to C and t2, is time to move from A to B, what is the relation between t1 and t2?

Ans. According to Kepler’s IInd law areal velocity for the planet is constant

18. Calculate the energy required to move a body of mass m from an orbit of radius 2R to 3R.
Ans. Gravitational PE of mass m in orbit of radius R

19. A man can jump 1.5 m high on earth. Calculate the height he may be able to jump on a planet whose density is one fourth that of the earth and Whose radius is one third of the earth.
Ans.

The gain in PE at the highest point will be same in both cases. Hence
mg’h’ = mgh

= 18 m

3 Marks Questions

1.A satellite is revolving is a circular path close to a planet of density P. find an expression for its period of revolution?

Ans.If satellite revolvers around the earth of radius r

T =

where  is orbital velocity

where

If a satellite is revolving near the plant’s surface then r = R radius of planet and

T =

2.How far away from the surface of earth does the value of g is reduced to 4% of its value on the surface of the earth Given radius of earth = 6400km

Ans.g h = g

h = 4% of g =

R = 6400km

2R+2h = 10R

2h = 8R

h = 4R = 4×6400 = 25,600km.

3.Obtain on expression showing variation of acceleration due to gravity with height?

Ans. Acceleration due to gravity at the surface of the earth

g =  (1)

If g h is the acceleration due to gravity at a pt situated at a height ‘h’ above the surface of the earth

g h =   (2)

Divide (2) by (1)

g h = g

If h <<< R then the above relation

g h = g

g h = g

Expanding Binomially and neglecting higher power

g h = g

4.Two uniform solid spheres of radii R and 2R are at rest with their surfaces just touching. Find the force of gravitational attraction between them if density of spheres be P?

Ans. Two spheres of density p and radii R and 2R

s = oo1 = 2R+R=3R

F =

F =

F =

5.Find expressions for (1) potential energy (2) kinetic energy (3) total energy for an artificial satellite.

Ans.Potential energy of a satellite

U=

U = GMm

U = GMm

U =GMm

U =

Kinetic energy KE =

But

K. E =

KE =

Total energy of satellite E =

E = –

E = –

6.  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?

Ans. Lesser by a factor of 0.63

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

= 1 year

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

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

Orbital radius of the planet =

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.

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

Ans. (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.

8. How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is  km.

Ans. Orbital radius of the Earth around the Sun, r = m

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

T = 1 year = 365.25 days

Universal gravitational constant, G =Thus, mass of the Sun can be calculated using the relation,

Hence, the mass of the Sun is kg.

9. A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is km away from the sun?

Ans. Distance of the Earth from the Sun, re =

Time period of the Earth =

Time period of Saturn,

Distance of Saturn from the Sun =

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.

10. 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?

Ans. 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

=

=

11.  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 =  kg).

Ans. 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,

Where,

M = Mass of the star =

m = Mass of the body

R = Radius of the star = 12 km = 1.2 ×104 m

Centrifugalforce,
= Angular speed = 2πv

v = Angular frequency = 1.2 rev

Since, the body will remain stuck to the surface of the star.

SHORT ANSWER TYPE QUESTIONS (3 MARKS)

1. Define gravitational potential at a point in the gravitational field. Obtain a relation for it. What is the position at which it is (i) maximum (ii) minimum.
2. Find the potential energy of a system of four particles, each of mass m, placed at the vertices of a square of side. Also obtain the potential at the centre of the square.
3. Three mass points each of mass m are placed at the vertices of an equilateral triangle of side. What is the gravitational field and potential at the Centroid of the triangle due to the three masses.
Ans.

