Unit VI: Gravitation

Gravitation: Kepler’s laws of planetary motion, universal law of gravitation.

Acceleration due to gravity and its variation with altitude and depth.

Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite, Geo-stationary satellites.

Chapter–8: Gravitation

1. INTRODUCTION

Have you ever wondered whether we would still be studying about with Gravitation if a stone had fallen on Newton’s head instead of an apple? Anyways, the real question is why does an apple fall down rather than go upward?

1 Gravitation:

The force of attraction between any two bodies in the universe is called gravitation or gravitational force. e.g.:- The attraction between sun and earth or the attraction between a book and table is due to gravitational force.

2 Newton’s Law of Gravitation: ^imp

According to this law, the force of attraction between any two bodies is directly proportional to product of masses and inversely proportional to square of distance between them.

Mathematically

And

Combining eq. (i) and (ii) we get

or                                

Where  is constant of proportionality called universal gravitational constant.

Its value is

And its dimensional formula is

Definition of G:

If                       and

then

Hence universal gravitational constant may be defined as the force of attraction between two unit masses separated by one meter distance apart.

1. Why G is called universal gravitational constant?

Ans.    Because the value of G does not depends upon nature, size, medium of the body and applicable at every place so it called universal constant.

• The Universal gravitational constant G is an experimental value calculated by Cavendish 71 years after the law was formulated.

QUESTION 1:    Two particles of masses 1.0 kg and 2.0 kg are placed at a separation of 50 cm. assuming that the  only forces acting on the particles are their mutual gravitation; find the initial accelerations of the two particles.

Sol: The force of gravitation exerted by one particle on the other is

The acceleration of 1 kg particle is  toward 2 kg particle.

Similarly The acceleration of 2 kg particle is  toward 1kg particle.

Question2:       A mass M is split into two parts m and (M – m), which are then separated by a certain distance. What ratio (M/m) maximizes the gravitational force between the parts?

Solution:            If ‘r’ is the distance between m and (M – m) , the gravitational force will be

F =  for F to be maximum, ,

i.e.,         or, M – 2m = 0      [ ]    or, M/m = 2 ,

i.e., the force will be maximum when two parts are equal.

• Evidences In support of Laws of Gravitation:

1. The rotation of earth around the sun and the rotation of moon around the earth can be explained by this law.
2. The formation of tides in oceans is due to gravitational force between moon and earth.
3. The limes of solar and lunar eclipses can be calculated by this law.
4. The orbits and periods of revolutions of artificial satellite can be predicted accurately by this law.
5. The value of varies from place on the surface of earth in accordance with this law.

4 Vector Form of the Law of Gravitation:

Suppose two bodies of mass  and  separated by  distance apart.

Again suppose that  is the force on body 1 due to 2 and  is

(From Newton’s Law of gravitation)

Where  is a unit vector from body 2 to body 1.

Similarly force on body 2 due to 1 may be given as

Where  is a unit vector acting from body 1 to 2.

Also we know that

Using in eq. (2) we get

Comparing eq. (i) and (iii) we get

This is Newton’s third law of motion. Hence Newton’s law of gravitation obeys Newton’s third law of motion.

   Some Important Features of Gravitational Force:

1. It is independent of inter mediums.
2. It obeys Newton’s third law of motion.
3. It is strictly holed by point masses.
4. It is a central force acts at centers.
5. It is a conservative force.
6. This force is independent of the presence of the other bodies. Hence we can sys it follow principle of superposition of forces.

5. Principle of Superposition of Gravitational force:

According to principle of superposition of gravitational force, the total force acting on a body is equal to vector sum of all the forces action on the body.

Suppose a body of mass . Again suppose that the bodies  are placed at  respectively.Then from Newton’s law of Gravitation, the force of attraction between  and  is

Again force of attraction between  and

Similarly

So the total force acting on the body  is

Or

Or

Or

Question 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 = 2  1030 kg, mass of the earth = 6  1024kg. Neglect the effect of other planets etc. (orbital radius = 1.5  1011 m).
Answer: Mass of Sun, M = 2  10³⁰ kg; Mass of Earth, m = 6  10²⁴ kg Distance between Sun and Earth, r = 1.5  10¹¹ m Let at the point P, the gravitational force on the rocket due to Earth= gravitational force due to sun.

