GRAVITATION.ppt

Chapter–8: Gravitation
Universal law of gravitation. Acceleration due to gravity
(recapitulation only) and its variation with altitude and
depth. Gravitational potential energy and gravitational
potential, escape velocity, orbital velocity of a satellite,
Geo-stationary satellites.
Revised syllabus -CORONA
PANDEMIC
2020
GRAVITATION

NEWTON’S UNIVERSAL LAW OF GRAVITATION
Every body in the universe attracts every other body with a force
which is directly proportional to the product of their masses and is
inversely proportional to the square of the distance between them.
The direction of force is along the line joining the centres of the
two bodies.
Let two bodies A and B of masses m1 and m2 are placed at a distance of ‘r’
from each other. Let ‘F’ be the force of attraction between the bodies.
Then,
(i) the force of attraction is directly proportional the product of their masses
i.e. F α m1 x m2
(ii) the force of attraction is inversely proportional the square of the distance
between them
i.e. F α
1
r2

F α
m1 x m2
r2
F = G
m1 x m2
r2
where G is a constant called as “universal
gravitational constant”
G = 6.67 x 10-11 Nm2kg-2
In vector form,

Properties of gravitational force
1. Gravitational force is the weakest force in nature.
2. It is an attractive force. (Unlike electrostatic and
magnetic force; they are both attractive and repulsive)
3. It is a mutual force. (First body attracts the second body
and the second body attracts the first body with equal
force)
4. It is a central force. (Acts along the line joining the centres
of the bodies)
5. It is mass and distance dependent.
6. It obeys inverse square law.
7. It is a long range force. (It decreases with distance as per
inverse square law and becomes zero only at infinite
distance – like electrostatic and magnetic force)
8. It does not depend on the medium between the interacting
bodies. (There is no gravitational shielding)

12/23/2023
Billiards, Anyone?
 Three 0.3-kg billiard balls
are placed on a table at
the corners of a right
triangle.
(a) Find the net gravitational
force on the cue ball;
(b) Find the components of
the gravitational force of
m2 on m3.
NUMERICAL

The red mass lies in space at the
center of an equilateral
triangular arrangement of three
identical grey masses. The net
gravitational force acting on the
central mass must
1. act upward in the diagram.
2. act downward.
3. act to the right.
4. act to the left.
5. be zero.
THINK

Center of Gravity
Newton found that his law would only work when
measuring from the center of both objects
This idea is called the center of gravity
Sometimes it is at the exact center of the object
Sometimes it may not be in the object at all
All forces must be from the CG of one object to the
CG of the other object

Acceleration due to gravity (g)
Acceleration due to gravity is defined as the uniform acceleration produced
in a freely falling body due to the gravitational force of the earth.
Acceleration due to gravity (g) = 9.8 m/s2 = 980 cm/s2.
Calculation of acceleration due to gravity (g)
Suppose a body of mass ‘m’ is placed on the earth of mass ‘M’ and radius ‘R’.
According to Newton’s universal law of gravitation,
Force exerted by the earth on the body is given by
R
m
M
F = G
M x m
R2
This force exerted by the earth produces an acceleration on the body.
Therefore, F = mg (g - acceleration due to gravity)
From the two equations, we have
mg = G
M x m
R2
or g =
G M
R2

2. Acceleration due to gravity decrease with depth.
g =
G M
R2
g’ =
G M’
(R-h)2
h
m
3. Acceleration due to gravity is greater at the poles
and less at the equator.
gp =
G M
Rp
2
ge =
G M
Re
2
m
M
Rp
m
Re
Earth is slightly flattened at the poles and bulging at the
equator. The radius of the earth at the poles is 21 km less
than that at the equator.
i.e. Rp < Re
Therefore, from the above equations, G and M being
same and g is inversely proportional to the square of the
radius,
gp > ge
gp = 9.823 m/s2 , ge = 9.789 m/s2 and average value of g = 9.8 m/s2
M
M’
R

VARIATION OF ACCELERATION DUE TO GRAVITY
1.WITH ALTITUDE
We know that at the surface of the earth,
At a height h above the surface of earth , acceleration due to gravity is,

2. With Depth
We know that at the surface of the earth,

Gravitational potential at any point inside the gravitational field region of the earth is equal
to the amount of work done in bringing an object of unit mass from infinity to that point.
PROOF:
Suppose an object of mass ‘m’ is kept at a height ‘h’ above the earth’s surface.
Let Re+ h = r _(1)
If at any instant, the object is at a point A, ‘x’ distance away from the centre of the earth,
then the force acting on the object, F = GMm/x2 _(2)
If the object is displaced through a short displacement ‘dx’, towards the earth’s surface, the
small amount of work done,
dW = F dx
⇒ dW = (GMm/x2)dx
∴Total amount of work done in transporting the object from infinity to point B, ‘r’
distance away from the centre of the earth,
∴The work done in bringing a unit mass, i.e., Gravitational potential, V = W/m
⇒V = -GM/r
Gravitational Potential

GRAVITATIONAL POTENTIAL ENERGY
Gravitational potential energy of an object is equal to the amount of work done in
bringing that object from infinity to that point.
PROOF:
Suppose an object of mass ‘m’ is kept at a height ‘h’ above the earth’s surface.
Let Re+ h = r _ (1)
If at any instant, the object is at a point A, ‘x’ distance away from the centre of
the earth, then the force acting on the object, F = GMm/x2 _(2)
If the object is displaced through a short displacement ‘dx’, towards the earth’s
surface, the small amount of work done,
dW = F dx ⇒dW = (GMm/x2)dx
∴Total amount of work done in transporting the object from infinity to point
B, ‘r’ distance away from the centre of the earth,
U=
This work done gets stored up in the form of gravitational potential energy.
1.The negative sign indicates that the potential energy is due to the gravitational attraction
between the earth and the body.
2.As distance between the two increases, the gravitational potential energy also increases and
becomes 0, i.e., maximum.

