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Physics
Session
Gravitation
Session Opener
“Planets revolve in elliptical orbits
with sun at its focus”. Have you
ever wondered what forces are
responsible?
Session Objectives
Session Objective
Newton's law of Gravity
Weight & Gravitational Force
Kepler's Law’s
Gravitational Field
Gravitational Potential
Relation between field and potential
Gravitational Potential Energy
Newton’s Law of Gravitation
Any particle of mass m1 attracts a
particle of mass m2 with a force given
by:
FG
m1m2
r2
G=6.67x10-11 N.m2/kg2
Where,
r is the distance between them
G is Universal Gravitational Constant
r
m1
F
m2
Newton’s Law of Gravitation
Important fact about this formula
• Applicable only for point masses
• Does not depend on the medium between
the masses
• Not valid for nuclear distances
• These forces are always added vectorially
Weight & Gravitational force
Earth attracts every body towards
itself by virtue of which the body
experience weight. The true weight of
nay body is only at the earths surface
at every other place it has apperant
weight.
Weight = mass x acceleration due to gravity.
W = mg
Kepler’s Laws
Kepler’s First Law:
All planet move in elliptical orbits with
the sun at one focus.
Sun
Kepler’s Second Law:
A line that connects a planet to the sun sweeps out
equal areas in equal time, i.e. the areal velocity of
the planet is always constant
Kepler’s Laws
dA
 cons tan t
dt
A
B
A’
Sun
B’
Kepler’s Third Law:
The square of the period of revolution of any planet is
proportional to the cube of the semi-major axis of the
orbit.
T2  a3
Gravitational Field Intensity
It is defined as force experienced
per unit mass acting on a test mass
supposed to be placed at that point.
m (Test mass)
F
E
m
[Gravitational Field]
 GMm  1 GM
E 2   2
 r m
r
r
M
Source
point
Field
point
Gravitational Potential
It is defined as negative of the work
done per unit mass in shifting a rest
mass from some reference point to
the given point.
 W 
V

m


U
V
m
V(r)  
GM
r
[Gravitational Potential]
Relation between Gravitational
Potential
F  mE
dW  F.dr
 mE.dr.
dU   dW   mE.dr.
dU
dV 
 E.dr.
m
Gravitational Potential
Energy
Gravitational potential energy at a point
is defined as the amount of work done
by an external agent in bringing any
body of mass (m) from infinity to that
point.
U(r)  
GMm
r
U

GMm
r
[For a point mass]
r
M
F(r) m

Expressions of potential for
different bodies
Gravitational potential V due to a
spherical shell of mass M and radius R
at a point distant r from the centre.
(a) When r > R
V
GM
r
(c) When r < R
GM
V
R
(b) When r = R
V
GM
R
V
(d) When r = 0
V
GM
R
R
r
Expressions of potential for
different bodies
Gravitational potential V due to a solid
sphere of radius R and mass M at a
point distant r from the centre.
(a) When r > R
GM
V
r
(c) When r < R
 3R2  r 2 
V   GM 

3
 2R

(b) When r = R
V
GM
R
(d) When r = 0
3  GM 
V 
2  R 
Expressions of gravitational
field for different bodies
Gravitational field E due to a spherical
shell of mass M and radius R at a point
distant r from the centre.
(a) When r > R
GM
E
r2
(b) When r = R
GM
E
R2
(c) When r < R
(d) When r = 0
E=0
E=0
E
R
r
Expressions of gravitational
field for different bodies
Gravitational field E due to a solid
sphere of radius R and mass M at a
point distant r from the centre.
(a) When r > R
E
GM
r2
(c) When r < R
E
GMr
R3
(b) When r = R
E
GM
R2
(d) When r = 0
E=0
E
R
r
Class Test
Class Exercise - 1
In order to find time, an astronaut
orbiting in an earth satellite can use
(a) pendulum clock
(b) spring -on trolled clock
(c) any one of above
(d) Neither of the two
Solution
As the acceleration for a satellite
continuously changes so it will give
wrong time. Where this is not in
case of spring-controlled clock.
Hence answer is (b).
Class Exercise - 2
Which of the following graphs represent
the motion of a planet moving about the
sun? T is the period of revolution and r is
the average distance (from centre to
centre) between the sun and the planet.
T2
T2
(a)
(b)
r3
r3
T2
2
(c)
(d
)
r
T
r3
Solution
By statement of Kepler’s law
Hence answer is (a).
Class Exercise - 3
A planet of mass M moves around the
sun along an ellipse so that its minimum
distance from the sun is r and maximum
is R. Using Kepler’s law, find its period of
revolution around the sun.
Solution
According to Kepler’s law
3
 r R 
3
T2  K 
  Kx
 2 
 Rr 
where x  
 and K  Constant
 2 
Mv 2 GMMs
Also

x
x2
2x
T

v
2x
GMs
x
v2 

GMs
;
x
2
v
GMs
x
3/2
r R


GMs  2 
Class Exercise - 6
If the radius of the earth were to shrink
by 1%, its mass remaining the same, the
acceleration due to gravity on the earth’s
surface would
(a) decrease
(b) remain unchanged
(c) increase
(d) Cannot say
Solution
g
GM
R2
but as R is decreased so g
would increase.
Hence answer is (c).
Class Exercise - 7
Two planets of radii r1 and r2 are made
from the same material. The ratio of the
g1
acceleration of gravity
at the surfaces
g2
of two planets is
(a)
r1
(b)
r2
 r1 
(c) 

r
 2 
2
r2
r1
 r2 
(d) 

r
 1 
2
Solution
4 3
G    r1
3
g1 
2r12
4
 G  r1
3
g1 r1
so,

g2 r2
Hence answer is (a).
Thank you