COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)

Gravitation
is denoted by ‘g’ and its S.I. unit is ms-2 or
GRAVITATION N/kg.

Gravitation is the name given to the force of ‘g’ is a vector. Its direction is towards the
attraction between any two bodies of the centre of the earth.
universe. It was discovered by Newton in year
1665. Relation between ‘g’ and ‘G’:

Newton’s Law of Gravitation: Let f be the force of

attraction between
It states that every body in this universe body and the earth.
attracts every other body with a force which is
directly proportional to the product of their According to Newton’s
masses and is inversely proportional to the law of gravitation
square of the distance between them. 𝑀𝑚
F=G 𝑟2

From gravity pull, F = Mg

𝑀𝑚
∴ mg = G 𝑟2

𝐺𝑀
Or g=
Consider two bodies A and B of masses m1 and 𝑅2
m2. Let r be the distance between their centres 𝐺𝑀
and F be the force of attraction between them. Mass of Earth: From g= 𝑅2

F∝
𝑚1 𝑚2
M = gR2/G .......(1)
𝑟2

Or F =G
𝑚1 𝑚2
Where G = We know that R = 6.4 X 106 m,
𝑟2
Universal Gravitational constant. G = 6.67 X 10-11 Nm2kg-2
Definition of G: And g = 9.8 ms-2
Let m1 = m2 =1 and r=1 9.8 𝑋 (6.4 𝑋 106 )
2
Then, M= 6.67 𝑋 10−11
= 6.018X1024 kg
F=G or G=F
Density of earth:
Thus Universal Gravitational Constant is equal
𝑀𝑎𝑠𝑠
to the force of attraction acting between two As we know, Density = 𝑉𝑜𝑙𝑢𝑚𝑒
bodies each of unit mass, whose centres are
placed unit distance apart. Density ρ= 4
𝑀
=
3𝑀
𝜋𝑅 3 4 𝜋 𝑅3
3
𝐹𝑟2 [𝑀𝐿𝑇 −2 ][𝐿2 ]
Dimensional formula for G = = [𝑀][𝑀] 3 𝑋 6.018 𝑋 1024
𝑚1 𝑚2
ρ= = 5.5 X 103 kgm-3
= [𝑀−1 3
𝐿𝑇 −2 ] 4 𝑋 3.14 𝑋 (6.4 𝑋 106 )3

Gravity: Shell Theorem:

Gravity is the force of attraction exerted by (i)If a point mass lies outside spherical shell/
earth towards its centre on a body lying on or sphere with a spherically symmetric internal
near the surface of earth. Gravity is merely a mass distribution, the shell/ sphere attracts the
special case of gravitation and is called earth’s point mass as if the entire mass of the
gravitational pull. shell/sphere were concentrated at its centre.

Acceleration due to Gravity:

Acceleration due to gravity is defined as the

constant acceleration produced in a body when
it falls freely under the effect of gravity alone. It

1 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376

COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
Take AG = BG = CG = 1 m

(ii) If a point mass lies inside a uniform

spherical shell, the gravitational force on the
point mass is zero. But if a point
mass lies inside a homogeneous
solid sphere, the force on the
point mass acts towards the
centre of sphere. This force is
exerted by the spherical mass 5. [NCERT] A rocket is fired from the earth
situated interior to the point towards the sun. At what point on its path is
mass. the gravitational force on the rocket zero? Mass
of the sun = 2 X 1030 kg, mass of the earth
= 6 X 1024 kg. Neglect the effect of other planet
tec. Orbital radius = 1.5 x 1011 m.
{Ans. 2.59 X 108 m}

Variation of acceleration due to Gravity:

(a) **Effect of Altitude:

Consider earth to be a sphere of mass M, radius

Practice Questions: R with centre O. Let g be the value of
acceleration due to gravity at a point A.
1.Calculate the force of attraction between two
balls each of mass 1 kg each, when their
centres are 10 cm apart. Given G = 6.67X10-11 𝐺𝑀
Nm2kg-2. g= ....(1)
𝑅2
{Ans. 6.67X10-9 N}
If g’ is the acceleration
2. Two particles, each of mass m, go round a due to gravity at a point
circle of radius R under the action of their B, at a height h above
mutual gravitational attraction. Find the speed the surface of earth,
of each particle. then
𝐺𝑀
{Ans. v = √ } 𝐺𝑀
4𝑅 g’ = (𝑅+ℎ)2
......(2)

