13.

GRAVITATION

Gravity is of negligible importance in the interactions of elementary particles, but

primary importance in the interactions of large objects. It is gravity that holds


the universe together.

Newton's law of gravitation: “every particle of matter in the universe attracts

every other particle with a force that is directly proportional to the product of the


masses of the particles and inversely proportional to the square of the distance
between them.”

m1 m2 m 1 m2 2
Fα ∨F=G whereG=6.67 ×10 ˉⁱⁱ N −m /kg ²
r² r²
Important points regarding the gravitational force :
i. Always attractive force.


ii. Independent of the medium between the particles.

iii. Conservative in nature.
iv. Valid for point masses. However, for external points of spherical bodies the
whole mass can be assumed to be concentrated at its centre of mass.
v. Act along line joining the centre.
vi. Independent of other masses in the vicinity
vii. For continuous body system, integration to be used.

Gravity: The force by which earth attracts a body towards its centre. It is a
special case of gravitation.


 Vector Form of Newton's law of gravitation

Force on 1 by 2: ⃗
F 12= ⃗ | |
F 12 ^
F 12

⃗ Gm ₁ m₂ (⃗ r 2−⃗
r 1) Gm ₁ m₂(⃗ r 2−⃗ r 1)
F 12= . =
(|⃗ r 1|) ² |r⃗2−⃗
r 1| 3
r 2−⃗ (|⃗ r 1|)
r 2−⃗
⃗
F 12=−⃗
F 21
 Null Point: Position, where net force is zero. If two forces m and nm are
separated by a distance d, then net force on a mass m 0 placed somewhere in
between them is zero at a distance :
Gm m0 Gnm m0
F 1= 2
∧F 2=
x ( d−x )2
¿ for netforce zero , F 1=F 2
d
on solving we get , x=
1+√ n
Broken Symmetry: In symmetrical mass distribution, if some masses are
removed, then consider a + mass and –mass at the removed point and calculate


the net force on a particle at the centre of symmetry.

 Some concepts:

1
 i. Gravitational force on the mass ‘m’ due to a uniform rod of length ‘ l’ and
mass ‘M’ at a distance ‘a’ from it.

Elemenatl mass of length dx=dM = ( Ml ) dx

F=∫ dF=∫
G dM m
=∫
a +l G ( Ml ) dx m = G M m
2 2
x a lx a(a+l)

ii. Gravitational force on the mass ‘m0’ placed at perpendicular bi-sector at

distance ‘x’ from centre due to two masses each of m separated by distance
‘d’.

2 Gm m0 x 2
F total=2 F sin θ= . =
( d4² + x ²) ( d4² + x ² ) ( d
1/ 2

iii. Value of ‘x’ for maximum force on the mass ‘m 0’ placed at perpendicular bi-
sector at distance ‘x’ from centre due to two masses each of m separated by
distance‘d’.

(( )
2 Gm m0 x
d
.
)
3 /2
d²
+ x²
dF 4 d
For F max , =0→ =0 → x=±
dx dx 2√ 2
2 Gm m0 x 2 Gm m0 x 2 Gm m0 x
F= = =
( ) ( ) ( )
3 /2 3/ 2 3/ 2
d² 3 d² 3 d²
+ x² x +1 x +1
4 4 x² 4x²
2 Gm m 0
¿ → Force is ZERO at x=o∧F → 0 at x →
( )
3/ 2
2 d²
x +1
4 x²

iv. Splitting ratio of a mass M for maximum force between them when separated
by ‘d’.
Gm( M −m) dF M
F= ; for Maximum force =0 → m=
d² dm 2

v. 3 masses each of ‘m’ are kept the corners of an equilateral triangle of side ‘l’.

Force on any mass at a corner Force at Centroid

2
 vi. Circular Symmetry

F net=F + F+ √ 3 F = (2+√3)F, F net=(2+√ 3)F ,

where F= where
Gm m0 2Gm m0
2 F= 2

vii. Angular Velocity of three masses each of ‘m’ are kept the corners of an
R l

equilateral triangle of side ‘l’ and revolve around Centroid under mutual
interaction.

Gm ²
F t=√ 3 F=√ 3
l²
¿ this F t acts as centripetal force

√3 Gm2 =mr ω2 → √ 3 Gm2 =m l ω 2

l
2
l
2
√3
→ ω=√ ( 3 Gm
l³ )

Acceleration due to Gravity: It is a force of attraction between a planet and a

body. If two bodies of different masses are allowed to fall freely, they have the


same acceleration, i.e. if they are allowed to fall from the same height; they will
reach the earth simultaneously.
F GM 4 4
g= ∨g= = πGρR since M = ρπR ³
m R² 3 3

Variation in the Value of g : The value of g varies from place to place on the
surface of earth as we go above or below the surface of earth and depends on
following factors:

i. Shape of the Earth: Earth is not a perfect sphere. It is somewhat flat at the
two poles. The value of g is minimum at the equator and maximum at the

poles. g α
1
2
.
R

ii. Height above the Surface of the Earth: The value of acceleration due to gravity g
goes on decreasing as we go above the surface of earth. If h→
infinity, g’→0.

