Gravitation | Physics Optional Notes for UPSC PDF Download

 Page 1


GRAVITATION 
THE DISCOVERY OF THE LAW OF GRAVITATION 
The way the law of universal gravitation was discovered is often considered the paradigm 
of modern scientific technique. The major steps involved were 
? The hypothesis about planetary motion given by Nicolaus Copernicus (1473-
1543). 
? The careful experimental measurements of the positions of the planets and the 
Sun by Tycho Brahe (1546-1601). 
? Analysis of the data and the formulation of empirical laws by Johannes Kepler 
(1571-1630). 
? The development of a general theory by Isaac Newton (1642-1727). 
NEWTON'S LAW OF GRAVITATION 
It states that everybody in the universe attracts every other body with a force that is 
directly proportional to the product of their masses and is inversely proportional to the 
square of the distance between them. 
F?m
1
 m
2
 and F?
1
r
2
 
so F?
m
1
 m
2
r
2
 
 
?F=-
Gm
1
 m
2
r
2
r ˆ[G= Universal gravitational constant ] 
{Note: This formula is only applicable for spherical symmetric masses or point masses.} 
VECTOR FORM OF NEWTON'S LAW OF GRAVITATION: 
Let  ???
12
= Displacement vector from ?? 1
 to ?? 2
 
r ?
21
= Displacement vector from m
2
 to m
1
 
F
??
21
= Gravitational force exerted on m
2
 by m
1
 
F
??
12
= Gravitational force exerted on m
1
 by m
2
 
Page 2


GRAVITATION 
THE DISCOVERY OF THE LAW OF GRAVITATION 
The way the law of universal gravitation was discovered is often considered the paradigm 
of modern scientific technique. The major steps involved were 
? The hypothesis about planetary motion given by Nicolaus Copernicus (1473-
1543). 
? The careful experimental measurements of the positions of the planets and the 
Sun by Tycho Brahe (1546-1601). 
? Analysis of the data and the formulation of empirical laws by Johannes Kepler 
(1571-1630). 
? The development of a general theory by Isaac Newton (1642-1727). 
NEWTON'S LAW OF GRAVITATION 
It states that everybody in the universe attracts every other body with a force that is 
directly proportional to the product of their masses and is inversely proportional to the 
square of the distance between them. 
F?m
1
 m
2
 and F?
1
r
2
 
so F?
m
1
 m
2
r
2
 
 
?F=-
Gm
1
 m
2
r
2
r ˆ[G= Universal gravitational constant ] 
{Note: This formula is only applicable for spherical symmetric masses or point masses.} 
VECTOR FORM OF NEWTON'S LAW OF GRAVITATION: 
Let  ???
12
= Displacement vector from ?? 1
 to ?? 2
 
r ?
21
= Displacement vector from m
2
 to m
1
 
F
??
21
= Gravitational force exerted on m
2
 by m
1
 
F
??
12
= Gravitational force exerted on m
1
 by m
2
 
F
??
12
=-
?? m
1
 m
2
r
21
2
r ˆ
21
=-
?? m
1
 m
2
r
21
3
r ?
21
 
 
Negative sign shows that: 
(i) The direction of ???
12
 is opposite to that ??ˆ
21
 
(ii) The gravitational force is attractive in nature 
Similarly F
??
21
=-
Gm
1
 m
2
r
12
2
r ˆ
12
 or F
??
21
=-
Gm
1
 m
2
r
12
3
r ?
12
?F
??
12
=-F
??
21
 
The gravitational force between two bodies are equal in magnitude and opposite in 
direction. 
GRAVITATIONAL CONSTANT "G" 
? Gravitational constant is a scalar quantity. 
? Unit:   S I: G=6.67×10
-11
 N-m
2
/kg
2
 
