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GRAVITATION
(For V SEMESTER)
(a) Newton’s 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 inversely
proportional to the square of the distance between their centres.
uur
Gm1m2 $
Force on m2 due to m1 in vector form is F21 = −
r 12 .
r122
In general F =
Gm1m2
.
r2
(b) Gravitational Field: Gravitational field intensity at a point is defined as gravitational
force experienced by a unit mass kept at that point.
ur F ur
[ E → intensity at P on m2 ]
E=
m2
F=−
Gm1m2 $
Gm
r 12 ⇒ E = − 2 1 r$ 12
2
r12
r12
(c) Gravitational potential: Gravitational potential at a point is the potential energy that a
unit mass would possess at the point in the gravitational field.
Potential difference ( dv ) between points A and B dV = − E dr .
Gravitational intensity E = −
dV
.
dr
Also gravitational field is the negative potential gradient at a point is V = −
Gravitational potential energy of a body at a point, PE = −
Gm
.
r
Gm1 m2
.
r
Expression for gravitational potential and field at a point due to a solid sphere:
(a) Point P outside the sphere:
Consider a solid sphere of mass M , radius R
and density ρ . Consider it to be made of large
number of thin spherical shells whose radii
vary from O to R . Let x be radius of one such
shell and dx its thickness.
Mass of the shell = ( 4 πx 2 dx ) ρ
(m = V ρ where V = Area × thickness= 4πx2 × dx )
The gravitational potential at point P due to shell is
G ( 4 πx 2 dx ) ρ
Gm
G 4 πρ 2
dV = −
=−
=−
x dx
a
a
a
The potential due to the entire sphere at P is
GRAVITATION
R
G 4 πρ 2
G 4 πρ  x 2 
V =−
x
dx
=
−
a ∫0
a  2  ο
R
R
V =−
G 4 πρ  x 3 
G 4 πρ 3
=−
R
a  3  ο
3a
V =−
G 4 3 
GM
4
πR ρ  or V = −
where M = πR 3 ρ

a 3
a
3

Gravitational field:
E=−
dV
dV
d  GM 
GM
=−
= − −
=− 2

dr
da
da  a 
a
E=−
GM
a2
(b) Point P on the surface of the sphere:
V =−
GM
GM
[Q a = R ] and E = − R 2  a2 = R 2 
R
(c) Point P inside the sphere:
Let P be a point inside the sphere of radius R on a imaginary spherical surface of radius
b . The gravitational potential at P due to all shells of radii between R and b (utside
point P ) s
R
V1 = − ∫
G ( 4 πx 2 dxρ )
x
b
mass of a shell = 4 πx 2 dxρ ]
R
G 4 πρ 2
 x2 
 R − b 2  --------- (1)
V1 = − G 4 πρ∫ x dx = −G 4 πρ   = −
2
2
 b
b
R
The potential at P due to shells within the sphere of radius b is
V2 = −
G 4 3 
π b ρ  ----------- (2)
b  3

4
[ πb 3ρ = m = mass of sphere of radius b ]
3
The potential at P due to entire sphere is
V = V1 + V2 = −
G 4 πρ 2
4
 R − b 2  − G πρb 2
2
3
 R2 − b2 b2 
 R2 b2 b2 
V = − G 4 πρ 
+  = − G 4 πρ 
− + 
3
2
3
 2
 2
G 4 πρ
 3R2 − b 2 
 3 R 2 − b 2  ------------- (3)
or V = −
V = − G 4 πρ 

6
6


or V = −
V =−
G
2 R3
4 3 
2
2
2
 3 πR ρ ( 3 R − b ) [by multiplying and dividing equation (3) by R ]
GM
 3 R 2 − b 2 
3 
2R
VIJAYA COLLEGE
4 3 