Again suppose  is the distance from earth as
Question. 4. How will you ‘weigh the sun’, that is, estimate its mass? The mean orbital radius of the earth around the sun is 1.5  108 km                                    
Answer: The mean orbital radius of the Earth around the Sun  R
Time period, T =  s Let the mass of the Sun be M and that of Earth be m
as  ….(1) and  Centripetal force

From (1) and (2)

6 Gravity:

It is defined as the force between earth and a object lying on or near the surface of earth. A body falls on surface of earth under the effect of gravity.

Acceleration due to Gravity:

The acceleration produced in a falling body due to gravity of earth is called acceleration due to gravity. It is a vector quality and denoted by .

The value of  does not depend upon mass, size and shape of the body but varies from place to place on the surface of earth due to shape of earth and position of the body.

Expression for acceleration due to gravity or Relation between  and :

Suppose earth is a sphere of mass  and radius  and a body of mass  is placed on the surface of earth, then according to Newton’s law of gravitation, the force of attraction between earth and body is

And the force experienced by the body due to its weight is

Comparing eq. (i) and (ii) we have

or

Thus value of g does not depend upon the mass of body but depends upon the mass of the planet and radius of the planet.

Question5:      Calculate the acceleration due to gravity (in cm/s²) at the surface of Mars if its diameter is 6760 km and mass one tenth that of the earth. (Given that the diameter of earth is 12742 km and acceleration due to gravity on its surface is 9.8 m/s²)

Solution:         We know that

\                      or,

\          g_M = 0.35 ´ g_E = 9.8 ´ 0.35 = 3.48 m/s² = 348 cm/s²

7 Mass and density of the earth:

Mass of Earth:   

Cavendish was the first which successfully determined  and .

As we know                                    ⟹

Where

So

Or

Average density of Earth:

If earth is supposed to be sphere of s  and    then average density of earth may be given as

As

Or                                                              also

⟹

Or

Where  &

⟹                          Or

Or

Density of earth on upper layer is smaller  while density at inner layers is larger then  

Question6:      The mass of moon is given as x ´ 10²⁰ kg and its density is y ´ 10² kg/m³. If acceleration due to gravity on its surface is 1.62 m/s² and its radius is 1.74 ´ 10⁶ m   (G = 6.67 ´ 10^–11 MKS units). Then find x and y.

Solution:         We know that

\             Or, M = 7.35 ´ 10²² kg  Þ x = 735

Now,

\          = Þ  y = 33

8 Variation of acceleration due to gravity:

The value of  varies due to (i) shape of earth (ii) at altitude (iii) at depth.

• Variation of g due to shape of Earth:

As we know earth is not a perfect sphere, it is elliptical shape. It has smaller radius at pole (R_p) and larger radius at equators (R_e).

Now suppose that a body of mass  is placed at the pole of earth, and then gravity at pole is

Now again if the body is placed at equator then gravity experienced by the body is

Now dividing eq. (i) by (ii) we get

Or

As  so

• Hence we can say that acceleration due to gravity at pole is larger than the acceleration due to gravity at equator.

Question 7. 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. Will an object placed on its equator remain stuck to its surface due to gravity?

(Mass of the sun = 2  1030 kg).
Answer: as the acceleration due to the gravity of the star   ……… (i)

So inward force due to gravity = N.

The centrifugal force acting on a body of mass m at the equator of the star

N.——-(ii)
Since the inward force due to gravity on a body at the equator of the star is about 2.2 million times more than the outward centrifugal force, the body will remain stuck to the surface of the star.