Gravity is a conservative force.
When two masses move apart or
come together, the work done by
gravity depends only on the overall
change in radial separation, not the
detailed path taken.
From F= –dU(r)/dr, we find the
gravitational potential energy
must have the form
U(r) = - G
m1m2
r
gravitational potential energy
http://hyperphysics.phy-astr.gsu.edu/HBASE/gpot.html
Source: Pearson Education, Inc.

Escape Velocity
~ The minimum velocity
given to the object on
the surface of the
planet so that it
ejects from the
gravitational field
region & never comes
back to the surface of
the planet is known as
escape velocity.

Escape Velocity
It is that minimum velocity with which if an object is projected
vertically upwards so that it escapes the earth’s gravitational pull of
earth.

Suppose an object of mass ‘m’ is projected with velocity‘ve’from the surface
of the earth.
Therefore the kinetic energy imparted, KEimparted = ½(m ve
2)
This kinetic energy will be utilised in moving the object from the surface of
the earth to infinity.
Suppose at any time, the object is at point A, ‘x’ distance away from the
centre of the earth.
∴The force acting on the object at point A, F = GMm/x2
The small amount of work done in shifting the object through a distance
‘dx’ away from the earth, dW = F dx
⇒dW = (GMm/x2)dx
Proof:
∴Work done in transporting the object from the surface to the infinity,
Hence,
Hence, we get,
On earth, g = 9.8 m/s2; R = 6.4 x 106m ∴ ve = 11.2 km/s

The force of gravity on the satellite is
The centripetal force required by the satellite to keep in its orbit is
Orbital velocity, Vo =
If g is the acceleration due to gravity on the surface of earth then
g=
Or, Vo =
= R
If a satellite is very close to earth’s surface, then
R + h ≈ R
∴ V =
As g = 9.8 ms-2 and R = 6.4x 106m, so vo =
=7.92 x103 ms-1 = 7.92 kms-1
or GM =gR2
It is the velocity required to put a satellite
into its orbit around the earth.
ORBITAL VELOCITY

Relation between Orbital velocity and Escape velocity
The escape velocity of a body from the earth’s surface is
ve =
The orbital velocity of a satellite revolving close to the earth’s
surface is vo =
ve = vo

A satellite orbits the Earth with
speed v in a circular orbit of
radius r. At what radius would
the satellite orbit with speed v/3?
1. 2r
2. 3r
3. 4r
4. 6r
5. 9r Source: NASA
THINK

GEOSTATIONARY SATELLITES
As the name suggests, a satellite which revolves around the earth in its equatorial
plane with the same angular speed and the same direction as the earth rotates
about its own axis is called geostationary satellite or synchronous satellite.

POLAR SATELLITE
It is a kind of satellite that revolves in polar orbit of the earth. It has a
smaller orbital radius of about 500-800 km. It virtually scans the entire
surface of the earth.

ENERGY ASSOCIATED WITH
SATELLITE
 The satellite has 2 types of energy
1. Kinetic Energy (due to its motion)
2. Potential Energy (due to presence of
gravitational force)

KEPLER’S LAWS OF PLANETARY MOTION
I LAW (Law of orbits):
Every planet moves around the sun in elliptical orbit with the sun at one of
the foci of the elliptical orbit.
PLANETS IN ORDER
F1
F2
Sun
Elliptical Orbit

F1
F2
Sun
S
F1
F2
Sun
S
D
C
A
B
Elliptical Orbit
II LAW (Law of areas):
The line joining the planet to the sun sweeps over equal areas in equal
intervals of time.
Apogee
Perigee
II law tells us that a planet does not move with constant speed around the
sun. It speeds up while approaching the nearest point called ‘perigee’ and
slows down while approaching the farthest point called ‘apogee’. Therefore,
distance covered on the orbit with in the given interval of time at perigee is
greater than that at apogee such that areas swept are equal.
Area of SAB = Area of SCD (For the given time)

III LAW (Law of periods):
The square of the time period of revolution of a planet around the sun is
directly proportional to the cube of the mean distance of a planet from the
sun.
Though Kepler gave the laws of planetary motion, he could not give a theory
to explain the motion of planets.
Only Newton explained that the cause of the motion of the planets is the
gravitational force which the sun exerts on them.
Newton used Kepler’s III law to develop the law of universal gravitation.
= constant
T2
r3
The law is mathematically expressed as
T2 α r3
or

h
m
M
R
CAVENDISH EXPERIMENT TO FIND GRAVITATIONAL CONSTANT

INTENSITY OF THE GRAVITATIONAL FIELD
Gravitational field intensity, at any point is defined as the gravitational
force experienced by a unit mass placed at that point.
Suppose, to find gravitational intensity at point ‘P’, ‘r’ distance away
from the centre O of mass ‘M’ place a test mass m at point P.
Thus, force acting on mass ‘m’, M m
F F
Gravitational field intensity, P
r
O
Intensity of gravitational field, I is a vector quantity. Its SI unit is Nkg-1 .

GRAVITATION.ppt

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