Q M = 3 πR ρ 
Page 2
GRAVITATION
Gravitational field:
E=−
E=
dV
d  GM

= − − 3 ( 3R 2 − b2 )
db
db  2 R

GM d
GM
3R 2 − b2 ) =
( −2b )
(
3
2 R db
2 R3
or
E=−
GMb
R3
Kepler’s Laws of Planetary Motion:
I law – Law of orbits: Every planet moves in an elliptical orbit with sun being at one of its foci.
II Law – Law of areas: The radius vector drawn from the sun to the planet sweeps equal areas in
equal intervals of time. i.e., the area velocity is constant.
III Law – Law of periods: The square of the time period of revolution of a planet around the sun
is proportional to the cube of the semi – major axis of the elliptical orbit.
Derivation of Kepler’s Second Law:
From Newton’s Law of gravitation
ur
GMm
F = − 2 r$ ------------- (1)
r
M → mass of sun
m → mass of planet
r → distance between them
r
ur
r
dv
The force experienced by the planet is F = ma = m
-------------- (2)
dt
r
r
r
dv
GMm $
r
$
$
From (1) and (2) m
= − 2 r as r = rr ⇒ r = .
dt
r
r
r
dv
GM r
∴
=− 3 r
dt
r
r
Taking left cross product on both sides of this equation with r .
r
r dv
GM r r
d r r
GM r r
r×
= − 3 r × r or
r ×v = − 3 r ×r
dt
r
dt
r
r r
r
r
r r
d
As r × r = 0
∴
r×v =0
or
r × v = constant
dt
r
r
r dr
 r dr 
= constant
Also r ×
Q v = 
dt
dt 

(
)
(
(
)
(
)
)
Multiplying and dividing LHS by 2, we get
r
 1  r dr  
2   r ×   = constant --------- (1)
dt  
 2 
VIJAYA COLLEGE
Page 3
GRAVITATION
In a interval of time ∆t , the planet covers an area ∆A , then from the diagram, area SPP′ is
ur 1 r
r
∆ A = r × ∆r .
2
ur
r
r
r
r 1  r dr 
∆A
1  ∆r 
The area velocity of the planet h = lim =
= lim  r ×
 or h = r ×  ---------- (2) or
∆t →0
∆t ∆ t → 0 2 
dt 
2  dt 
r 1 r r
h =  r × v 
2
(
)
Comparing (1) and (2) we get h =
1
× constant or h = constant.
2
Thus area velocity of the planet is a constant. Hence IInd law.
Derivation of Kepler’s First Law:
From second law, areal velocity = constant
r 1 r r
r r r
or h = r × v ⇒ 2 h = r × v
2
(
)
(
r
As r = r r$
)
r
 r dr d

=
r ⋅ r$ 
Q v =
dt dt


r
d
∴ 2 h = r r$ +
r ⋅ r$
dt
( )
( )
r
 dr$ dr 
2 h = r r$ × r + r$ 
 dt dt 
r
dr$
dr $ $
2 h = r 2 r$ × + r
r×r
dt
dt
(
)
r 2  $ dr$ 
 r ×  ---------- (1)
2 
dt 
r
Taking right cross product on both sides of (2) with h
r
r
r
 dv

GM $
dv r
GM
× h = − 2 r$ × h − − − − − (3)
 = − 2 r − − − − − (2) ⇒
r
dt
r
 dt

r
Substituting for h from (1) in RHS of equation (3)
r
dv r
GM
r 2  dr$ 
GM  $  $ dr$  
× h = − 2 r$ ×  r$ ×  = −
r ×  r ×   --------- (4)
dt
r
2
dt 
2  
dt  
ur ur ur
ur ur
ur ur ur
From vector identity A × B × C = A ⋅ C B − A ⋅ B C
r
dr$
2 h = r 2 r$ ×
dt
Q r$ × r$ = 0 