• Variation of with Altitude: (Height)

Suppose earth is a sphere of  and . Again suppose that a body of mass  is placed on the surface of earth. Then according to Newton’s law of gravitation, the force between earth and the body is

And

Again if body is moved to height  above the surface of earth then acceleration due to gravity becomes

Dividing eq. (ii) by (i) we get

Or

Or

Or

Appling binomial theorem we get

Or

Clearly

Hence acceleration due to gravity decreases when we move above the surface of earth.

• Here in numerical eq. (iii) is used if and eq. (iv) is used if .

Question 8. A body weighs 63 N on the surface of the Earth. What is the gravitational force at a height equal to half the radius of the Earth?
Answer:  Let g_h be the acceleration due to gravity at a height (h = R/2) and g its value on Earth’s surface.

We know that

Suppose that  is weight on earth and is weight on h height. So

• Variation of with Depth:

Suppose that earth is sphere of mass  and radius . Again suppose that a body of mass  is placed on the surface of earth. Then acceleration due to gravity on the body is

Also

So

or

Now again if the body is moved at a depth d in the earth then acceleration due to gravity becomes.

Where

So

Now dividing eq. (ii) by (i) we get

Or

Or

• Clearly

         Hence acceleration due to gravity decreases when we go inside the earth.

• At the centre of earth so

Hence acceleration due to gravity at the centre of earth becomes zero. So the weight of the body  also becomes zero.

9 Relation between height  and depth  for the same change in :

As acceleration due to gravity at a height  is

And acceleration due to gravity at depth  is

If

Then

Or

Or

Hence acceleration due to gravity at a height  is same to as that of depth  providing .

Question 9.  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: as  so

10 Gravitational Field:

The space around a body within which its gravitational force of attraction can be experienced is called gravitation field.

e.g.:- The earth has a gravitation al field in this field earth attract the bodies.

11 Intensity of Gravitational Field or Gravitational Field intensity:

The force experience by a unit mass in the field of a other massive body is called gravitational field intensity. It is a vector quantity and denoted by  or .

If F is the amount of force experienced by unit mass then gravitational field intensity

But

So

Or

Also we know that

Hence gravitational field intensity is equal to amount of acceleration produced in the unit mass due to the field intensity.

• The gravitational field intensity becomes zero when object is placed at .
• Unit and dimensional formula of gravitation field intensity are same as that of acceleration.

Example 10: The distance between earth and moon is 4×10⁵ km and the mass of earth is 81 times the mass of moon. Find the position (take 10⁴ km as unit) of a point on the line joining the centers of earth and moon, where the gravitational field is zero. 

Solution: Let x be the distance of the point of no net field from earth. The distance of this point from moon is
(r – x), where km.

The gravitational field due to earth  and that due to moon

For the net field to be zero these are equal and opposite.

But \    so = 9

9r – 9x = x

10x = 9 r

= 3.6 × 10⁵ km = 36 ´ 10⁴ km   Þ  x = 36 unit

12 Gravitational Potential Energy:

It is defined as the amount of work done to displace a body from ∞ to a required point in gravitational field of earth.

Expression for Gravitational Potential Energy:

Suppose a body of mass  is placed at a distance  from the centre of earth of mass .

Then force experienced by the body is

Now we want to move the body small distance  to Q i.e. . Then amount of work done to move the body is

Or

So the total amount of work done to displace body from ∞ to  is

Or

Or

Or

Or

This work is store in the body, in form of potential energy.

So potential energy is

Here  sign indicate that gravitational potential energy is due to gravitational force of attraction between earth and the body. The maximum value of gravitational potential energy is zero at ∞.When the body is brought to near earth then U decreases.

13 Gravitational Potential:

Gravitational potential is defined as the amount of work done to displace a body of unit mass from ∞ to a point in gravitational field.

Suppose a unit mass body is placed at a distance  from the centre of the earth of mass .

Then force experienced by the body is

Now if we want to move the body small distance  from  to . Then the amount of work done to move body is

Or

The total amount of work done to move body from to  is

or

Or

Or

This work is stored in the body in form of gravitational potential, so gravitational potential is

Here  sign indicate that gravitational potential of the body is due to gravitational force of attraction between earth and the body.