[ ]
−2

( )
2
' R h ' 2h
g =g =g 1+ ∨g ≈ g 1− if h ≪ ≪ R
( R +h ) 2
R R
3
 % change in ‘g’ for small change in height: we have

( )
'
g 2h g−g ' 2 h 2h
= 1− → = → %g= ∗100
g R g R R

iii. Depth below the Surface of the Earth: According

to Gauss theorem the gravitational force of
attraction on a mass inside a spherical shell is
always zero. Therefore, the object experiences
gravitational attraction only due to inner solid
sphere. The mass of this sphere is

[( ) ] ()
M 4 ( R−d ) ³. M
M’=
π ( R−d ) ³=
4 3 R³
πR ³
3
3
G M ' m GMm ( R−d ) GMm ( R−d )
F= 2
= 3 2
=
( R−d ) R ∗( R−d ) R3
F GM ( R−d ) GM (R−d )
→ = → g' = .
m R³ R² R
'
→ g =g .
(R−d )
R
d
( )'
=g 1− ; i. e . g < g
R
 g depth∈terms of density at a distance ' x ' ¿ centre of earth:
4 3
M =density∗volume=ρ πR ;
3
GM ( R−d ) G 4 4
gdepth = = ρ πR ³ x = πGρx , where x=R−d
R³ R³ 3 3
iv. Axial Rotation of the Earth: g' =g−Rω ² cos ² ∅

At Equator,∅ =0 , g' =g−Rω ²and at poles ∅ =90 º → g ' =g .

Pseudo force on the particle is mrω² in outward
direction.
True acceleration g is towards the centre O of the
earth. Thus,
effective acceleration g′ is the resultant of g and rω².
 Conclusion:
i. Effective value of g is not truly vertical passing through the centre O.
ii. Effect of centrifugal force due to rotation of earth is to reduce the effective
value of g.
iii. Latitude angle is measured from equator and co-latitude from poles.

 Variation in the value of g with r (the distance from the centre of earth)

For r ≤ R , g ' =g 1− ( dR ) as R−d =r ; g ∝r '

' g gR ² ' 1
For r ≥ R , g = = as R+ d=r ; g ∝
[ ] d
2
r² r²
1+
R

4
 Gravitational Field Strength (E) or Gravitational field intensity(I) : The space
around a mass or system of masses in which any other test mass experiences a


gravitational force is called gravitational field. When a test mass is moved from
one point to another point, some work is done by this gravitational force.
The force experienced on a unit test mass placed at a distance r in a gravitational
field is gravitational field intensity 0r strength.

⃗I = F = GMm = GM .
⃗
m R² m R²

Acceleration due to gravity g is also F/ m. Hence, for the earth’s gravitational

field g and E are same. The E versus r (the distance from the centre of earth)
graph is same as that of g′ versus r graph.

(i) Gravitational Field (E) due to a Uniform Circular Ring at some Point on its
Axis:

' M
Elementary mass ' d m =λdl= dl
2 πR
G dm G . λdl G λdl G M dl
dE= = = 2 2= ;
R +r 2 πR ( R2 +r 2 )
2 2
x x
2 πR
G M dl r
E=∫ dE cos θ= ∫ .
0 2 πR ( R +r ) √ ( R2 +r 2 )
2 2
cos θ=
r
2 πR √( R + r )
2 2

GMr GMr
¿
2 3 /2
∫ dl= 3/2
2 πR ( R +r ) ( R2 +r 2 )
2
0
R
This is directed towards the centre of the
sin θ=
√ ( R ²+r ² )
ring. It is zero at the centre of the ring and maximum at r =R/√2 (can be
obtained by putting dE/dr = 0).