or
(
h=
) (
r
r  r dr   r dr$ 
dr$
∴ r ×  r ×  =  r ⋅  r$ − r$ ⋅ r$
dt   dt 
dt

r
dr
As r$ is perpendicular to
(velocity)
dt
r
r  r dr 
dr$
∴ r ×r ×  = −
dt 
dt

) (
)
( )
VIJAYA COLLEGE
r
r dr
∴ r ⋅ = 0, r$ .r$ = 1
dt
Page 4
GRAVITATION
r
dv r
GM  dr$ 
Thus equation (4) becomes
×h = −
− 
dt
2  dt 
r
dv r GM dr$
d r r GM dr$
or
×h =
or
v× h =
------------ (5)
dt
2 dt
dt
2 dt
(
)
Integrating equation (5)
r r GM
r
v×h =
r$ + c
2
r
Taking dot – product on both sides of the above equation with r , we get,
r r r GM r
r r
r ⋅ v×h =
r ⋅ r$ + r ⋅ c
2
GM $ $ $
rr ⋅ r + rc cos θ
=
2
r r r GM
r ⋅ v×h =
r + rc cos θ ----------- (6)
2
ur ur ur ur ur ur ur ur ur
using vector identity A ⋅ B × C = B ⋅ C × A = C ⋅ A × B
(
)
(
)
( ) ( )
r r r r r r
r ⋅ ( v × h) = h ⋅ (r × v)
r
= h ⋅ ( 2h )
(
)
r r
Q 2 h = r × v 


= 2h2
Thus equation (6) can be written as
2h2 =
GM
1 GM c cos θ
 GM

r + r c cos θ = r 
+ c cos θ  or = 2 +
2
r 4h
2h2
 2

Multiplying throughout by
4h 2
GM
4h 2
4h 2
2c
 2C 
=1+
,e =
 cos θ let l =
GMr
GM
GM
M
Then
l
= 1 + e cos θ ----------- (7)
r
This is the equation of a conic section with l → semi latus rectum and e → eccentricity. If
θ < 1 , then the above equation is the equation of ellipse with sun at one of its foci. This
proves Kepler’s first law.
Derivation of Kepler’s third law:
The path of the planet with sun at one of the foci is as shown
a → semi major axis, b → semi minor axis, l → semi latus
rectum of the ellipse.
For an ellipse b 2 = a 2 ( 1 − e 2 ) …….(1) and l = a ( 1 − e 2 ) ---------- (2)
Dividing (2) / (1)
l 1
=
b2 a
VIJAYA COLLEGE
or
l=
b2
a
Page 5
GRAVITATION
As l =
4h 2
4h 2 b 2
⇒l=
=
GM
GM a
or
b2 =
4h 2 a
----------- (3)
GM
The period of revolution of the planet round the sun is T is given by T =
h=
[Q areal velocity
or
T2 =
πab
.
h
area of ellipse
period of revolution
πab
=
T
]
π2 a 2 b 2
------------- (4)
h2
substituting for b 2 from (3) in (4)
T2 =
or
π2 a 2  4 h 2 a  4 π2 a 3
=
h 2  GM  GM
T 2 4π2
=
= constant
a3 GM
T 2 ∝ a3
or
Hence the Kepler’s third law.
Inertial mass and gravitational mass:
ur
r
If a force F acts on a body of mass m , and the acceleration of the body is a , then from
ur
r
Newton’s second law F = mi a , where mi is called the inertial mass.
This body is also acted upon by gravitational force. Then the mass of the body is called
gravitational mass mg . The gravitational force on the body by the earth is
ur GM e mg
F=
--------- (1)
R 2e
From Newton’s second law F = mi g ---------- (2)
Comparing (1) and (2)
mi g =
GM e mg
R
2
e
or
mi GM e
=
----------- (3)
mg
gRe2
At a place, RHS of above equation is a constant.
∴ ratio
mi
= K = constant
mg
If K = 1 , then G = 6.67 × 10 −11 Nm2 kg −2 . For any other value of K , G is different. Thus
mi = m g .
Thus inertial mass and gravitational mass are equivalent.
Satellite motion: Objects that move around a planet is called a satellite. Moon is a natural
satellite of earth. Man made satellite that revolve around the planets are called artificial
satellites.
VIJAYA COLLEGE
Page 6
GRAVITATION
Elements of satellite motion:
1. Orbital velocity: The velocity with which a satellite revolves round a planet is called
orbital velocity. Consider a satellite of mass m , moving around the earth in an orbit of
radius r . Let M be the mass of the earth and vο be the orbital velocity of the satellite.
The necessary centripetal force acting on the satellite is provided by the gravitational
force of attraction on the satellite.