⟹Dimensional formula of  is  and unit is

 at ∞ is  and a scalar quantity.

⟹The dimension formula of  is same to as that of square of velocity.

14 Relation between Gravitational field intensity and Gravitational Potential:

As we know gravitational field is the amount of force experienced by a unit mass i.e. . The direction of this force is toward . Now the amount of work done to move unit mass is

Here E and  are in opposite direction so

⟹                  But

So

Or

So we can say that gravitation field intensity is equal to the  gradient of gravitational potential.

Question 11. 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  1024 kg; mean radius of the earth = 6.4  106 m; G = 6.67  10-11 N m2 kg-2.
Answer:  Initial kinetic energy of rocket =
At distance r from centre of earth, kinetic energy becomes zero.

Change in kinetic energy = 1.25  10⁷ – 0 = 1.25  10⁷ m J. This energy changes into potential energy.
Initial potential energy at the surface of earth =

Final potential energy

So change in potential energy –  .

Using law of conservation of energy –

Question 12 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:  (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.

Question 13. 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   1024 kg; radius of the earth = 6.4   106 m;                         G = 6.67   10-11 N m2 kg-2.
Answer:

Total energy of orbiting satellite at a height .  =

Energy expended to rocket the satellite out of the earth’s gravitation field= – total energy of the orbiting satellite

Question 14. Two stars each of one solar mass (=2   1030 kg) are approaching each other for a head on collision. When they are at 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.

 Answer: Here, mass of each star, M = 2   10³⁰ kg Initial distance between two stars, r =  10⁹ km =  10¹² m.
Initial potential energy of the system =
Total K.E. of the stars where  is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, r’ = 2 R.
:. Final potential energy of two stars =
since gain in K.E. is at the cost of loss in P.E   so

So        or

Question 15. 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 field and potential at the midpoint 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: Here G = 6.67  10^-11 Nm² kg^-2; M = 100 kg; R = 0.1 m, distance between the two spheres, d = 1.0 m
suppose that the distance of either sphere from the mid-point of the line joining their centre is r. Then  The gravitational field at the mid-point due to two spheres will be equal and opposite. Hence resultant gravitational field at the midpoint is zero. And gravitational potential at the midpoint =

The object placed at this point would be in stable equilibrium.

15 Escape speed or Escape Velocity:

In over daily life when we through a object in upward direction then it return back to the earth, but at a certain speed the object does not came on the earth. This speed is called Escape speed and may be defined as the minimum amount of speed required to eject a body from the surface of earth to ∞. So that body never returns to earth. Or

The minimum amount of speed required to escape a body from the field of earth is called Escape speed or Escape Velocity.

Expression for Escape Speed:

Suppose that a body of mass  is at a distance  from the centre of earth of mass . Then force experienced by the body is

Now small amount of work is to be done to displace the body through small distance  as

or

Now total amount of work done to displace the body from  to ∞ is

Or

This work will be provide the maximum K.E. to the body as

Or

Or

But

So

So

⟹

Which is the required expression for escape speed

As        and

⟹

⟹

Hence if a body is thrown with a speed of 11.2 Km/sec. will escape from the gravitation field of the earth.

• The escape speed of a body from the Earth does not depend on the mass of the body.
• The escape speed does not depend on the location from where a body is projected.
• The escape speed does not depend on the direction of projection of a body
• The escape speed of a body depends upon the height of the location from where the body is projected, because the escape velocity depends upon the gravitational potential at the point from which it is projected and this potential depends upon height also.

Value of Escape speed from Moon:

In case of moon

⟹

Question16 :  What will be the acceleration due to gravity on the surface of the moon if its radius were th the radius  of

the earth and its mass  the mass of the earth.        Given g = 10 m/s².

Solution: As on the surface of planet,      we have,    \  = 2 m/s²

Question 17. 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: as    . if

Than

16 Natural and Artificial Satellite:

Satellite:

A satellite is a body which revolves around a much larger body in stable orbits. Satellite may be of two types.