CGS: 6.67×10
-8
 dyne- -cm
2
/g
2
 
Dimensions: M
-1
 L
3
 T
-2
 
? Its value is the same throughout the universe, G does not depend on the nature 
and size of the bodies, it also does not depend upon the nature of the medium 
between the bodies. 
? Its value was first found out by the scientist "Henry Cavendish" with the help of 
the "Torsion Balance" experiment. 
? Value of G is small therefore gravitational force is weaker than electrostatic and 
nuclear forces. 
Example. Two particles of masses 1 kg and 2 kg are placed at a separation of 50 cm . 
Assuming that the only forces acting on the particles are their mutual gravitation, find 
the initial acceleration of heavier particle. 
Solution: Force exerted by one particle on another ?? =
Gm
1
 m
2
r
2
=
6.67×10
-11
×1×2
(0.5)
2
=5.3×
10
-10
 N 
Acceleration of heavier particle =
?? m
2
=
5.3×10
-10
2
=2.65×10
-10
 ms
-2
 
This example shows that gravitation is very weak but only this force keep bind our solar 
system and also this universe, all galaxies, and another interstellar system. 
Page 3


GRAVITATION 
THE DISCOVERY OF THE LAW OF GRAVITATION 
The way the law of universal gravitation was discovered is often considered the paradigm 
of modern scientific technique. The major steps involved were 
? The hypothesis about planetary motion given by Nicolaus Copernicus (1473-
1543). 
? The careful experimental measurements of the positions of the planets and the 
Sun by Tycho Brahe (1546-1601). 
? Analysis of the data and the formulation of empirical laws by Johannes Kepler 
(1571-1630). 
? The development of a general theory by Isaac Newton (1642-1727). 
NEWTON'S LAW OF GRAVITATION 
It states that everybody in the universe attracts every other body with a force that is 
directly proportional to the product of their masses and is inversely proportional to the 
square of the distance between them. 
F?m
1
 m
2
 and F?
1
r
2
 
so F?
m
1
 m
2
r
2
 
 
?F=-
Gm
1
 m
2
r
2
r ˆ[G= Universal gravitational constant ] 
{Note: This formula is only applicable for spherical symmetric masses or point masses.} 
VECTOR FORM OF NEWTON'S LAW OF GRAVITATION: 
Let  ???
12
= Displacement vector from ?? 1
 to ?? 2
 
r ?
21
= Displacement vector from m
2
 to m
1
 
F
??
21
= Gravitational force exerted on m
2
 by m
1
 
F
??
12
= Gravitational force exerted on m
1
 by m
2
 
F
??
12
=-
?? m
1
 m
2
r
21
2
r ˆ
21
=-
?? m
1
 m
2
r
21
3
r ?
21
 
 
Negative sign shows that: 
(i) The direction of ???
12
 is opposite to that ??ˆ
21
 
(ii) The gravitational force is attractive in nature 
Similarly F
??
21
=-
Gm
1
 m
2
r
12
2
r ˆ
12
 or F
??
21
=-
Gm
1
 m
2
r
12
3
r ?
12
?F
??
12
=-F
??
21
 
The gravitational force between two bodies are equal in magnitude and opposite in 
direction. 
GRAVITATIONAL CONSTANT "G" 
? Gravitational constant is a scalar quantity. 
? Unit:   S I: G=6.67×10
-11
 N-m
2
/kg
2
 
CGS: 6.67×10
-8
 dyne- -cm
2
/g
2
 
Dimensions: M
-1
 L
3
 T
-2
 
? Its value is the same throughout the universe, G does not depend on the nature 
and size of the bodies, it also does not depend upon the nature of the medium 
between the bodies. 
? Its value was first found out by the scientist "Henry Cavendish" with the help of 
the "Torsion Balance" experiment. 
? Value of G is small therefore gravitational force is weaker than electrostatic and 
nuclear forces. 
Example. Two particles of masses 1 kg and 2 kg are placed at a separation of 50 cm . 
Assuming that the only forces acting on the particles are their mutual gravitation, find 
the initial acceleration of heavier particle. 
Solution: Force exerted by one particle on another ?? =
Gm
1
 m
2
r
2
=
6.67×10
-11
×1×2
(0.5)
2
=5.3×
10
-10
 N 
Acceleration of heavier particle =
?? m
2
=
5.3×10
-10
2
=2.65×10
-10
 ms
-2
 