The maximum value is GMAX =

2 GM
3 √3 R ²
Circular Symmetry: In a circular path maximum force occurs when a mass is kept

at a distance equals to from the centre of the circle. A point mass ‘mo’ is
Radius
√2
kept at a distance ‘x’ perpendicularly from Centroid of an equilateral triangle for
equal masses can be solved by assuming the vertices as points on circumference
of a circle and centre at Centroid. Similarly for square and other symmetrical
bodies also.

l Similarly for square

Here Radius=
√3 l
Radius= where l=side
∴ F=
3 GM mo x √2

(( ) )
2 3 /2
l 2 4 GM mo x
+r ∴ F=
√3
(( ) )
2 3/ 2
l 2
+r
R l √2
F max occur at x = = R l l
√ 2 √6 F max occur at x = = =
√ 2 √2∗2 2

5
 Some Concepts:

Arc (Linear Mass Density :λ) dm λ Rd ∅ λd∅

dI =G =G =G
R² R² R
θ/ 2
λd ∅
I =∫ dI cos ∅ = ∫ G cos ∅ d ∅
(sin Components cancel out )
−θ / 2 R

λ ( 2)
θ

I =G [−sin∅ ] −θ
R (2)
λ
¿ 2 G sin θ /2
R

Semi-circular wire (length : ꙥR ) Quarter Ring (length :ꙥR/2)

λ θ 2GM π λ θ 4 GM π
I =2 G sin = sin I =2 G sin = sin
R 2 πR 2
2 R 2 πR 2
4

¿
2GM 2 √ 2 GM
¿
πR ² πR ²

M λ π M π GM
λ 1= I 1=2G sin =2 G sin =
π .2 R 2R 2 2R .π 2R 2 2 πR ²

M λ π M π 2 GM
λ 2= I 2=2G sin =2G sin =
π .R R 2 R . πR 2 πR ²

λ θ 2 GM GM 3 GM ^
I =2 G sin I =I 2−I 1= − = −j
R 2 πR
2
2 π R 2 πR ²
2

Gravitational field strength due to thread: Linear Mass Density= λ

Horizontal Component:

| Gλ
r |
( cos θ 2−cos θ 1 )

Vertical Component:
Gλ
( sin θ 2+sin θ 1 )
r

Infinite Wire Semi-Infinite Wire

Gλ Gλ Gλ ⃗
E =⃗E x + ⃗y
E x= ( cos 90 º−cos 90 º )=0 E x= ( cos 0 º−cos 90 º )=
r r r Gλ
¿ √2
r
6
 Gλ 2 Gλ Gλ Gλ
E y= ( sin 90º +sin 90 º ) = E y= ( sin 0º +sin 90 º )=
r r r r

λ θ λ 90 √2 Gλ E=0
E=2G sin =2G sin =
R 2 R 2 r

E x 1=
Gλ
r
( sin 45 º +sin 90º )=
Gλ
r
1+
[
1
√2 ]
Gλ Gλ 1
E y1= ( cos 90 º +cos 45 º )=
r r √2

Can be considered Gλ Gλ 1
as two separate
E x 2= ( cos 90 º + cos 45 º ) =
r r √2
lines→
E y2=
Gλ
r
( sin 45º +sin 90 º ) =
Gλ
r
1+
[ 1
√2 ]
E=
Gλ 1
[
r √2
−1−
1
√2
iˆ +
]
Gλ 1
r √2
−1−
1
[
√2
jˆ=
Gλ
r ]
(−iˆ − jˆ ) → Magnitude=√ 2
Gλ
r

A tunnel is dug along a chord at a perpendicular distance R/2 from centre.

Tunnelis frictionless. Then the normal force exerted by wall of tunnel

√[( ) ]
2
4 4 R
g= πGρ r = πGρ +x ²
3 3 2

[ ]
R

√[( ) ] [( ) ]
2
4 R 2
N=mgCos α =m πGρ +x ² .

√
3 2 2
R
+x²
2
4 πGρmR 2 πGρmR
¿ =
3∗2 3

Gravitational Field due to a Point Mass: Force acting on a particle of mass m

placed at P.


, Therefore,
GM m F GM
F= E= =
r² m r²

7
 Shell Theorem: To find force between spherical
bodies, first draw a shell from where ‘E’ or ‘F’ to be


calculated and then take mass inside shell as point

mass and apply Newton’s law of gravitation.
Draw a shell with the point mass on the circumference and
the larger body at the centre of sphere and take whole mass
of larger body as point mass. Now apply Newton’s law.