3. A mass M is broken into two parts of masses Dividing (2) by (1)

m1 and m2. How are m1 and m2 related so that
force of gravitational attraction between the 𝑔′ 𝐺𝑀 𝑅2 𝑅2
= X =
two parts is maximum? 𝑔 (𝑅+ℎ)2 𝐺𝑀 (𝑅+ℎ)2

𝒈′ 𝑹𝟐 𝑹𝟐
{Ans. m1 = m2 = M/2} = ⇒ g’ = g ........(3)
𝒈 (𝑹+𝒉) 𝟐 (𝑹+𝒉)𝟐

4. [NCERT] Three equal mass of m kg are fixed 𝑔′ 𝑅2 ℎ −2 2ℎ

at the vertices of an equilateral triangle ABC, as = ℎ 2
= (1 + ) =1–
𝑔 𝑅 2 (1+ ) 𝑅 𝑅
shown in figure. 𝑅

𝒈′ 𝟐𝒉 𝟐𝒉
(a) What is the force acting on a mass 𝒈
=1– 𝑹
⇒ g’ = g(𝟏 – 𝑹
) .......(4)
2m placed at the centroid G of the triangle?
Hence, the value of g decreases with height.
(b) What is the force if the mass at the
vertex A is doubled? # From g’ = g(𝟏 –
𝟐𝒉
)
𝑹

2 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376

COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
g – g’ =
𝟐𝒉𝒈
Thus the value of acceleration due to gravity is
𝑹
maximum at the earth’s surface and becomes
Fractional decrease in the value of g zero at the centre of earth.
𝑑
𝒈−𝒈′
=
𝟐𝒉 # From g’ = g (1 − 𝑅)
𝒈 𝑹

% decrease in the value of g, Fractional decrease in the value of g

𝒈−𝒈′ 𝒅
𝒈−𝒈′
x 100 =
𝟐𝒉
X 100 𝒈
= 𝑹
𝒈 𝑹

(b) ** Effect of depth: % decrease in the value of g,

𝒈−𝒈′ 𝒅
Consider earth to be a homogeneous sphere of 𝒈
x 100 = 𝑹
X 100
radius R and mass M with centre at O. Let g be
the value of acceleration due to gravity at point (c) Effect of shape of earth
A on the surface of earth.
Earth is not a perfect sphere. It is flattened at
𝐺𝑀
g= the pole and bulges out at the equator.
𝑅2
Equatorial radius Re of the earth is about 21 km
If ρ be the uniform density of earth, then greater than the pole radius Rp
𝐺𝑀
Now g = 𝑅2
G, M are constant
Mass of earth
1
4 g∝
M= 3
𝜋 𝑅3 𝜌 𝑅2