This example shows that gravitation is very weak but only this force keep bind our solar 
system and also this universe, all galaxies, and another interstellar system. 
Example. Two stationary particles of masses ?? 1
 and ?? 2
 are at a distance 'd' apart. A 
third particle lying on the line joining the particles, experiences no resultant 
gravitational forces. What is the distance of this particle from?  
Solution: The force on ?? towards ?? 1
 is ?? 1
=
?? ?? 1
?? ?? 2
 
 
The force on ?? towards ?? 2
 is ?? 2
=
GM
2
 m
( d-r)
2
 
According to question net force on ?? is zero i.e. ?? 1
=?? 2
 
?
GM
1
 m
r
2
=
GM
2
 m
( d-r)
2
?(
d-r
r
)
2
=
M
2
M
1
 
?
d
r
-1=
vM
2
vM
1
 ?r=d[
vM
1
vM
1
+vM
2
] 
Example. Three particles, each of mass ?? , are situated at the vertices of an equilateral 
triangle of side ' ?? '. The only forces acting on the particles are their mutual gravitational 
forces. It is desired that each particle moves in a circle while maintaining their original 
separation 'a'. Determine the initial velocity that should be given to each particle and 
time period of circular motion. 
 
Solution: The resultant force on particle at A due to other two particles is 
F
A
=vF
AB
2
+F
AC
2
+2 F
AB
F
AC
cos 60
°
=v3
Gm
2
a
2
… (i) [?F
AB
=F
AC
=
Gm
2
a
2
] 
Radius of the circle r=
2
3
×asin 60
°
=
a
v3
 
If each particle is given a tangential velocity v, 
So that F acts as the centripetal force, 
 Now 
?? ?? 2
?? =v3
?? ?? 2
?? (???? )
 
Page 4


GRAVITATION 
THE DISCOVERY OF THE LAW OF GRAVITATION 
The way the law of universal gravitation was discovered is often considered the paradigm 
of modern scientific technique. The major steps involved were 
? The hypothesis about planetary motion given by Nicolaus Copernicus (1473-
1543). 
? The careful experimental measurements of the positions of the planets and the 
Sun by Tycho Brahe (1546-1601). 
? Analysis of the data and the formulation of empirical laws by Johannes Kepler 
(1571-1630). 
? The development of a general theory by Isaac Newton (1642-1727). 
NEWTON'S LAW OF GRAVITATION 
It states that everybody in the universe attracts every other body with a force that is 
directly proportional to the product of their masses and is inversely proportional to the 
square of the distance between them. 
F?m
1
 m
2
 and F?
1
r
2
 
so F?
m
1
 m
2
r
2
 
 
?F=-
Gm
1
 m
2
r
2
r ˆ[G= Universal gravitational constant ] 
{Note: This formula is only applicable for spherical symmetric masses or point masses.} 
VECTOR FORM OF NEWTON'S LAW OF GRAVITATION: 
Let  ???
12
= Displacement vector from ?? 1
 to ?? 2
 
r ?
21
= Displacement vector from m
2
 to m
1
 
F
??
21
= Gravitational force exerted on m
2
 by m
1
 
F
??
12
= Gravitational force exerted on m
1
 by m
2
 
F
??
12
=-
?? m
1
 m
2
r
21
2
r ˆ
21
=-
?? m
1
 m
2
r
21
3
r ?
21
 
 
Negative sign shows that: 
(i) The direction of ???
12
 is opposite to that ??ˆ
21
 