(ii) Gravitational Field ( E) due to a Uniform Solid

Sphere:

a) External: May be treated as a single particle of same mass placed at its

centre for calculating the gravitational field at an external point.
Gdm G GM 1
dE= →∫ dE= ∫ dm→ E ( r )= for r ≥ R∨E ( r ) α
r² r² r² r²
Here, r is the distance of the point from the centre of the sphere and R the
radius of sphere.

b) Internal: The gravitational field at an internal point is proportional to the

distance of the point from the centre of the sphere. At the centre itself, it is

zero and at surface it is , where R is the radius of the sphere. ’d’ is the
GM

depth from surface and r =R−d

R²

Mass of sphere with radius ‘r’ is

[( ) ] ( 43 ) π ( R−d )
M
M’= ∫ dm=¿ =
3

4
π R3
3

[( ) ] ()
M 4 Mr ³
¿M’= 4
π ( r ) ³=
3 R³
πR ³
3

∴dE=
3
Gdm G Mr
2
→∫ dE= 2 ∫ 3
r r R
GM
→ E ( r )= r for r ≤ R∨E ( r ) α r
R³
(iii) Gravitational Field (E) due to a Uniform Spherical Shell:
a) External: May be treated as a single particle of same mass placed at its centre for
calculating the gravitational field at an external point

GM
E ( r )= 2
for r ≥ R ; for r=R ( at the surface )
r
GM 1
E= ∧otherwise E α
R² r²
b) Internal: The field inside a uniform spherical
shell is zero. i.e. field strength (E) = 0.The
reason for this zero field is that the vector
sums of fields due to each point on the shell cancel each other.

Gravitational Potential (V): A scalar quantity. Gravitational potential at any point is

defined as the negative of work done by gravitational force in moving a unit test


8
 mass fr=-om infinity (where potential is assumed to be zero) to that point, such that
no velocity imparted to the body.

Work done by External Force=−F ext dx

GMm
Here F ext = 2
x
−GMm
Small work , dW = 2
dx
x

[ ]
r r
−GMm −1
Work done ¿ move the particle ¿∞ ¿ ‘ r ’=W T =∫ 2
dx=−GMm
∞ x x ∞
1 1
¿ GMm − =
x ∞
GMm
r[ ]
We have W C +W nC + W Ext =K f −K i becomes here W C +0+ W Ext=0−0 → W C =−W Ext

−GMm
→ Work done by gravitation=
r

work Potential energy (U ) −GMm −GM

Potential= = = =
Mass(moving body) Mass(movingbody ) r∗m r

Note : Potential is w . r .t unit mass∧Potentional energy is w . r . t some mass .

We have, F=
−∂ U
∂r
=−
∂r[ (
∂ −GM m
r
=
∂r r
= )] (
∂ GM m −GM m
r² )
Potential due a number of sources is equal to scalar sum of individual potentials.

Here, negative sign implies that this force is towards ‘M’ or towards negative x-
direction (attractive force). This is a variable force (a function of x).

Gravitational Potential Difference: It is defined as work done to bring unit mass from one point to
another point without acceleration(Change in velocity)

Work done by External Force=−F ext dx

GMm
Here F ext = 2
x
−GMm
Small work , dW = 2
dx
x

[ ]
rf rf
−GMm −1
Work done ¿ move t h e particle ¿ r i ¿ ‘ r f ’=W T =∫ 2
dx=−GMm
ri x x ri

¿ GMm
[ 1 1
−
rf ri ]
We have W C +W nC + W Ext =K f −K i

becomes here W C +0+ W Ext=0−0 → W C =−W Ext =−GMm

[ 1 1
−
rf ri ]
→ W C =−[ U f −U i ] → W Ext =[ U f −U i ]

[
Note : At ∞ Potential V =
work
Mass ]
∧Potential Energy [ ∆ U ] are Zero

9
 (i) Potential due to a Uniform Solid Sphere:
a) Potential at some External Point: The gravitational potential due to a
uniform sphere at an external point is same as that due to a single particle
of same mass placed at its centre.

r
GM −GM
V =−∫ dE →−∫ 2
dr → V ( r )= for r ≥ R .
∞ r r
At the surface, r=R, V =
−GM

b) Potential at some Internal Point: At some internal point, potential at a distance r

R

from the centre is given by

[ ][ ]
r R r
V =−∫ −E dr= −∫ −Edr + −∫ −E dr
∞ ∞ R

[ ] [∫ ]
R r
→ v=GM ∫ r12 dr + GM
R
3
r dr
∞ R

R
→ V =GM [ ]
−1
r ∞+
2 R³
r
GM [ r ² ] →−
R
GM GM
+
R 2 R³
[ r ²−R ² ]

