physics is my favorite subject from childhood so i got truely satisfaction when any thing do in physics.

1. Gravitation
Prof. Mukesh N. Tekwani
Department of Physics
Ismail Yusuf College,
Mumbai
mukeshtekwani@hotmail.com

2. Isaac Newton
Mr. N. Singh

3. Newton’s Law of Gravitation
Every particle of matter attracts every other
particle with a force which is directly
proportional to the product of the masses and
inversely proportional to the square of the
distance between them.

4. Newton’s Law of Gravitation
G is called the Universal Gravitational Constant
G = 6.67 x 10-11 N m2/kg2
G is constant throughout the Universe and G does
not depend on the medium between the masses

5. Difference between G and g
G g
G is the Universal
Gravitational Constant
g is acceleration due to
gravity
G = 6.67 x 10-11 N m2/kg2 Approx value g = 9.8 m / s2.
Value of g varies from one
place to another on the
Earth.
Constant throughout the
Universe
Changes every place on a
planet. E.g., on the Moon,
the value of g is 1/6th of that
on the Earth’s surface.

6. Relation between G and g
Let M = mass of the Earth
m = mass of an object on the surface of
the Earth
g = acceleration due to gravity on the
Earth’s surface
R = radius of the Earth

7. Relation between G and g
m
M

8. Relation between G and g
m
M
Weight of the object is the
gravitational force acting on it.
………………………….(1)

9. Relation between G and g
At height h from the surface of the
Earth’s surface, acceleration due to
gravity is gh
At height h,
Weight of object = gravitational force
m
h
M
………………………….(2)

10. Relation between G and g
Dividing (2) by (1) we get,
m
h
M
Thus, g is independent of the mass of the
object.

11. Projection of a Satellite
Why is it necessary to have at least a two stage rocket to launch
a satellite?
A rocket with at least two stages is required to launch a satellite
because
The first stage is used to carry the satellite up to the desired
height.
In the second stage, rocket is turned horizontally (through 90
degrees) and the satellite is fired with the proper horizontal
velocity to perform circular motion around the earth.

12. Critical Velocity of a Satellite
The horizontal velocity
with which a satellite
should be projected from a
point above the earth's
surface, so that it orbits in
a circular path around the
earth is called the orbital
velocity or critical velocity
(Vc) of the satellite.

13. Kepler’s Laws of Motion
Born:
December 27, 1571
Died:
November 15, 1630
German
Mathematician,
Astronomer
Astrologer.

14. Kepler’s First Law – Law of Orbit
Every planet revolves in an elliptical
orbit around the Sun, with the Sun
situated at one focus of the ellipse.

15. Kepler’s Second Law
or Law of Equal Areas
The radius vector drawn from the Sun
to any planet sweeps out equal areas in
equal intervals of time. This law is called
the law of areas.
The areal velocity of the radius vector is
constant.

16. Kepler’s Law of Equal Areas

17. Kepler’s Laws

18. Kepler’s Third Law - Laws of Period
The square of period of revolution of
the planet around the Sun is directly
proportional to the cube of the semi-major
axis of the elliptical orbit.
T2 α r3
According to this law, when the planet is
closest to the Sun, its speed is maximum
and when it is farthest from the Sun, its
speed is minimum.

19. Critical Velocity of a Satellite
vc Let
R
M
r
h
M = mass of the Earth
R = radius of Earth
m = mass of satellite
h = height of the
satellite above Earth’s
r = R + h, where r is
the distance of the
satellite from the
center of the Earth
Vc = critical velocity
of the satellite

20. Critical Velocity of a Satellite
The centripetal force necessary for the circular motion
of the satellite around the Earth is provided by the
gravitational force of attraction between the Earth and
the satellite.
Centripetal force = gravitational force

21. Critical Velocity of a Satellite
……………………………………………….(1)
Factors on which Critical Velocity of a satellite depends:
1. Mass of the planet
2. Radius of the planet
3. Height of the satellite
Critical velocity is not dependent on the mass of the
satellite as m does not appear in the above equation

22. Critical Velocity of a Satellite
But we know that
Substituting this value in eqn (1), we get,
…………………………………..(2)

23. Critical Velocity of a Satellite
Assignment 1:
Modify eqn (2) to find the critical velocity of a satellite
orbiting very close to the surface of the Earth (h << R)
Assignment 2:
How does the critical velocity (or orbital velocity) of a
satellite vary with an increase in the height of the
satellite above the Earth’s surface?