(ii) The gravitational force is attractive in nature 
Similarly F
??
21
=-
Gm
1
 m
2
r
12
2
r ˆ
12
 or F
??
21
=-
Gm
1
 m
2
r
12
3
r ?
12
?F
??
12
=-F
??
21
 
The gravitational force between two bodies are equal in magnitude and opposite in 
direction. 
GRAVITATIONAL CONSTANT "G" 
? Gravitational constant is a scalar quantity. 
? Unit:   S I: G=6.67×10
-11
 N-m
2
/kg
2
 
CGS: 6.67×10
-8
 dyne- -cm
2
/g
2
 
Dimensions: M
-1
 L
3
 T
-2
 
? Its value is the same throughout the universe, G does not depend on the nature 
and size of the bodies, it also does not depend upon the nature of the medium 
between the bodies. 
? Its value was first found out by the scientist "Henry Cavendish" with the help of 
the "Torsion Balance" experiment. 
? Value of G is small therefore gravitational force is weaker than electrostatic and 
nuclear forces. 
Example. Two particles of masses 1 kg and 2 kg are placed at a separation of 50 cm . 
Assuming that the only forces acting on the particles are their mutual gravitation, find 
the initial acceleration of heavier particle. 
Solution: Force exerted by one particle on another ?? =
Gm
1
 m
2
r
2
=
6.67×10
-11
×1×2
(0.5)
2
=5.3×
10
-10
 N 
Acceleration of heavier particle =
?? m
2
=
5.3×10
-10
2
=2.65×10
-10
 ms
-2
 
This example shows that gravitation is very weak but only this force keep bind our solar 
system and also this universe, all galaxies, and another interstellar system. 
Example. Two stationary particles of masses ?? 1
 and ?? 2
 are at a distance 'd' apart. A 
third particle lying on the line joining the particles, experiences no resultant 
gravitational forces. What is the distance of this particle from?  
Solution: The force on ?? towards ?? 1
 is ?? 1
=
?? ?? 1
?? ?? 2
 
 
The force on ?? towards ?? 2
 is ?? 2
=
GM
2
 m
( d-r)
2
 
According to question net force on ?? is zero i.e. ?? 1
=?? 2
 
?
GM
1
 m
r
2
=
GM
2
 m
( d-r)
2
?(
d-r
r
)
2
=
M
2
M
1
 
?
d
r
-1=
vM
2
vM
1
 ?r=d[
vM
1
vM
1
+vM
2
] 
Example. Three particles, each of mass ?? , are situated at the vertices of an equilateral 
triangle of side ' ?? '. The only forces acting on the particles are their mutual gravitational 
forces. It is desired that each particle moves in a circle while maintaining their original 
separation 'a'. Determine the initial velocity that should be given to each particle and 
time period of circular motion. 
 
Solution: The resultant force on particle at A due to other two particles is 
F
A
=vF
AB
2
+F
AC
2
+2 F
AB
F
AC
cos 60
°
=v3
Gm
2
a
2
… (i) [?F
AB
=F
AC
=
Gm
2
a
2
] 
Radius of the circle r=
2
3
×asin 60
°
=
a
v3
 
If each particle is given a tangential velocity v, 
So that F acts as the centripetal force, 
 Now 
?? ?? 2
?? =v3
?? ?? 2
?? (???? )
 
 
 
Time period T=
2?? r
v
=
2?? a
v3
v
a
Gm
=2?? v
a
3
3Gm
 
? Fractional forces are central forces as they act along the line joining the centers of 
two bodies. 
? The gravitational forces are conservative forces so work done by gravitational 
force does not depends upon path and therefore if any particle moves along a 
closed path under the action of gravitational force then the work done by this 
force is always zero. 
? The total gravitational force on one particle due to a number of particles is the 
resultant of forces of attraction exerted on the given particle due to individual 
particles i.e. ???
=???
1
+???
2
+???
3
+ means the principle of superposition is valid. 
GRAVITATIONAL FIELD 
The gravitational field is the space around a mass or an assembly of masses over which it 
can exert gravitational forces on other masses. 
 