−GM
V ( r )= [ 3 R ²−r ² ]= −GM ( 1.5 R2−0.5 r 2 ) for r ≤ R
2 R³ R
3

At surface, r=R, V = ; while at r = 0,V =

−GM −1.5 GM
R R

(ii) Potential due to a Uniform Thin Spherical Shell

a) Potential at an External Point: A uniform spherical shell
may be treated as a point mass of same magnitude at its
centre.
−GM
V ( r )=
for r ≥ R
r
At the surface, r=R , V =
−GM

b) Potential at an Internal Point: Potential due to a uniform spherical shell is

R

constant at any point inside the shell and this is equal to.
−GM
R
Force inside a shell is zero. We have F=
−∂ U −∂ U
→ 0=
∂r ∂r
−GM
→ Potential energy =constant → Potential=Constant=
R
(iii) Potential due to a Uniform Ring at some Point
on its Axis: The gravitational potential at a
distance r from the centre on the axis of a
ring of mass M and radius R is given by

Here dM =λdl=
[ ]
M
2 πR
dl∧l=√ ( r ²+ R ² )

10
 [ ]
2 πR
−G dM GM dl
dV = → V =∫
R 0 2 πR √( r ²+ R ² )

=
−GM −GM
[ l ]20 πR = 2 πR
2 πR ( √ ( r ²+ R ² ) ) 2 πR ( √ ( r ²+ R ² ) )
−GM
V ( r )= 2 2
0≤r≤
√ (R +r )

At r =0,
−GM
V= :
R

;
GM
i .e . at the centre of the ring gravitational potential is−
R

 Potential at a distance a from one end of a uniform rod of length l and mass M.

Here dM =λ dx=
M
dx
l

−G dM −G λ dx −G M dx
dV = = =
x x lx
a=l
−G M
V=
l
∫ dxx =−Gl M [ log x ] a+l
a
=
−G M
l
[ (log ( a+l )−log ( a ) )]
a

GM a+l G M a
¿− log = log
l a l (a+l)

 Relation between Gravitational Field and Potential:

, (where U=potential energy, E is gravitational field and F is force.)

−∂ U
F=
∂r

(Where E is potential energy and V is potential)

−∂ (mV )
→⃗
F =m. E=
∂r
−∂ V
→ E=
Conversion of V Function into E Function:
∂r


(i) For more than one variable. E=−

∂ V ^ ∂ V ^ ∂V ^
i+
∂x ∂ y
j+
∂z
k
[ ]
(ii) Only one variable: In this case, E= =− or
−dv dv
dx dr

E=(−slope ofV −x )∨(−Slo pe ofV −r graph)

 Conversion of E Function into V Function :

(i) For more than one variable:V b −V a=−

b

∫ E . dr , where dr=dx i+dy

^ ^j+dz k^
a

(ii) Only one variable: V b −V a=−

b

∫ E dr
Gravitational Potential Energy: The change in potential energy (dU) of a system
a

corresponding to a conservative force is given by



11
 f rf rf
dU =−F . dr →∫ dU=−∫ F . dr → U f −U i =−∫ F . dr

We generally choose the reference point at infinity and assume potential energy
i ri ri

to be zero there, i.e. if we take r i=∞ (infinite) and U i=0, then we can write,
r
U =−∫ F . dr
(i) Gravitational Potential Energy of a two Particle System: The gravitational
❑

potential energy of two particles of masses m1 and m2 separated by a distance r

is given by,
−G m1 m 2
U=
r
This is actually the negative of work done in bringing those masses from infinity
to a distance ‘r’ by the gravitational forces between them.

(ii) Gravitational Potential Energy for a System of Particles: The gravitational

potential energy for a system of particles (say m1 , m2 , m3
and m4) is given by

U =−G
[ m ₄₃ m₂₄ m₄₁ m ₃₂ m₃₁ m₁₂
+ + + + +
r ₄₃ r ₂₄ r ₄₁ r ₃₂ r ₃₁ r ₁₂ ]
Example: For 3 particles placed corners of an equilateral
triangle

( )( , PE Total =)
−Gm m −Gm m −Gm m −3Gm m
PE1=0 , PE2= , PE 3= +
l l l l
(iii) Gravitational Potential Energy of a Body on Earth’s Surface: The
gravitational potential energy of mass m in the gravitational
field of mass M at a distance r from it is,
−GMm
U=
r

The earth behaves for all external points as if its mass M were
concentrated at its centre. Therefore, a mass m near earth’s surface may be considered
at a distance R (the radius of earth) from M. Thus, the potential energy of the system

will beU =
−GMm
R

(iv) Two particles m1 & m2 are at rest initially at infinite distance. Find their
relative velocity of approach when the separation is ‘d’ due to gravitational
attraction.

