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.
Prof. Mukesh N Tekwani
3
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 Mukesh N Tekwani the
between
Prof.
4
5.
Difference between G and g
G
G is the Universal
Gravitational Constant
G = 6.67 x 10-11 N m2/kg2
Constant throughout the
Universe
g
g is acceleration due to
gravity
Approx value g = 9.8 m /
s2.
Value of g varies from one
place to another on the
Earth.
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.
Prof. Mukesh N Tekwani
5
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
Prof. Mukesh N Tekwani
6
7.
Relation between G and g
m
M
Prof. Mukesh N Tekwani
7
8.
Relation between G and g
m
M
Weight of the object is the
gravitational force acting on
it.
………………………….(1)
Prof. Mukesh N Tekwani
8
9.
Relation between G and g
At height h from the surface of the
Earth’s surface, acceleration due to
gravity is gh
m
h
M
At height h,
Weight of object = gravitational force
………………………….(2)
Prof. Mukesh N Tekwani
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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.
Prof. Mukesh N Tekwani
10
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.
Prof. Mukesh N Tekwani
12
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.
Prof. Mukesh N Tekwani
13
13.
Kepler’s Laws of Motion
Born:
December 27, 1571
Died:
November 15, 1630
German
Mathematician,
Astronomer
Astrologer.
Prof. Mukesh N Tekwani
14
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.
Prof. Mukesh N Tekwani
15
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.
Prof. Mukesh N Tekwani
16
16.
Kepler’s Law of Equal Areas
Prof. Mukesh N Tekwani
17
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 semimajor 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.
Prof. Mukesh N Tekwani
19
19.
Critical Velocity of a Satellite
Let
vc
h
R
M
r
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
Prof. Mukesh N Tekwani
of the satellite
20
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
Prof. Mukesh N Tekwani
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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
Prof. Mukesh N Tekwani
22
22.
Critical Velocity of a Satellite
But we know that
Substituting this value in eqn (1), we get,
…………………………………..(2)
Prof. Mukesh N Tekwani
23
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?
Prof. Mukesh N Tekwani
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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.
Prof. Mukesh N Tekwani
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25.
Time Period of a Satellite
Let
vc
h
R
M
r
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
Prof. Mukesh N Tekwani
of the satellite
26
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
Prof. Mukesh N Tekwani
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27.
Time Period of a Satellite
As
Squaring both sides, we get
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.
Prof. Mukesh N Tekwani
28
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.
Prof. Mukesh N Tekwani
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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
Prof. Mukesh N Tekwani
30
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
Prof. Mukesh N Tekwani
31
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?
Prof. Mukesh N Tekwani
32
32.
Binding Energy of a Satellite
Let
vc
h
R
M
r
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
Prof. Mukesh N Tekwani
of the satellite
33
33.
Binding Energy of a Satellite
The critical velocity is given by
……………………………………………………...(1
)
Kinetic energy of motion KE =
…..……...(2)
Substituting (1) in (2), we get,
KE =
..………………………………………………..(3
)
Prof. Mukesh N Tekwani
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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
)
Prof. Mukesh N Tekwani
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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.
Prof. Mukesh N Tekwani
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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. N Tekwani
Prof. Mukesh
37
37.
Weightlessness in a Satellite
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.
Prof. Mukesh N Tekwani
38
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
Prof. Mukesh N Tekwani
astronaut. This causes the “feeling” of
39
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.
Prof. Mukesh N Tekwani
40
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
Prof. Mukesh N Tekwani
41
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.
Prof. Mukesh N Tekwani
42
42.
Escape Velocity of a Satellite
Therefore, KE of satellite = BE
KE =
R
Therefore,
=
R
Prof. Mukesh N Tekwani
43
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?
Prof. Mukesh N Tekwani
44
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?
Prof. Mukesh N Tekwani
45
45.
Variation of g with Altitude
Weight of the object is the
gravitational force acting on
it.
m
M
………………………….(1)
Prof. Mukesh N Tekwani
47
46.
Variation of g with Altitude
At height h from the surface of the
Earth’s surface, acceleration due to
gravity is gh
m
h
M
At height h,
Weight of object = gravitational force
………………………….(2)
Prof. Mukesh N Tekwani
48
47.
Variation of g with Altitude
Dividing (2) by (1) we get,
m
h
M
Thus, gh 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.
Prof. Mukesh N Tekwani
49
48.
Variation of g with Altitude
By Binomial expansion:
Since the higher powers
of h/R are neglected.
Prof. Mukesh N Tekwani
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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
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52
50.
Variation of g due to depth
Mass = Volume x Density
Mass =
………………………………………….
Prof. Mukesh N Tekwani
(1)
53
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)
Prof. Mukesh N Tekwani
54
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)
Prof. Mukesh N Tekwani
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53.
Variation of g due to depth
Mass = volume x density
……………………………………………(2)
Prof. Mukesh N Tekwani
56
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.
Prof. Mukesh N Tekwani
57
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.
Prof. Mukesh N Tekwani
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56.
Variation of g due to depth
g’
depth
altitude
Prof. Mukesh N Tekwani
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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)
Prof. Mukesh N Tekwani
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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
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59.
Variation of g due to latitude
NP-North Pole
SP-South Pole
Prof. Mukesh N Tekwani
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60.
Variation of g due to latitude
Consider a body at point P on the
surface of the earth.
Let the latitude of point P be
The body at point P moves in a
circular path whose center is at Q
and radius is PQ.
PQ = r
XOP =
Prof. Mukesh N Tekwani
63
61.
Variation of g due to latitude
Therefore,
OPQ = (alternate
angles – transversal cutting
two parallel lines)
Centripetal acceleration
needed for a body at point P is
ar = rω2
Consider
OPQ
cos =
Prof. Mukesh N Tekwani
64
62.
Variation of g due to latitude
cos =
cos =
r = R cos
Therefore, ar = R cos x ω2
ar = R ω2 cos ………………. (1)
Prof. Mukesh N Tekwani
65
63.
Variation of g due to latitude
From OPQ, the radial component
of centripetal acceleration is
a = ar cos
= R ω2 cos cos
a
= R ω2 cos2 ……………… (2)
a
effective acceleration g’ due to
The
gravity at point P is directed towards
the centre of the earth and is given
by:
Prof. Mukesh N Tekwani
66
64.
Variation of g due to latitude
= g - R ω2 cos2
g’
latitude increases, cos
As
decreases, so g’ will increase.
value of g’ increases as we
The
move from the equator to the pole
due to rotation of the earth.
Prof. Mukesh N Tekwani
67
65.
Variation of g due to latitude
Assignment 1: Obtain an expression / value for the
acceleration due to gravity at the equator.
Assignment 2: Obtain an expression / value for the
acceleration due to gravity at the poles.
Prof. Mukesh N Tekwani
68
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