• Natural Satellite:
• A satellite which is created by nature (God) is called natural satellite. e.g.:- moon is a natural satellite of earth and earth also a satellite of sun. The planet Jupiter and Saturn have four fourteen and twelve satellites respectively.
• Artificial Satellite:
• A man made satellite is called an artificial satellite. Russians launched the first man made satellite SPUTNIK-I in 4Oct.1957. India lunched its first satellite in 19 April 1975 named Aryabhatta from Russian soil.

Later India Launched satellites like INSAT-1A, INSAT-IB,INSAT-2B,IRS-IC and INSAT-2D etc. other satellite launched by India are Bhaskara, Rohini-I, Apple, Bhaskara-II.

17 Launching of a Satellite:

Suppose a body is placed at a long tower on the surface of earth. Now if body is projected with a velocity then it fall at point A, if velocity is increased then body fall at B. But velocity may be increased up to a certain limit that is does not fall on the earth this velocity is called orbital velocity and the body becomes satellite of earth. Thus to launch a satellite we need to give two velocities to the body as

1. A vertical velocity (escape velocity) to take the body at a suitable height.
2. After suitable vertical velocity, a suitable horizontal velocity is given to the satellite so that it revolves around the earth in a definite orbit.

18 Orbital Velocity of Satellite:

A velocity required to put a satellite in a definite orbit around the earth is called orbital velocity.

Suppose a satellite of mass  is at height  from the surface of earth of mass  and radius . And satellite is given a horizontal velocity .

Then force of gravitation experienced by satellite is

And the centripetal force due to this force is

Comparing eq. (i) and (ii) we get

Or

Or        orbital velocity

Also

So

If satellite revolve close to the surface of earth then  so  can be neglected

⟹

But

So

19 Relation between orbital velocity and escape velocity:

As we know                  and

So

⟹

Hence escape velocity is  times the orbital velocity of the satellite i.e.

20 Time period of a satellite:

The time taken by the satellite to complete one revolution around the earth is called time period of the satellite. As

Or

But

So

If satellite revolves close to the surface of earth then , so  can be neglected

⟹

Here                                   and

⟹

21 Height of the satellite from surface of earth:

As we know

Squaring both sides, we get

⟹

Or

or

22 Angular Momentum of the Satellite:

The angular momentum of a satellite of mass m moving with velocity v in an orbit of radius

Is given by

23 Geostationary Satellites:

The satellite whose position is fixed with a certain point on the surface of earth and having time period equal to earth i.e. 24 hour is called geostationary satellite. It is also called geo synchronous satellite.

The orbit of a geostationary satellite is called ‘Parking orbit’.

We know that,

For geostationary satellite,           T = 24 hours.

Putting this value of T in the above equation, we get    r » 42000 km or, h » 36000 km.

Where h is the height of the satellite from the surface of the earth.

Conditions for a Geostationary Satellite:

1. Time period of geostationary satellite must be 24 hour.
2. It should move in same sense as earth moves i.e. from west to east.
3. The height of a geostationary satellite must be 36000 Km.
4. It should revolve in an orbit concentric and coplanar with equatorial plane of earth.

Uses of Geostationary Satellites:

1. In communication ratio, TV, and telephone signals across the world are sent though geostationary satellite.
2. In studying upper region of earth.
3. In weather for casting.
4. In finding exact shape and dimension of earth.
5. In studying solar radiations and cosmic rays.
6. Telstar was first geostationary satellite launched by America (USA) in 1962.

24 Polar Satellite:

A satellite which revolves in a polar orbit around the earth is called polar satellite.

It revolve in a plane ⊥ to equatorial plane of earth and passes over north and south poles of the earth a height nearly 500 to 800 Km above the surface of earth. Polar satellites scene almost entire surface of earth.

e.g.:- European spot and Indian earth resources satellite (IRS).

Uses of Polar Satellite:

1. Polar satellite is used in weather and environment monitoring.
2. Used for military purposes.
3. British polar satellite first detected hole in ozone layers.