Initially PE=KE=0 ,∴ TEinitial =0( At infinite distance PE=0)

−G m1 m2 1 1
PEfinal= ∧KE final= m1 V 1 ²+ m2 V 2 ²
d 2 2

−G m 1 m 2 1 1
→ TEfinal= + m1 V 1 ²+ m2 V 2 ²
d 2 2

12
 = m1 V 1 ²+ m2 V 2 ² (since No External force)
Gm1 m2 1 1
TE final=TEinitial →
d 2 2

= ¿)→ (Since m 1 V 1 =m 2 V 2= p )
G m1 m2 p ² p² G m1 m2 p ² GMm p ²
→ = + → =
d 2m1 2 m2 d 2 d 2μ

P= (√ 2 Gmd m μ ) → m V =m V =√( 2dGm

1 2
(m +m ) )
1
²m ²
1 2 2
1

1
2

Example: Gravitational potential energy of system, each of mass m has radius are placed
at corners of square. Find velocity with which they collide when released from rest.
2
−G m (
PEinitial= 4 +√ 2 ) ,
L
2
−G m (
PEfinal= 4 + √ 2 )∧¿
2R

KE final=4 ( 12 m v )=2m v
2 2

2 2
Gm ( −G m (
4+ √ 2 ) = 4+ √ 2 )
2
TE final=TEinitial →2 m v −
2R L

→ v=
√ Gm ( 4 + √ 2 ) 1 1
2
−
2R L ( )
(v) Work done ¿ change the sha pe ofsystem of particles ¿ of
˚ radius R' . Each mass ‘ m
and separation is d=4R.

2 2 2 2 2 2 2
G m G m G m Gm Gm G m −13G m
PEinitial=0− − − − − − =
4R 8R 4 R 12 R 8 R 4R 12 R

2 2 2 2 2 2
Gm G m G m G m G m G m
PE final =0 – − − − − −
R √2 2 R R √2 R √2 R √ 2 2 R
(2 √ 2+1)G m
2
¿−
R

W ext =PEf −PEi=U f −U i = −(2 √ 2+1)G m - ( −13G m ¿= = −(24 √ 2+1)G m

2 2 2

R 12 R 12 R
A ring of mass ‘M’ and radius ‘R’ is fixed. A small mass ‘m’ is kept on an axis of
the ring and released from a distance ‘R’ from axis. Fined velocity when mass


reaches centre.
PEi= Mass * Potential of ring (at x=R) =
−GMm −GMm −GMm
= =
√ ( R ²+ x ² ) √ ( R ²+ R ² ) R √2
PEf= Mass * Potential of ring (at x=0) =
−GMm
R

TE f =TEi → PEi + KE i=PE f + KE f

13
 →
−GMm 1
R √2
= m v ²−
2
GMm
R
→ v=
GM (2−√ 2)
R √
(vi) Difference in Potential Energy (∆U):Let us find the difference in potential
energy in two positions. The potential energy when the mass is on the surface of
earth (at B) is

and
−GMm
Uʙ=
R
Potential energy when the mass m is at height h above the surface of earth (at A)
is,

(Uᴀ >Uʙ )
−GMm
Uᴀ=
R+h

∆ U =Uᴀ−Uʙ =
−GMm −GMm
R+ h
−
R [ ]
¿ GMm
[ 1
−
1
=
]
GMmh
R R+h R(R+ h)
GMmh mgh
and for h<<R, ∆ U =mgh
∆U= =
2
R 1+[ ][ ]
h
R
1+
h
R
With what minimum velocity, a ball must be thrown so that the ball escapes out
from the field of sphere


Potential inside sphere:

−GM
3
( 3 R 2−r 2 )
2R

( )
2
R −GM 2 R −11GM
¿ at r= , PE= 3
3R − =
2 2R 4 8R

P.E= Mass * potential= ¿−

11GMm
8R
Ei= mv ²− and Ef= PE + KE =0 at infinity
1 11GMm
2 8R
1
2
mv ²=
11GMm
8R
→v=
11GM
4R √
→ escape velocity
Maximum height attained by a particle: If a particle of mass m is projected vertically
upwards with a speed v and attains maximum height h, then by conservation of


mechanical energy, i.e. Decrease in kinetic energy = increase in gravitational

potential energy of particle. Then,

On solving this, we get

1 1 mgh
mv ²=∆ U → mv ²=
2 2 h
1+
R
v₂
h=
v₂
2 g−
R
Another Method:
GMm 1 GMm
TE f =PE f + KE f =– TE i=PEi + KE i= mv ²−
R+h 2 R
TE f =TEi (Since there is no disturbance force )

