Download 27. Gravitation

Physics
Session
Gravitation - 2
Session Opener
How much velocity do you need to
impart a stone such that it escapes
the gravitational field of the earth?
Session Objectives
Session Objective
1. Energy Conservation in Planetary
& Satellite Motion
2. Escape Velocity
3. Orbital Velocity of satellite
4. Variation of g with height & depth
5. Calculation of gravitational field due to
some common systems
6. Satellite Motion and weightlessness
Energy Conservation in Planetary
& Satellite Motion
Gravitational force is a conservative
force. Therefore, work done by
gravitational force is a independent of the
path followed by a body under the force.
Thus mechanical energy is conserved.
Work done by gravitational force is independent
of the path followed
Escape Velocity
It is the minimum velocity with which
a body must be projected vertically
upward in order that it may just escape
the gravitational pull of earth.
Important Fact :
Escape velocity depends on mass of planet and radius
of planet and not on mass of the object which has to be
given this velocity.
Escape Velocity
From Conservation of Energy
K.Ei+P.Ei=K.Ef+P.E.f

GM em
1
mV 2 
0
2
Re
V
2GM e
Re
1
mV2
2
GMem
P.E.  
Re
K.E. 
Re
Escape Velocity 
2GMe
Re
P.E.=0
K.E.=0
Orbital Velocity of satellite (Vorb)
Orbital velocity is the velocity which
is given to an artificial earth’s satellite
so that it may start revolving round
the earth in a fixed, circular orbit.
Orbital Velocity of Satellite (Vorb)
Fg 
mV 2
Fa 
r
GMm
r2
m
V2
a
r
For motion in an orbit
r
Fa=Fg

Orbital Velocity Vorb 
GMe
r
GMem
r2
V
mV2

r
GMe
r
g With Depth and Height
Height ‘h’ above the surface of the
earth.
F

GMm
(R  h)2
gh 
GM
2
2
R (1  h / R)
F
GM

m (R  h)2

g
(1  h / r)2
GM 
2h 
 g  2 1 
 for h  R e
Re 
Re 
Depth ‘d’ below the surface of the earth.
GM 
d 
g  2 1 

Re 
Re 
m
R =Re+h
F
R
M
Re
h
Time Period of Satellite (T)
T
r
t=T
t=0
[Kepler’s Third Law]
Dis tan ce 2r
2r


Velocity
Vorb
GM
r
r3
T  2
GM
2

4

T2  
r3

 GM 


Hence,
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
Satellite Motion and weightlessness
The motion of satellite is based on the
gravitational force. The planet applies
necessary centripetal force by applying
gravitational force.
Inside the satellite astronauts feel weightlessness
because the centrifugal force there is balanced by
gravitational force.
Important Facts :
•Neither the speed nor the time period depend upon
the mass, size and shape of the satellite.
•Speed and time period depend upon radius (r)
Class Test
Class Exercise - 1
A planet revolves in an elliptical orbit
around the sun. Then out of the following
physical quantities the one which remains
constant is
(a) velocity
(b) kinetic energy
(c) total energy
(d) potential energy
Solution
As there is no loss of energy in the
motion of the planet in the orbits
so total energy remains constant
Hence answer is (c).
Class Exercise - 2
Energy supplied to change the radius of
the orbit of a satellite of mass m from r
to 2r is
(a)
GMm
4r
(c) zero
(b)
GMm
2r
(d) None of these
Solution
Initial energy
Ei 
GMm 1
GM
GMm
 m

r
2
r
2r
Final energy
GMm
GMm 1
GM


Ef 
 m
4r
2r
2
2r
Energy supplied
GMm
E = E f – Ei 
4r
Hence, answer is (a).
Class Exercise - 3
A planet with uniform density  and
radius r shrinks in size to a new radius of
R
, but same density. What is the ratio
2
of the new and old escape velocities?
(a) 2 : 1
(b) 2 : 1
(c) 1 : 2
(d) 1: 2
Solution
Ve 
GM
R
4R3
G 
G 4
3

R

R
R
3
So Ve  R
So (Ve) new : (Ve)old = 1 : 2.
Hence, answer is (c).
Class Exercise - 4
The escape velocity of a body when
projected vertically upward from the
surface of earth is 11.2 km/s. If the body
is projected in a direction making an
angle of 30o with the vertical, then its
escape velocity will be
11.2
(a)
km/s
(b) 11.2 km/s
2
3
(c) 11.2 
3
km/s
2
(d) 11.2 km/s
Solution
Escape velocity does not depend on
angle projection.
Hence, answer is (d).
Class Exercise - 5
The escape speed for a projectile in
the case of earth is 11.2 km/s. A
body is projected from the surface of
the earth with a velocity which is
equal to twice the escape speed. The
velocity of the body, when at infinite
distance from the centre of the earth, is
(a) 11.2 km/s
(b) 22.4 km/s
(c) (11.2) 3 km/s
(d) (11.2) 2 km/s
Solution
Total initial energy is
GMm 1
(T.E.)i 
 m(2ve )2
R
2
GMm
GMm GMm

 2

R
R
R
Total final energy
1 2
(T.E.)f = 0  mv
2
1 2
mv  m  GM  m  Ve2
2
R
or V = 2v e =
 2  (11.2)
Hence, answer (d).
Class Exercise - 6
R is the radius of the earth. At a height h
above the surface of the earth the
acceleration due to gravity is
1 of its
100
value on the surface of the earth. Then h
is
(a) h = 100R
(c) h =10R
(b) h = 99R
(d) h = 9R
Solution
g at surface is gs 
GM
R2
GM
1
GM


At height h gh 
2 100
(R  h)
R2
or 10R = R + h
h = 9R
Hence, answer is (d).
Class Exercise - 7
As we go from equator to poles, the
value of g
(a) remains unchanged
(b) decreases
(c) increases
(d) decreases and then increases
Solution
g´ = g - R 2 cos2
cos  decreases
So g´ increases.
Hence, answer is (c).
Class Exercise - 8
If the earth suddenly stops rotating, then
acceleration due to gravity would not
change at
(a) equator
(b) poles
(c) a latitude 45o
(d) None of these
Solution
We know that
g´ = g - 2 cos2
At poles,  

2
So g´ = g for any value of
Hence, answer is (b).
.
Class Exercise - 9
A body of mass m is released from the
surface of earth into a tunnel passing
through the centre of the earth. What is
the kinetic energy of the particle as it
passes through the centre?
(a)
3
mgr
2
5
(c) mgr
2
(b)
1
mgr
2
1
(d) mgr
3
Solution
Potential energy at the surface
P.E. 
GMm
R
At the centre of earth it has
3 GMm
P.E.  
2 R
1
2
and K.E. = mv
2
Then
GMm 3 GMm 1 2

 mv
R
2 R
2
1 2 1 GMm 1
mv 
 mgR
2
2 R
2
Hence, answer is (b).
Class Exercise - 10
The gravitational field in a region is given
  
by E  10  i  j  N/kg . What is the work




done by an external agent to slowly shift
a particle of mass 2 kg from point (0, 0)
to point (5 m, 4 m)?
(a) –180 J
(b) 180 J
(c) 20 J
(d) –20 J
Solution
Work done
W  mE  (5iˆ  4ˆj  0iˆ  0ˆj)
 2  10(iˆ  ˆj)  (5iˆ  4 ˆj)
 20(5  4 )  180 Joule
Hence, answer is (b).
Thank you