25 Total Energy and Binding Energy of a Satellite:

Total Energy:–

As we know potential energy of a satellite is

And Kinetic energy of the satellite is

So

Or

Here total energy of satellite is  which indicate that satellite is bounded to the earth and to separate from earth we have to give some more energy.

Binding Energy of Satellite:

The energy required by a satellite to leave its orbit around the earth and escape to infinity called binding energy of satellite.

As total Energy of satellite is            so binding energy of satellite is            =

Question 18. A spaceship is stationed on Mars. How much energy must be expended on the spaceship to rocket it out of the solar system? Mass of the spaceship = 1000 kg, Mass of the Sun = 2  1030 kg. Mass of the Mars =6.4  1023 kg, Radius of Mars = 3395 km. Radius of the orbit of Mars = 2.28  1011 m, G = 6.67  10-11 N m2 kg-2.
Answer:  Let  be the radius of orbit of Mars and  be the radius of the Mars. M be the mass of the Sun and  be the mass of Mars. If m is the mass of the space-ship, then Potential energy of space-ship due to gravitational attraction of the Sun  Potential energy of space-ship due to gravitational attraction of Mars

Since the K.E. of space ship is zero, so total energy of spaceship=

Question 19. 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 m2kg-2
Answer: total initial energy  

Since 20%  is lost, only 80% remains to reach the height.

So total initial energy available=   .

At highest point KE is zero and potential energy

Using law of conservation of energy we get          

Or

Question 20.Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientation problem.
Answer: (a) The blood flow in feet would be lesser in zero gravity. So, the astronaut will not get swollen feet.
(b) In the conditions of weightlessness, the face of the astronaut is expected to get more supply. Due to it, the astronaut may develop swollen face.
(c) Due to more blood supply to face, the astronaut may get headache.
(d) Space also has orientation. We also have the frames of reference in space. Hence, orientation problem will affect the astronaut in space.

26 Theory about Planetary Motion:

Geocentric Model:-Around 100 A.D. Ptolemy proposed geocentric model of earth according to him earth is fixed and the Planets, Sun, Moon revolves around it.

Aryabhatta Contribution:-In 498 A.D., the great Indian mathematician and astronomer Aryabhatta proposed that earth revolves around Sun along with other Planet and also rotate about its own axis.

Heliocentric Model:-In 1543, Nicolas Copernicus suggested that Sun is at the centre of solar system and earth and other Planets revolve around it.

Contributions of Brahe and Kepler:

Ty Cho Brahe was the first to make the extra ordinary observations by studying the motions of planets and stars without the aid of a telescope. His data were analyzed by Kepler and gave the Laws of Planetary motion.

27 Kepler’s Law of Planetary Motion:

Kepler proposed three laws to explain the motion of planets around sun as given below:

(i)  Kepler’s First Law(Law of Orbit):-

According to Kepler’s first law, every planet revolves around the sun is an elliptical orbit and sun is at any one of its foci. As shown in fig.

 (ii)    Kepler’s Second Law (Law of Area):-

According to Kepler’s second law of planetary motion, every planet swept out equal area in equal interval of time.

Suppose a planet is at position  and moves up to  in time  then area swept out by the planet is

In the same time  is planet move from  to  then area swept out by the planet

As from diagram

⟹

Clearly                  ⟹

Hence if a planet is near the sun then its velocity beams larger and is planet is away from sun then its velocity decreases.

(iii)   Kepler’s Third Law(Law of Period):-

According to Kepler’s third law of planetary motion, the square of time period is equal to the cube of semi major axis.i.e.

or

Where K is a constant of proportionality for two different planets, we can write

Thus the larger the distance of planet from Sun larger will be the period of rotation.

Question 21. A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.50 108 km away from the Sun?
Answer: we know that                        

Also                                  so

28 Proof of Kepler’s First Law of Planetary Motion:

As we know

So that torque exerted on the planet  about the sun is

Also                 ⟹

Thus angular momentum of the planet about sun remains constant also it can be shown that central force varies as square of distance between the planets and sun and this orbit is an ellipse.