4
𝐺𝑋 𝜋 𝑅3 𝜌
g= 3
𝑅2
So, we can conclude
4 that the value of g is least at the equator
g= 𝜋𝐺𝑅𝜌 ...(1)
3 and maximum at the pole. It means, the
value of acceleration due to gravity increases as
Let g’ be the acceleration due to gravity at the
we go from equator to the pole.
point B at a depth d below the surface of earth.
𝐺 𝑀′ 4 Practice Questions:
g’ = (𝑅−𝑑)2
and M’ = 𝜋 (𝑅 − 𝑑)3 𝜌
3
1.[NCERT] The acceleration due to gravity at
𝐺𝑋
4
𝜋 (𝑅−𝑑)3 𝜌 4 the moon’s surface is 1.67 ms-2. If the radius of
g’ = 3
= 𝜋 𝐺 (𝑅 − 𝑑) 𝜌 ...(2)
(𝑅−𝑑)2 3 the moon is 1.74 x 106 m, calculate the mass of
the moon. Use the known value of G.
Dividing eqn (2) by (1) {Ans. 7.58 X 1022 kg)
4
𝑔′
= 3
𝜋 𝐺 (𝑅−𝑑) 𝜌
4 =
𝑅−𝑑
=1-
𝑑 2. A body weighs 90 kg f on the surface of the
𝑔
3
𝜋𝐺𝑅𝜌 𝑅 𝑅
earth. How much will it weigh on the surface of
𝑑
Mars whose mass is 1/9 and the radius is ½ of
g’ = g (1 − ) that of the earth?
𝑅
{Ans. 40 kg f}
Hence, the value of acceleration due to gravity
decreases with depth. 3. The radius of earth shrinks by 2.0% mass
remaining constant, then how would the value
At the centre of earth, of acceleration due to gravity change?
{Ans. 4%}
𝑅
d = R, g’ = g (1 − ) = 0
𝑅
4. At what height from the surface of earth, will
Therefore the weight of the body of mass m at the value of g be reduced by 36% from the
the centre of earth = mg = m(0) = 0 value at the surface? Radius of the earth =
6400 km. {Ans. 1600 km}

3 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376

COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
5. Find the percentage decrease in the weight r = PC = R cosλ
of a body when taken to a height of 32 km
above the surface of earth. Radius of the earth The centrifugal force acting on the particle is
is 6400 km. {Ans. 1%}
f = mrω2, acting along PA
6. Find the percentage decrease in weight of a
body, when taken 16 km below the surface of This force has two components: mrω2 cosλ and
earth. Take radius of the earth as 6400 km. mrω2 sinλ. The component mrω2 sinλ acts
{Ans. 0.25%} perpendicular to mg and has no effect. The
component mrω2 cosλ acts opposite to mg. So,
7. Imagine a tunnel dug along a diameter of the apparent weight of the particle P is
earth. Show that a particle dropped from one
end of the tunnel executes simple harmonic mg’ = mg - mrω2 cosλ
motion. What is the time period of this motion?
or g’ = g - rω2 cosλ
Assume the earth to be a sphere of uniform
mass density (equal to its known average or g’ = g - R ω2 cos2 λ [r= Rcosλ]
density = 5520 kgm-3). G = 6.67X10-11
Nm2kg-2. Neglect all damping forces. As λ increases, cos λ decreases and g’
[Hint: g∝ y;
𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
T = 2 𝜋 √ 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 ] increases. So as we move from equator to pole,
accceleration due to gravity g increases.
{Ans. 5059.77 s}
Special cases:
8. Suppose the gravitational force varies
inversely as the nth power of distance. Then, (i) At equator, λ = 00, cosλ = 1, Hence
find the expression for the time period of a ge = g - R ω2
planet in a circular orbit of radius r around the (ii) At the pole, λ = 900, cosλ = 0, Hence
sun. {Ans. T =
2𝜋
.𝑟
𝑛+1
2 } gp = g - R ω2 X 0 = g
√𝐺 𝑀
Thus acceleration due to gravity is minimum at
9. A simple pendulum has a time period T1 the equator and maximum at the pole. The
when on earth’s surface, and T2 when taken to difference in two values is
a height R above the earth’s surface, where R is
the radius of the earth. Then what will be value gp – ge = g – (g - R ω2) = R ω2
of T2/T1. {Ans. 2}
Graph showing the variation of
Variation of ‘g’ with Latitude (or Rotation acceleration due to gravity g with distance
of Earth): r from the centre of the earth.

The latitude of a place is defined as the angle For points lying outside the earth (r>R)
which the line joining the place to the centre of
the earth makes with the equatorial plane. 𝑔′ 𝑅2 𝑅2 𝑔𝑅 2
𝑔
= (𝑅+ℎ)2
= (𝑟)2
or g’ = 𝑟2

1
g’ ∝ 𝑅2

For the points inside the earth

𝑔(𝑅−𝑑) 𝑔𝑟 𝑔𝑟
g’ = 𝑅
= 𝑅
or g’ = 𝑅

g’ ∝ r

Hence, the graph showing the variation of

acceleration due to gravity g with distance r
from the centre of earth is as shown in the
figure.
Consider aparticle of mass m lying at point P,
whose latitude is λ. The partical P sescribes a
horizonatl circle of radius

4 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376

COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
𝐺𝑀
But 𝑟2
is equal to the acceleration due to
gravity at the point P. Hence the gravitaional
field intensity of earth at any point is equal to
acceleration produced in the freely falling body
at that point.