mvο2
F
=
 c
r

Fc = Fg
i.e.,
mvο2 GMm
GM
=
⇒ vο2 =
r
r
r
Fg =
GMm 
r 2 
vο =
or
GM
--------- (1)
r
If g → acceleration due to gravity on the satellite near the surface of earth, then
GMm
mg =
.
R2
gR 2 = GM ----------- (2)
or
Substituting for GM from (2) in (1), vο =
gR 2
r
If the satellite is close to earth, then r = R (radius of earth)
vο = gR
2. Time period of the satellite orbit:
The orbital velocity of the satellite is vο =
GM
---------- (1)
r
If ω is the angular velocity of the satellite, then vο = r ω -------- (2)
Comparing (1) and (2) r ω =
ω=
or
Thus
GM
r
GM
2π
by definition ω =
where T is the time period of the satellite.
r3
T
2π
GM
r3
=
or T = 2 π
3
T
r
GM
If h is the height of the satellite from the surface of earth and R → radius of earth, then
r =R+h
Thus
As GM = gR 2
T = 2π
T = 2π
( R + h )3
GM
(R + h)
3
gR 2
If the satellite is very near surface of earth h << R
∴ T = 2π
R
g
VIJAYA COLLEGE
Page 7
GRAVITATION
Escape velocity: It is defined as the minimum velocity with which an object has to be
projected from the surface of the planet so that it escapes the planet’s gravitational force of
attraction.
If m is mass of an object projected from surface of earth, and M → mass of earth, R →
radius of earth and v is the velocity of projection, then total energy of the object at the
surface of the earth (just after projection) is
TE = ( KE + PE )r = R =
where KE =
1
GMm
----------- (1)
mv 2 −
2
R
1
−GMm
mv 2 and PE =
2
R
If ve is the velocity of the object just enough to take the object to infinity ( v = ve ) , as
both kinetic and potential energies are zero at infinity, then
TE = ( KE + PE )r =∞ = 0 --------- (2)
From the law of conservation of energy (1) = (2)
GMm
1
mve2 −
=0
2
R
or
ve2 =
2GM
R
or
or
ve =
1
GMm
mve2 =
2
R
2GM
R
In general ve =
2GM
r
The condition for launching artificial satellites:
(1) When v = vο =
v>
(2) When
ve =
(3) When v = ve =
(4) When v >
GM
⇒ Satellite goes in a circular orbit.
r
GM
r
but
less
than
escape
velocity
2GM
, then satellite goes in a elliptical path.
r
2GM
, the satellite takes a parabolic path and escapes to infinity.
r
2GM
, the satellite takes a hyperbolic path and escapes to infinity.
r
Geostationary satellite: A satellite that appears to be fixed at a position above a certain
distance from earth having same period of rotation of earth (24 hr) is called geo stationary
satellite.
The satellite rotates in the plane of the equator with the same rotation as that of earth and
thus appears to be stationary to an observer on the earth.
The angular velocity of the satellite is ω =
2π
.
T
where T → time period of geo stationary satellite.
As T = 24 hrs = 86,400 sec, ω =
VIJAYA COLLEGE
2π
= 7.3 × 10 −5 rad s −1 .
86, 400
Page 8
GRAVITATION
The orbital velocity vο =
As vο = r ω =
GM
.
r
GM
GM
GM
⇒ω=
or ω2 = 3
3
r
r
r
1
GM
or r = 2
ω
3
or
 GM  3
r= 2 
 ω 
G = 6.67 × 10 −11 Nm2 kg −2 , M = 5.98 × 10 24 kg and ω = 7.3 × 10 −5 rad s −1
1
 6.67 × 10 −11 × 5.98 × 10 24  3

r=
−5 2


7.3
×
10
(
)


r = 42 × 106 m
The height of the geo stationary satellite above the surface of the earth is given by
h=r−R
[Q r = R + h ]
R → radius of earth
R = 6.37 × 106 m
h =  42 × 10 6 − 6.37 × 106 
h = 35.63 × 106 m
VIJAYA COLLEGE
or
h = 35,630 km
Page 9