Theoretically speaking, the gravitational field extends up to infinity. However, in actual 
practice, the gravitational field may become too weak to be measured beyond a particular 
distance. 
Gravitational Field Intensity [g or E
g
 ] 
Gravitational force acting per unit mass at any point in the gravitational field is called 
Gravitational field intensity. 
?? =
?????? ?? 2
/?? =
????
?? 2
  Vector form :??? =
???
?? or  ???=-
????
?? 2
??ˆ 
Page 5


GRAVITATION 
THE DISCOVERY OF THE LAW OF GRAVITATION 
The way the law of universal gravitation was discovered is often considered the paradigm 
of modern scientific technique. The major steps involved were 
? The hypothesis about planetary motion given by Nicolaus Copernicus (1473-
1543). 
? The careful experimental measurements of the positions of the planets and the 
Sun by Tycho Brahe (1546-1601). 
? Analysis of the data and the formulation of empirical laws by Johannes Kepler 
(1571-1630). 
? The development of a general theory by Isaac Newton (1642-1727). 
NEWTON'S LAW OF GRAVITATION 
It states that everybody in the universe attracts every other body with a force that is 
directly proportional to the product of their masses and is inversely proportional to the 
square of the distance between them. 
F?m
1
 m
2
 and F?
1
r
2
 
so F?
m
1
 m
2
r
2
 
 
?F=-
Gm
1
 m
2
r
2
r ˆ[G= Universal gravitational constant ] 
{Note: This formula is only applicable for spherical symmetric masses or point masses.} 
VECTOR FORM OF NEWTON'S LAW OF GRAVITATION: 
Let  ???
12
= Displacement vector from ?? 1
 to ?? 2
 
r ?
21
= Displacement vector from m
2
 to m
1
 
F
??
21
= Gravitational force exerted on m
2
 by m
1
 
F
??
12
= Gravitational force exerted on m
1
 by m
2
 
F
??
12
=-
?? m
1
 m
2
r
21
2
r ˆ
21
=-
?? m
1
 m
2
r
21
3
r ?
21
 
 
Negative sign shows that: 
(i) The direction of ???
12
 is opposite to that ??ˆ
21
 
(ii) The gravitational force is attractive in nature 
Similarly F
??
21
=-
Gm
1
 m
2
r
12
2
r ˆ
12
 or F
??
21
=-
Gm
1
 m
2
r
12
3
r ?
12
?F
??
12
=-F
??
21
 
The gravitational force between two bodies are equal in magnitude and opposite in 
direction. 
GRAVITATIONAL CONSTANT "G" 
? Gravitational constant is a scalar quantity. 
? Unit:   S I: G=6.67×10
-11
 N-m
2
/kg
2
 
CGS: 6.67×10
-8
 dyne- -cm
2
/g
2
 
Dimensions: M
-1
 L
3
 T
-2
 
? Its value is the same throughout the universe, G does not depend on the nature 
and size of the bodies, it also does not depend upon the nature of the medium 
between the bodies. 
? Its value was first found out by the scientist "Henry Cavendish" with the help of 
the "Torsion Balance" experiment. 
? Value of G is small therefore gravitational force is weaker than electrostatic and 
nuclear forces. 
Example. Two particles of masses 1 kg and 2 kg are placed at a separation of 50 cm . 
Assuming that the only forces acting on the particles are their mutual gravitation, find 
the initial acceleration of heavier particle. 
Solution: Force exerted by one particle on another ?? =
Gm
1
 m
2
r
2
=
6.67×10
-11
×1×2
(0.5)
2
=5.3×
10
-10
 N 
Acceleration of heavier particle =
?? m
2
=
5.3×10
-10
2
=2.65×10
-10
 ms
-2
 