29 Proof of Kepler’s Second Law:

Let a planet moves in an elliptical orbit from  to  in time .

Then area swept out by planet by position vector  is

Area of triangular reason

Dividing by  we get

As        we have

Or

As angular momentum

⟹

So

Here  and  are constant so  which is Kepler’s second law of planetary motion.

30  Proof of Kepler’s Third Law of Planetary Motion:

Suppose a planet of mass  moves around sun in radius  with orbital speed . Let  is mass of sun. Then gravitational force between sun and planet is

⟹

But            So

⟹

Or

This is Kepler’s third law of planetary motion.

31 Deduction of Newton’s law of Gravitation from Kepler’s Law of Motion:

Suppose a planet of mass  moves around sun in a radius .Let  is the mass of sun and v is the velocity of planet.

Then centripetal force required moving planet is

But orbital velocity

⟹

Now from Kepler’s third law

Or

⟹

As the force between the sun and planet is mutual so   or

⟹

This is Newton’s law of gravitation.

Question 22 . 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: here   and   also  so using Kepler’s third law

Question 23 . Let us assume that our galaxy consists of 2.5  1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky way to be 105 ly.
Answer: Here, r = 50000 ly = 50000  9.46 10¹⁵ m = 4.73  10²⁰ m
M = 2.5  10¹¹ solar mass = 2.5  10¹¹   (2  10³⁰) kg = 5.0  10⁴¹kg
We know that

question 24. 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) The linear speed of the comet is variable in accordance with Kepler7s second law. When comet is near the sun, its speed is higher. When the comet is far away from the sun, its speed is very less.
(b) Angular speed also varies slightly.   (c) Comet has constant angular momentum.
(d) Kinetic energy does not remain constant.  (e) Potential energy varies along the path.
(f) Total energy throughout the orbit remains constant.

Question 25 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  1024 kg, radius = 6400 km.
Answer: Distance of satellite from the centre of earth = R = r + x= 6400 + 36000 = 42400 km = 4.24  10⁷ m

Using potential

32 Weightlessness: 

A body is said to be in a state of weightlessness if the apparent weight of the body becomes zero. This situation arises in the following conditions.

1. In a freely falling lift.
2. Inside a space craft.
3. At null points in space i.e. the point where gravitational forces due to various masses cancel out. At that point
4. At the centre of earth

Problems Related to Weightlessness:

1. Eating and drinking becomes difficult.
2. It affects the human organism.
3. While walking in aircraft injury may occur due to collision.
4. Some phenomenon like oscillation of simple pendulum cannot occur in weightlessness.

33 Inertial and Gravitational Mass:

The mass of the body which measure inertia of the body called inertial mass

i.e.

Properties:-

1. It is directly proportional to matter possessed by the body.
2. It is independent on size, shape and state of body.
3. It is conserved in both physical and chemical processes.
4. It is not affected by presence of other bodies.
5. Inertial mass increase with speed as

Gravitational Mass:

The mass of a body which determine the gravitational pull of earth acting upon it is called gravitational mass.      As

⟹

Larger the  larger is the gravitation pull of the earth on it.   Both  and  are equals.

Similarity:-

1. Both represent quantity of matter of the body.
2. Both do not depend on shape and state of matter.
3. Both are equal in magnitude.
4. Both are not affected by presence of other bodies.
5. Both are scalar quantity.

Difference:-

1. Inertial mass is determined by Newton’s second law of motion and gravitation mass is determined by Newton’s law of gravitation.
2. Inertial mass can be measured only when body is motion which is neither convenient nor practical while gravitational mass can be measured using common balance.

SOME IMPORTANT MCQ.