For any point on the surface of the earth, r = R,

so
𝐺𝑀
Esurface = 𝑅2
=g

This is the acceleration due to gravity at the

Practice Questions: surface of the earth.
1.Calculate that imaginary angular velocity of Unit of E: In S.I. system: N kg-1
the earth for which effective acceleration due to
gravity at the equator becomes zero. In this In c.g.s. system : dyne g-1
condition what will be the length (in hours) of
the day? Given radius of the earth = 6400 km Dimension of E : As E = F/m ;
and g = 10 ms-2.
[𝑀 𝐿 𝑇 −2 ]
{Ans. 1.4 h} Dimension of E = = [𝐿 𝑇 −2 ]
[𝑀]

2. Determine the speed with which the earth

Gravitational Potential:
would have to rotate on the axis so that a
person on the equator would weigh 3/5th as The gravitaional potential at a point in the
much as at present. Take the equatorial radius gravitaional field of a body is defined as the
as 6400 km. amount of work done in bringing a body of unit
{Ans. 7.8X10-4 rads-1} mass from infinity to that point.
3. If the earth were a perfect sphere of radius Gravitational potential; V =
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
=
𝑊
6.37X106 m, rotating about its axis with a 𝑀𝑎𝑠𝑠 𝑚

period of 1 day (=8.64X104 s), how much

It is scalar quantity. S.I. unit is J kg-1 and c.g.s
would the acceleration due to gravity differ
unit is erg g-1.
from the pole to the equator?
{Ans. 3.37 X 10-2 ms-2} *Gravitaional Potential at a point due to
earth:
Gravitation Field: The space surrounding a
material body within which its gravitational Small amount of work done inP bringing a body
force of attraction can be experienced is called of mass m through the displacement dx
its gravitational field. towards the centre of earth; W = Force X
displacement
Intensity of Gravitational field due to
Earth: ⃗⃗⃗⃗⃗ = F dx cos 00 = F.dx
dW = 𝐹⃗ . 𝑑𝑥
According to Newton’s law of gravitation, the 𝑀𝑚
dW = G . dx
force of attraction on test mass m is 𝑥2

F=G
𝑀𝑚 Total amont of work done from infinity to r
𝑟2
𝑟 𝑀𝑚
W = ∫∞ G 𝑥2
. dx
𝑟
𝑟 1 𝑥 −2+1
W = G M m ∫∞ 𝑥 2 dx = G M m [ −2+1 ]
∞

1 𝑟
= - G M m [𝑥]
∞
The gravitational field intensity at point P will be
1 1 𝐺𝑀𝑚
𝐹 𝐺𝑀 W = - G M m [𝑟 − ]=-
E= 𝑚
= 𝑟2
∞ 𝑟

5 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376

COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
Hence the gravitaional potential due to earth at Gravitational Potential Energy = Gravitational
distance r from its centre is Potential X mass

V=
𝑊
=-
𝐺𝑀
Impotant Points:
𝑚 𝑟

At the surface of earth, r = R, therefore 1.If a body of mass m is moved from a point at
distance r1 to a point at distace r2, then the
Vsurface = -
𝐺𝑀 chanege in potential energy of the body will be
𝑅
𝑟 𝑀𝑚 1 𝑟2
*Gravitaional Potential Energy: ∆U = ∫𝑟 2 G 𝑥2
. dx = G M m [− 𝑥]
1 𝑟1