14
 1 GMm −GMm ½ mv ² 1 −1 v² 1 −1
mv ²− = → − = → − =
2 R R+h GMm R R+h 2GM R R+h
2
R v −2 GM −1 −2 GMR 2GMR
= → R+h= 2 → h= 2
−R
2 GMR R+h R v −2 GM 2 GM −R v
Example: Show that the work done to move a mass along 2x + 3y =5 is zero in
gravitational field of potential 2x + 3y.
−∂ V ⃗ 3
V =2 x +3 y → ⃗I = → I =−2 i^ −3 ^j → ⃗
F =m∗⃗I → ⃗
F =−2 m i^ −3 m ^j → slope=
∂r 2
5 2 −2
Line equation is2 x+ 3 y=5 →3 y=5−2 x → y= − x → slope=
3 3 3
3
∗−2
2
Produ ct of both slopes= =−1 , hence work done is zero
3
Binding Energy: Total mechanical energy (potential + kinetic) of a closed system is
negative. The modulus of this total mechanical energy is known as the binding


energy of the system. This is the energy due to which system is closed or different
parts of the system are bound to each other.
Suppose the mass m is placed on the surface of earth. The radius of the earth is R
and its mass is M. Then, the kinetic energy of the particle, K = 0 and the potential
energy is
U =−GMm/ R

Therefore, the total mechanical energy is, E= KE + PE = 0 – =–

GMm GMm
R R

Therefore, binding energy¿− . If minimum this much energy is given to the

GMm
R
particle in any form (normally kinetic) the particle goes out of the gravitational field
of earth.

Satellite Motion: The path of the satellites is elliptical with the centre of earth at a
focus. However the difference in major and minor axes is so small that they can be


treated as nearly circular.

a) Orbital Speed: The necessary centripetal force to the satellite is provided by
the gravitational force exerted by the earth on the satellite.

. Therefore, V0=
M v 0 ² GMm
=
r r²

√ √
GM
r
=
GM
R+h
→ v0∝
1
√r

Orbital speed of a satellite close to the earth’s

surface is

V0= ( since r=R at surface )

√ GM V
=√ gR= e
r √2
Substituting Escape velocity, Ve =11.2 km/s ⇒V0 = 7.9 km/s

b) Period of Revolution: The period of revolution (T ) is given by

15
 T=
2 πr
V0
∨T =

√
2 πr
GM
r
→ T =2 π
r³
GM √
=2 π
r³
gR ² √
as GM =gR ²

Time period of a satellite very close to earth’s surface (r≈R ): T =2 π

=84.6 min.
√ R
g

c) Angular Velocity (ω):

ω=
2π
T
=
2π
2π

√ √
r³
GM
√
=
GM
r 3
=
GM
( R+ h )
3

d) Energy of Satellite:

The potential energy of the system is U =

−GMm
r
The kinetic energy of the satellite is,
1
2
1
m v 0 ²= m
2
GM
r ( )
=
GMm
2r

The total energy is, E=K +U= .

−GMm
2r
This energy is constant and negative, i.e. the system is closed.
Farther the satellite from the earth the greater its total energy.

e) Parking Energy in any Orbit: If a satellite is projected from ground, such that it
becomes a revolving satellite of orbit (r =R +h), then

Initial energy = PE + KE =
−GMm 1 2
+ m v ∧¿
R 2
−GMm −GMm
Final Energy=Total Energy= =
2r 2( R +h)
v=√ 2GM ( 1=2 h )

f) Extra Energy to be given to change orbit :

−GMm
Initial energy=PE+ KE =TE= ∧¿
2r1
−GMm
Final Energy=PE+ KE=TE=
2r2
Extra Energy : E f −Ei
g) Escape Velocity (Ve): Minimum velocity needed to take a particle to
infinity ( PE=0 & KE = 0)from the gravitational field of earth is called the
escape velocity. Depends on mass & radius of the planet. On the surface of
earth its value is 11.2 km/s.Escape velocity is independent of the direction
in which it is projected

1 2 GMm 1 2 GMm
Ei =Ef → KE i+ PEi =KE f + PE f → mV e − =0+0 → m V e =
2 R 2 R

→ V e=
√ 2 GM
R [
=√ 2 gR=√ 2∗Orbital speed as g=
GM
R² ]
16
 If velocity of the particle is less than Ve then total mechanical energy is
negative and it does not escape to infinity.
If velocity of the particle is more than V e then total mechanical energy is
positive. Even at infinity some kinetic energy and speed are left in the particle.
Although its potential energy becomes zero.