From ∆ ODB Cos 30°

Gravitational field at O due to m at A, B and C is say

along OD
Is equal and opposite to
∴ net gravitational field = zero
As gravitational potential is scalar
V = V+ V2 + V3

4. Briefly explain the principle of launching an artificial satellite. Explain the use of multistage rockets in launching a satellite.
5.  In a two stage launch of a satellite, the first stage brings the satellite to a height of 150 km and the 2 stage gives it the necessary critical speed to put it in a circular orbit. Which stage requires more expenditure of fuel? Given mass of earth = 6.0 x 1024 kg, radius of earth = 6400 km
Ans. Work done on satellite in first stage = W1 = PE at 150 km – PE at the surface

Work done on satellite in 2nd stage = W2 = energy required to give orbital Velocity vo

∴ W2 > W1 so second stage requires more energy
6. The escape velocity of a projectile on earth’s surface is 11.2 kms-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.
Ans. ve = 11.2 kms-1, velocity of projection = w = 3ve Let m be the mass of projectile and v the velocity after it escapes gravitational pull.
By law of conservation of energy

= 31.68 kms-1
7. A satellite orbits the earth at a height ‘R’ from the surface. How much energy must be expended to rocket the satellite out of earth’s gravitational influence?
Ans. The energy required to pull the satellite from earth influence should be equal to the total energy with which it is revolving around the earth.
The K.E of satellite
The P.E of satellite
∴ Energy required will be
8. Deduce the law of gravitation from Kepler’s laws of planetary motion.
Ans. Suppose a planet of mass m, moves around the Sun in a circular Orbit of radius ‘r’ with velocity v.
Then centripetal force
But

According to kepler’s IIIrd law

The force between planet and sun must be mutual, so must be proportional to mass of Sun.

This is Newton’s law of gravitation
9. Mention at least three conditions under which weight of a person can become Zero.
Ans.
(i) When the person is at centre of earth.
(ii) When the person is at the null points in space (at these points the gravitational forces due to different masses cancel each other)
(iii) when a person is standing in a freely falling lift.
(iv) When a person is inside a space craft which is orbiting around the earth.

4 Marks Questions

1.  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 – is more/less accurate than the formula for the difference of potential energy between two points and distance away from the centre of the earth.

Ans.

(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,

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 r2 and r1 distance away from the centre of the Earth is respectively given by:

Difference in potential energy,

Hence, this formula is more accurate than the formula .

2. Let us assume that our galaxy consists of  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  ly.

Ans. Mass of our galaxy Milky Way, M =  solar mass

Solar mass = Mass of Sun =  kg

Mass of our galaxy,

Diameter of Milky Way, d =  ly

Radius of Milky Way, r =  ly

1 ly =  m

r =

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

3. 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?

Ans. 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

Re = Radius of the Earth

For,

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

=28 N

4. The escape speed of a projectile on the earth’s surface is 11.2 km. 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.

Ans. Escape velocity of a projectile from the Earth, = 11.2 km/s

Projection velocity of the projectile,

Mass of the projectile = m

Velocity of the projectile far away from the

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

5.  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.

Ans. (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.

6. 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.

Ans.(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.

7. 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 = kg; radius of the earth = m; G =.

Ans. Mass of the Earth, M = kg

Mass of the satellite, m = 200 kg

Radius of the Earth, Re = m

Universal gravitational constant, G =

Height of the satellite, h = 400 km =

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)

5 Marks Questions

1.Define Gravitational potential energy Hence deduces an expression for gravitational potential energy of a body placed at a point sear the surface of earth?

Ans.It is defined as the work done in bringing a body from infinity to that point.

A body of mass (m) lying at a distance x from earth of mass (m)

F =

If the body is displaced through a distance dx then

dw = Fdx =

Total work done

W =

w = GMm

w = GMm

w = -GMm

w =

This work done is equal to the gravitational potential energy

i.e.

2.  Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is. Show that the mass of Jupiter is about one-thousandth that of the sun.

Ans. Orbital period of

Satelliteis revolving around the Jupiter

Mass of the latter is given by the relation:

…………..(i)

Where,

= Mass of Jupiter

G = Universal gravitational constant

Orbital period of the Earth,

Mass of sun is given as:

…………(ii)

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

3.  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 =  kg, mass of the earth = kg. Neglect the effect of other planets etc. (orbital radius =  m).

Ans. Mass of the Sun, kg

Mass of the Earth,  kg

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:

4.  A rocket is fired vertically with a speed of 5 km from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth =  kg; mean radius of the earth = m; G=.