The gravitaional potential energy of a body at a = - G M m [𝑟 −

1 1
]=GMm[ −
1 1
]
point in a gravitational field of another body is 2 𝑟1 𝑟 1 𝑟2

defined as the amount of work done in bringing

If r1 > r2, the ∆U be negative. So whena body is
the given body from infinity to that point
without acceleration. brought closer to the earth, its gravitational
potential energy (P.E.) decreases.
**Expression for gravitational potential
energy: 2. If a body is moved from the surface of earth
to a point at a height h above the surface of
Gravitaional force of earth, then r1 = R and r2 = R + h. The change
attraction on the bady of in potential energy will be
mass m at point A is 1 1 𝐺𝑀𝑚 𝑅
∆U = G M m [𝑅 − ]= [1 − ]
𝑀𝑚 𝑅+ℎ 𝑅 𝑅+ℎ
F=G
𝑥2
𝐺𝑀𝑚 1
= [1 − ]
Small amount of work done 𝑅 (1 + )
ℎ
𝑅
in bringing the body without
acceleration through a very 𝐺𝑀𝑚 ℎ −1
= [1 – (1 + ) ]
small distance AB is given 𝑅 𝑅

by 𝐺𝑀𝑚 ℎ
= [1 – (1 − + … . )]
𝑅 𝑅
𝑑𝑥 = F dx cos 00
dW = 𝐹⃗ . ⃗⃗⃗⃗⃗ (Expending by Binomial Theorem)

= F . dx =
𝐺𝑀𝑚ℎ 𝐺𝑀
= ( 𝑅2 ) m h = gmh
𝑅2
𝑀𝑚
dW = G . dx
𝑥2 ∆U = mgh
Total amount of work done in bringing the 3. Graviatation potential energy of a point mass
body from infinity to point P is given by m relative to infinity at the surface of earth will
𝐺𝑀𝑚
𝑟 𝑀𝑚 𝑟 1 be U=- 𝑅 (r=R)
W = ∫∞ G . dx = G M m ∫∞ 2 dx
𝑥2 𝑥

𝑟 4. The potential energy of a point mass m, at

𝑥 −2+1 1 𝑟
=GMm[ ] =-GMm[ ] the centre of earth relative to infinity will be
−2+1 ∞ 𝑥 ∞
given by
1 1 𝐺𝑀𝑚
W=-GMm[ − ]=- 3 𝐺𝑀𝑚 3
𝑟 ∞ 𝑟 U=- [ VC = VS ]
2 𝑅 2

This work done is stored in the body as its

Practice Questions:
gravitational potential energy U. Therefore
𝐺𝑀𝑚 1.Two bodies of masses 10 kg and 1000 kg are
U=- 𝑟 at a distace 1 m apart. At which point on the
line joining them will the gravitational field
Relation bbetween gravitational potential intensity be zero?
1
{Ans. 11 m}
energy and gravitational potential:
𝐺𝑀𝑚 𝐺𝑀 2. Three mass points each of mass m are
U=- = (− ) m = VP X m
𝑟 𝑟 placed at the vertices of an equilateral triangle
of side 𝑙. What is the gravitational field and
6 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376
COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
potential due to three masses at the centroid of small distance dx
the triangle?
𝑀𝑚
𝐺𝑚
{Ans. 0, -3√3 𝑙 } W = F.dx= G 𝑥2
. dx

3. [NCERT] Find the potential energy of a The total work done in moving the body from
system of four particles, each of mass m, the surface x=R to x=∞.
placed at the vertices of a square of side l. Also ∞ 𝑀𝑚 ∞
obtain the potential at the centre of the square. W = ∫ 𝑑𝑊 = ∫𝑅 G . dx = G M m ∫𝑅 x −2 . dx
𝑥2
𝐺 𝑚2 4√2 𝐺 𝑚
{Ans. – 5.41 ,- } ∞
𝑙 𝑙 𝑥 −2+1 −1 ∞
= G M m[ ] = G M m[ ]
−2+1 𝑅 𝑥 𝑅
4. Two bodies of masses m1 and m2 are placed
at a distance r apart. Show that at the position = G M m [−
1
+ ]=G
1 𝑀𝑚