24. Time Period of a Satellite
The time taken by a satellite to complete
one revolution around the earth is called its
periodic time or time period.

25. Time Period of a Satellite
vc Let
R
M
r
h
M = mass of the Earth
R = radius of Earth
m = mass of satellite
h = height of the
satellite above Earth’s
r = R + h, where r is
the distance of the
satellite from the
center of the Earth
Vc = critical velocity
of the satellite

26. Time Period of a Satellite
Distance covered by the satellite in 1 revolution
= Circumference of the circle
Time taken to cover this distance is the time period
Critical speed Vc =
Vc =
But

27. Time Period of a Satellite
Squaring both sides, we get
As is a constant,
so we get,
T2 α r3
Thus, the square of the
period of revolution is
directly proportional to
the cube of the radius of
its orbit.

28. Time Period of a Satellite
Factors on which Time Period of a satellite depends:
1. Mass of the planet
2. Radius of the planet, and
3. Height of the satellite from the planet’s surface
Period of the satellite does not depend on the mass of
the satellite.
Assignment:
(1)Obtain an expression for the time period of a
satellite in terms of gh.
(2)For a satellite close to the earth, calculate the period
of revolution in minutes.

29. Binding Energy of a Satellite
Definition:
The minimum amount of energy required to
remove a satellite from the earth’s gravitational
influence is called as binding energy of the satellite.
A satellite revolving around the Earth has
 Kinetic energy, and
Potential energy

30. What is kinetic energy?
The energy possessed by a body due to
its motion is called its kinetic energy.
If m = mass of an object, and
v = its velocity
K.E = (1/2) mv2

31. What is potential energy?
The energy possessed by a body due to its
position is called its potential energy.
If m = mass of an object, and
h = its height above the surface
P.E. = mgh
Quiz: Does a body in motion have potential
energy?

32. Binding Energy of a Satellite
vc Let
R
M
r
h
M = mass of the Earth
R = radius of Earth
m = mass of satellite
h = height of the
satellite above Earth’s
r = R + h, where r is
the distance of the
satellite from the
center of the Earth
Vc = critical velocity
of the satellite

33. Binding Energy of a Satellite
The critical velocity is given by
Kinetic energy of motion KE =
Substituting (1) in (2), we get,
KE =
……………………………………………………...(1)
…..……...(2)
..………………………………………………..(3)

34. Binding Energy of a Satellite
The gravitational potential at a distance r from
the centre of the Earth is given by:
GP =
Potential energy = gravitational potential x mass
of the satellite
Therefore, PE = ..…….………………………..(4)

35. Binding Energy of a Satellite
The total energy of the satellite is given by
TE = KE + PE
TE = +
TE =
..…….…………………………………………..(5)
The negative sign indicates that the satellite is
bound to the Earth due to the gravitational force
of the Earth.

36. Binding Energy of a Satellite
To free the satellite from the Earth’s gravitational
influence, an amount of energy equal to its total
energy must be supplied. This is called the binding
energy of the satellite.
Therefore, BE =
Where r = R + h
Assignment: Calculate the BE of a satellite at
rest on the surface of the Earth.

37. Weightlessness in a Satellite
38
1. The weight of a body is the gravitational force
exerted on it by the Earth.
2. When a person stands on a floor, he exerts a
force on the floor. The floor in turn exerts a
force (normal reaction) on the person.
3. This normal reaction is equal to the weight of
the person.
4. The person has a feeling of weight due to this
normal reaction.