Ans.  m from the centre of the Earth

Velocity of the rocket, v = 5 km/s =  m/s

Mass 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, V=0

And, Potential energy =

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

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

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

5. Two stars each of one solar mass (=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).

Ans. Mass of each star, M = kg

Radius of each star, R =

Distance between the stars, r =

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

…….(i)

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

Total potential energy of both stars

Total energy of the two stars =  ……..(ii)

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

6.  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?

Ans. 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 midpoint 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.

7. 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 = kg; mass of mars =  kg; radius of mars = 3395 km; radius of the orbit of mars =;.

Ans. Mass of the spaceship, ms = 1000 kg

Mass of the Sun, M = kg

Mass of Mars, mm = kg

Orbital radius of Mars, R = kg =m

Radius of Mars, r = 3395 km =

Universal gravitational constant,

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

Energy required for launching the spaceship out of the solar system

= – (Total energy of the spaceship)

8.  A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km. 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 =  kg; radius of mars = 3395 km; .

Ans. Initial velocity of the rocket, v = 2 km/s = m/s

Mass of Mars, M = kg

Radius of Mars, R = 3395 km =  m

Universal gravitational constant,

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:

NUMERICALS

1. The mass of planet Jupiter is 1.9 x 1027 kg and that of the sun is 1.99 x 1030kg. The mean distance of Jupiter from the Sun is 7.8 x 1011m. Calculate gravitational force which sun exerts on Jupiter, and the speed of Jupiter.
Ans.

v = 1.3 × 104 ms -1
2. A mass ‘M’ is broken into two parts of masses m1 and m2. How are m2 and m related so that force of gravitational attraction between the two parts is maximum.
Ans. Let m = m then m = M – m
Force between them when they are separated by distance “r

For F to be maximum, differentiate F W.r.t m and equate to zero

3. If the radius of earth shrinks by 2%, mass remaining constant. How would the value of acceleration due to gravity change?
Ans. increases by 4%
4. A body hanging from a spring stretches it by 1 cm at the earth’s surface. How much will the same body stretch at a place 1600 klm above the earth’s surface? Radius of earth 6400 km.
Ans. In equilibrium mg = kg,
At height h mg’ = kx’,

5. Imagine a tunnel dug along a diameter of the earth. Show that a particle dropped from one end of the tunnel executes simple harmonic motion. What is the time period of this motion?
Ans.

The acceleration due to gravity at a depth below the earth’s surface is given by

where y is distance from centre of earth

As acceleration is proportional to displacement and is directed towards mean position, the motion would be S.H.M
T = Time period =

6. The gravitational field intensity at a point 10,000 km from the Centre of the earth is 4.8 N kg-1. Calculate gravitational potential at that point.
Ans. Gravitational intensity
Gravitational potential

Or, V = -E × R
Or V = -4.8 × 10,000 × 103 = 4.8 × 107 J kg-1
7. A geostationary satellite orbits the earth at a height of nearly 36000 km. What is the potential due to earth’s gravity at the site of this satellite (take the potential energy at a to be zero). Mass of earth is 6 x 1024 kg, radius of earth is 6400 km.
Ans. U = potential at height
8. How much faster than the present speed should the earth rotate so that bodies lying on the equator may fly off into space.

Ans. The speed of earth
at present

The gravitational force should be equal to the centripetal force so that centrifugal force, given by

The velocity should become 17 times the present velocity

9. The distance of Neptune and Saturn from the sun is nearly 1013m and 1012m respectively. Assuming that they move in circular orbits, then what will be the ratio of their periods.
Ans. By kepler’s IIIrd law

10. Let the speed of the planet at perihelion P in fig be vp and Sun planet distance SP be rp Relate (rp, vp) to the corresponding quantities at the aphelion (rA, vA). Will the planet take equal times to traverse BAC and CPB?

Ans. The magnitude of angular momentum at P is Lp = mp rp vp
Similarly magnitude of angular momentum at A is LA = mA rA vA
From conservation of angular momentum

area bounded by SB & SC (SBAC > SBPC)
∵ By 2nd law equal areas are swept in equal intervals of time.Time taken to transverse BAC > time taken to traverse CPB
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