1. In the formula , the quantity G:
2. depends on the local value of g             B.    is used only when Earth is one of the two masses
3. is greatest at the surface of Earth           D.     Is a universal constant of nature *
4. Earth exerts a gravitational force on the Moon, keeping it in its orbit. The reaction to this force, in the sense of Newton’s third law, is:
5. the centripetal force on the Moon               B. the nearly circular orbit of the Moon
6. the gravitational force on Earth by the Moon*    D. the tides due to the Moon
7. Let F1 be the magnitude of the gravitational force exerted on the Sun by Earth and F2 be the magnitude of the force exerted on Earth by the Sun. Then:
8. F1 is much greater than F2                                       B.    F1 is slightly greater than F2
9. F1 is equal to F2*                                                    D.     F is slightly less than F
10. The magnitude of the acceleration of a planet in orbit around the Sun is proportional to:
11. the mass of the planet B. the mass of the Sun *   C. the distance between the planet and the Sun
12. the reciprocal of the distance between the planet and the Sun
13. Let M denote the mass of Earth and let R denote its radius. The ratio g/G at Earth’s surface is:
14.                        B.             *                    C.                    D.
15. The mass of an object:
16. is slightly deferent at deferent locations on Earth
17. is independent of the acceleration due to gravity*
18. is the same for all objects of the same size and shape D. is a vector
19. An astronaut on the Moon simultaneously drops a feather and a hammer. The fact that they land together shows that:
20. no gravity forces act on a body in a vacuum
21. the acceleration due to gravity on the Moon is less than on Earth
22. in the absence of air resistance all bodies at a given location fall with the same acceleration*
23. the feather has a greater weight on the Moon than on Earth
24. The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600N on Earth, what would he weigh on this planet?
25. 24N                   B   48N                    C        96N*                           D       192N
26. Of the following where would the weight of an object be the least?
27.  2000miles above Earth’s surface                     B at the North Pole
28. At the equator                                                   D.   At the center of Earth*

10. An astronaut in an orbiting spacecraft feels “weightless” because she:
11. is beyond the range of gravity                          B. is pulled outward by centrifugal force
12. has no acceleration                         D. has the same acceleration as the spacecraft*
13. The escape speed at the surface of Earth is approximately 8 km/s. What is the mass, in units of Earth’s mass, of a planet with twice the radius of Earth for which the escape speed is twice that for Earth?
14. 2                                B. 4                                C. 8 *                                   D. 1/2
15. Neglecting air resistance, a 1.0-kg projectile has an escape velocity of about 11 km/s at the surface of Earth. The corresponding escape velocity for a 2.0 kg projectile is:
16. 3.5km/s                       B. 5.5km/s                     C. 7.1km/s                          d. 11 km/s  *
17. An artificial satellite of Earth releases a bomb. Neglecting air resistance, the bomb will:
18. strike Earth under the satellite at the instant of release
19. strike Earth under the satellite at the instant of impact
20. strike Earth ahead of the satellite at the instant of impact       D. never strike Earth *
21. A planet is in circular orbit around the Sun. Its distance from the Sun is four times the average distance of Earth from the Sun. The period of this planet, in Earth years, is:
22. 4                                     B. 8*                             C. 16                                         D. 64
23. The period of a satellite in a circular orbit around a planet is independent of

(a) The mass of the planet                                     (b) the radius of the orbit

(c) The mass of the satellite*                               (d) all of three parameters a, b and c

16. Which of the following is not true for a geostationary satellite?

(a) Its time period is 24 hr                 (b) Its angular speed is equal to that of earth about its own axis

(c) It is stationary in space*            (d) It revolves from west to east

17. A satellite is orbiting a planet at a certain height in a circular orbit. If the mass of the planet is suddenly reduced to half, the satellite would

(a)    Continue to revolve round the planet at the same speed        (b)    Falls freely on the planet

(c)    Orbit the planet at lesser speed                                                (d)   Escape from the planet*

18. A uniform spherical shell gradually shrinks maintaining its shape. The gravitational potential at the centre

(a) Increases                      (b) decreases*            (c) remains constant                (d) oscillates

19. A satellite revolving around earth experiences a small, air resistance due to earth atmosphere. Due to this satellite follows a spiral path towards the earth

(a)  Its potential energy decrease*                                       (b) its kinetic energy decreases

(c)   Satellite will eventually fall on earth*                        (d) its speed increases*