38. Weightlessness in a Satellite
5. Consider an astronaut of mass m, in a satellite
that is moving around the Earth in a circular
orbit.
6. There is a centripetal force on the satellite and
the astronaut. Thus, both are attracted towards
the Earth with the same acceleration, due to the
Earth’s gravitational force.
7. So the astronaut is not able to exert a force on
the floor of the satellite & the satellite in turn
cannot exert normal reaction on the astronaut.
This causes the “feeling” of weightlessness.

39. Weightlessness in a Satellite
8. We must remember that the mass of the
astronaut DOES NOT become zero.
9. This condition of weightlessness is also known
(incorrectly) as zero gravity condition.
10. But, weightlessness does not mean the absence
of gravity.

40. Escape Velocity of a Satellite
The minimum
velocity with which
a body should be
projected from the
surface of the earth
so that it escapes
the gravitational
field of the earth is
called the escape
velocity of the body.

41. Escape Velocity of a Satellite
Consider a satellite of mass m, stationary on the
surface of the Earth.
The binding energy of the satellite, on the surface
of the Earth, is given by
BE =
R
To escape from the Earth’s influence, energy must
be provided to the satellite in the form of kinetic
energy.

42. Escape Velocity of a Satellite
Therefore, KE of satellite = BE
KE =
R
Therefore, =
R

43. Numerical Problems
Obtain an equation for the escape
velocity of a body from the surface of a
planet of radius R and mean density ρ.
What would be the duration of the year if
the distance between the Earth and Sun
gets doubled?

44. Numerical Problems
Calculate the height of a communications satellite
from the surface of the Earth. Values of G, M and R
are as given in the text book. (These values will also
be provided in the question paper)
A body weighs 4.5 kgwt on the surface of Earth.
How much will it weigh on the surface of a planet
whose mass is (1/9)th that of the Earth’s mass and
radius is half that of the Earth?

45. Variation of g with Altitude
m
M
Weight of the object is the
gravitational force acting on it.
………………………….(1)

46. Variation of g with Altitude
At height h from the surface of the
Earth’s surface, acceleration due to
gravity is gh
At height h,
Weight of object = gravitational force
m
h
M
………………………….(2)

47. Variation of g with Altitude
Dividing (2) by (1) we get,
m
h
M
Thus, g is independent of the mass of the
object.
From this eqn. it is clear that the acceleration
due to gravity decreases as altitude of the
body from the earth’s surface increases.

48. Variation of g with Altitude
By Binomial expansion:
Since the higher powers of
h/R are neglected.

49. Variation of g due to depth
The acceleration due to gravity on the surface of the
earth is given by:
Consider the earth as a sphere of density ρ
Mass = Volume x Density

50. Variation of g due to depth
Mass = Volume x Density
Mass =
…………………………………………. (1)

51. Variation of g due to depth
Consider a point P at the depth d below the
surface of the earth. At this point, let the
acceleration due to gravity be gd.
Distance of point P from centre of Earth is
(R – d)

52. Variation of g due to depth
The acceleration due to gravity at point P
due to sphere of radius (R –d) is
Here, M’ is the mass of inner solid sphere of
radius (R - d)

53. Variation of g due to depth
Mass = volume x density
……………………………………………(2)

54. Variation of g due to depth
Dividing (2) by (1), we get,
This is the expression for the acceleration due to gravity
at a depth d below the surface of the earth.
Therefore, acceleration due to gravity decreases with
depth.

55. Variation of g due to depth
At the centre of the earth, d = R, therefore
gd = 0
So if a body of mass m is taken to the centre of the earth, its
weight will be equal to zero (since w= mg). But its mass will
not become 0.

56. Variation of g due to depth
g’
Inside the earth
Outside the earth
depth altitude

57. What is latitude
The latitude of a location on the Earth is the
angular distance of that location south or
north of the Equator.
Latitude of equator is 0o
Latitude of North pole is : 90o north (+90o)
Latitude of South pole is : 90o south (-90o)

58. Variation of g due to latitude
The Earth is rotating from west to east and the
axis of rotation passes through the poles.
Let angular velocity of earth be ω.
Every point on the surface of the earth is moving
in a circle, i.e. every point is in an accelerated
motion

59. BY : N. SINGH
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