This example shows that gravitation is very weak but only this force keep bind our solar 
system and also this universe, all galaxies, and another interstellar system. 
Example. Two stationary particles of masses ?? 1
 and ?? 2
 are at a distance 'd' apart. A 
third particle lying on the line joining the particles, experiences no resultant 
gravitational forces. What is the distance of this particle from?  
Solution: The force on ?? towards ?? 1
 is ?? 1
=
?? ?? 1
?? ?? 2
 
 
The force on ?? towards ?? 2
 is ?? 2
=
GM
2
 m
( d-r)
2
 
According to question net force on ?? is zero i.e. ?? 1
=?? 2
 
?
GM
1
 m
r
2
=
GM
2
 m
( d-r)
2
?(
d-r
r
)
2
=
M
2
M
1
 
?
d
r
-1=
vM
2
vM
1
 ?r=d[
vM
1
vM
1
+vM
2
] 
Example. Three particles, each of mass ?? , are situated at the vertices of an equilateral 
triangle of side ' ?? '. The only forces acting on the particles are their mutual gravitational 
forces. It is desired that each particle moves in a circle while maintaining their original 
separation 'a'. Determine the initial velocity that should be given to each particle and 
time period of circular motion. 
 
Solution: The resultant force on particle at A due to other two particles is 
F
A
=vF
AB
2
+F
AC
2
+2 F
AB
F
AC
cos 60
°
=v3
Gm
2
a
2
… (i) [?F
AB
=F
AC
=
Gm
2
a
2
] 
Radius of the circle r=
2
3
×asin 60
°
=
a
v3
 
If each particle is given a tangential velocity v, 
So that F acts as the centripetal force, 
 Now 
?? ?? 2
?? =v3
?? ?? 2
?? (???? )
 
 
 
Time period T=
2?? r
v
=
2?? a
v3
v
a
Gm
=2?? v
a
3
3Gm
 
? Fractional forces are central forces as they act along the line joining the centers of 
two bodies. 
? The gravitational forces are conservative forces so work done by gravitational 
force does not depends upon path and therefore if any particle moves along a 
closed path under the action of gravitational force then the work done by this 
force is always zero. 
? The total gravitational force on one particle due to a number of particles is the 
resultant of forces of attraction exerted on the given particle due to individual 
particles i.e. ???
=???
1
+???
2
+???
3
+ means the principle of superposition is valid. 
GRAVITATIONAL FIELD 
The gravitational field is the space around a mass or an assembly of masses over which it 
can exert gravitational forces on other masses. 
 
Theoretically speaking, the gravitational field extends up to infinity. However, in actual 
practice, the gravitational field may become too weak to be measured beyond a particular 
distance. 
Gravitational Field Intensity [g or E
g
 ] 
Gravitational force acting per unit mass at any point in the gravitational field is called 
Gravitational field intensity. 
?? =
?????? ?? 2
/?? =
????
?? 2
  Vector form :??? =
???
?? or  ???=-
????
?? 2
??ˆ 
 
Gravitational field intensity is a vector quantity having dimension [LT
-2
] and unit N/kg . 
? Since the force between two point masses is having the similar expression as that 
of force between two point charges, we can write the gravitational field & 
gravitational potential in the same manner as the electric field & electric 
potential. 
Gravitational Potential: 
Gravitational field around a material body can be described not only by gravitational 
intensity ?? , but also by a scalar function, the gravitational potential V. Gravitational 
potential is the amount of work done in bringing a body of unit mass from infinity to that 
point without changing its kinetic energy. 
V=
W
m
 
This work done is the measure of gravitational potential at point (P) 
?V
P
=-
GM
r
 
 
(I) 
(1) Dini Charge 
(a) E=
kQ
r
2
 
r 
(2) Uniform charged ring 
(a) =
kQx
(i
2
+x
2
)
3/2
 on axis ?? when ?? =
?? v2
 
(b) ?? =
kQ
vy
2
+x
2
 on axis, 
kQ
r
 at center 
(3) Uniform linear charge