1 2 GMm 1 2
Ei =Ef → KE i+ PEi =KE f + PE f → m v − = m V intersteller +0
2 R 2
1 1 1
→ m v − mV e = mV intersteller → V intersteller =√ v −V e
2 2 2 2 2
2 2 2

Geostationary satellite: Time period of the satellite is 24 h and will always appear to
be stationary from earth and orbit above the


equator from west to east.A single geostationary

satellite is on a line sight with about 40 percent of
the earth's surface. Three such satellites, each
separated by 120 degrees of longitude, can
provide coverage of the entire planet. The height is 36000 km from surface of earth
& 42400 km from centre of earth.
GMm
r²
=mrω ² → ω=
GM
r³
→
√2π
T
→T =2 π
r³
GM
=2 π
√
(R+ h)³
GM √
 Trajectory of a body projected from point A in the direction AB with different
initial velocities:Let a body be projected from point A with velocity vo in the
direction AB. For different values of v the paths are different. Here, are the
possible cases.

Total M.E= negative in case (i) to

(iv), Zero in case (v) & Positive in
case (vi)

(i) If v = 0, path is a straight line from A to M.

(ii) If 0 < v < v0, path is an ellipse with centre O of the earth as a focus.
(iii) If v = v0, path is a circle with O as the centre.
(iv) If v0 < v < ve, path is again an ellipse with O as a focus.
(v) If v = ve , body escapes from the gravitational pull of the earth and path is
a parabola
(vi) If v > ve , body again escapes but now the path is a hyperbola.

Kepler's Laws of Planetary Motion: The three laws are

(i) Law of elliptical orbits: Each planet moves in an elliptical orbit, with the sun at


one focus of the ellipse.

17
(ii) Law of areas: The radius vector, drawn from the sun to a planet, sweeps out equal
areas in equal time, i.e. its areal
velocity (or the area swept out by it per
unit time) is constant and gives the
relationship between the orbital speed
of the planet and its distance from the sun. This is actually the law of conservation
of angular momentum.
2
1 1 2 dA r dθ 1 2
Areaof triangle=dA= r rdθ= r dθ → = →= r ω
2 2 dt 2 dt 2
2 2
dA r ω m dA mr ω Angular Momentum L
= . → = = =
dt 2 m dt 2m 2m 2m
Gravitational force is a central force (passing through centre) and hence no
torque,
Therefore, Initial angular momentum=Final momentum, i.e. Li = Lf
(iii) Law of time period: The Square of the planet’s time period is proportional to the
cube of the semi-major axis of its orbit and gives the relationship between the
size of the orbit of a planet and its time of revolution.


GMm
r²
=mrω ² → T =2 π
Maximum and Minimum Velocity of a Planet:
r³
gR ² √
→T ∝ √ r ³ →T ² ∝r ³

Angular Momentum(L)=I ω=mr ² ω=mvr

→ mV min r max=mV max r min
→ V min r max =V max r min

Some Examples:
1. A test particle (mass =1 kg) is in a uniform circular motion of radius ‘R’ in a


central force inversely proportional to nth power of R. then time period is

proportional to.

→ F= =mrω ² , (Where ‘k’ is constant & mrω² is centripetal

1 k
Given F ∝
Rⁿ Rⁿ
force)

Concept
→
√ k
m . Rⁿ ⁺¹
=ω=
2π
T
( )
→T ∝ ( R ) n+1 / 2

2. An electron is rotating in a circular orbit of potential energy(PE)=k log r.find

velocity of electron.
Given PE=U =k log r
−∂ U −∂ ( k log r ) −k m v k
2
F= → F= → F= → =
∂r ∂r r r r

→ v=
m
k
√
3. Volume calculation of a solid sphere (radius R):
We take a shell as an elementary element of thickess ‘ dr ’ ,then
Volume of the element=Circular area x Thickness → dV =4 πr ². dr
R
4
Volume of solid sphere , V =∫ dV =∫ 4 πr ² dr= πr ³
0 3

18
 4. A test particle is moving in a circular orbit in the gravitational field produced
by mass density ρ(r) = ρ=k/r², . Identify the correct relation between the
radius R of the particle’s orbit and its period T.
Volume of shell=Curved surface area∗thicknes=4 πr ². dr
dM =ρ dV = 4πr².dr =4k π dr
k
r²
R
M =∫ dM =∫ 4 k π dr =4 k π R
0

(4k π R)m 1 2π 1 T
F=G =mRω ² → ω ² ∝ 2 → = → =2 π =Constant
R² T R R
13.11 Summary:
R

Parameter Formula At surface of earth

Gravitational Force Gm1 m2 GM m
F= F=
R²
Acceleration due to Gravity
r²
F GM
g= =
Gravitational Field
m R²

Gravitational Field intensity

F GM F GM
E= = E= =
Gravitational Field strength
m r² m R²
Gravitational potential −GM −GM
V= V=
Gravitational potential
r R

energy
−G m1 m 2 −G M m
U= =−W U=
R
Gravitational
r

19