where the gravitational field due to them is ∞ 𝑅 𝑅

zero, the potential is given by Tis work done is at the cost of kinetic energy
𝐺 given to the body at the surface of earth. Let V e
V=- [𝑚1 + 𝑚2 + 2 √𝑚1 𝑚2 ]
𝑟 be the escape velocity of the body projected
from the surface of earth, then
5. A spherical cavity is made inside a sphere of
density ρ. If its centre lies at a distance l from Kinetic Energy of the body =
1
m 𝑣𝑒2
the centre of the sphere, show that the 2

gravitational field strength of the field inside the 1 𝑀𝑚

4𝜋 m 𝑣𝑒2 = G
cavity is E= 3 Glρ 2 𝑅

2𝐺𝑚
Or 𝑣𝑒2 =
6.The distance between the centres of two stars 𝑅
is 10a. The masses of these stars are M and
16M and their radii a and 2a respectively. A Or ve = √
𝟐𝑮𝒎
....(i)
body of mass m is fired straight from the 𝑹
surface of the larger star towards the smalle
𝐺𝑀
star. What should be its minimum initial speed As g= ⇒ G M = g R2
𝑅2
to reach the surface of the smaller star? Obtain
the expression in terms of G, M and a. 2 𝑔 𝑅2
ve = √ = √𝟐 𝒈 𝑹 ....(ii)
3 5𝐺𝑀 𝑅
{Ans. √ }
2 𝑎
if ρ be the density of material of earth, then
**Escape Velocity:
4
M= 3
π R3 ρ
Escape velocity is the minimum velocity with
which a body must be projected vertically 2𝐺 4 8 π ρ G R2
upwards in order that it may just escape the ve = √ 𝑅 . 3
π R3 ρ = √ 3
.....(3)
gravitational field of the earth.
Important points:
**Expression for escape velocity:
1.For Earth, g = 9.8 ms-2, R = 6.4 X 106 m
Consider the earth to be a
sphere pf mass M and radius R Ve = √2 𝑋 9.8 𝑋 6.4 𝑋 106 = 11.2 X 103 ms-1
with centre O. Suppose a body = 11.2 kms-1.
of mass m lies at point P at
dsitance x from its centre, as 2. If a body is projected from a planet with a
shown in figure. velocity v which is smaller than the escape
velocity ve (v < ve), then thebody will reach
The gravitational force of a certain height may either move in an orbit
attraction on the body at P around the planet or may fall back to the
𝑀𝑚 planet.
F=G
𝑥2
3. If v>ve then the body will escape out from
Small amount of work done in the graviatational field of that planet and will
moving the body through of that planet and will move in the

7 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376

COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
interstaller spacei with velocity v’. According
to the law of conservation of energy
2. Second Law (Law of areas): The radius
1
m v2 + (−
𝐺𝑀𝑚
)=
1
m v’2 + 0 vector drawn from the sun to a planet sweeps
2 𝑅 2
out equal areas in equal intervals of time i.e.,
Or v’2 = v2 –
2𝐺𝑀
= v2 - 𝑣𝑒2 the areal velocity (area covered per unit time)
𝑅 of a planet around the sun is constant.
Or v’2 = √v 2 − 𝑣𝑒2

4. Escape velocity of sun is 618 km/s.

Practice Questions:

1.Determine the escape velocity of a body form

the moon. Take the moon to be a uniform
sphere of radius 1.76 X 106 m, and mass 7.36 X 3. Third Law ( Law of period): The square
1022 kg-2. Given G = 6.67 X 10-11 Nm2kg-2. of the revolution of a planet around the sun is
{Ans. 2.375 kms-1} proportional to the cube of the semi-major axis
of its elliptical orbit.
2. A black hole is a body from whose surface
nothing may escape. What is the condition for a i.e. T2 ∝ R3
uniform spherical body of mass M to be a black
hole? What should be the radius of such a black or T 2 = k R3
hole if its mass is 9 times the mass of the
earth? {Ans. nearly 8 cm.} Where T is the period of revolution of a planet
and R is the length of semi-major axis of the
3. [NCERT] Two uniform solid sphere of equal elliptical orbit and K is proportionality constant.
radii R, but mass M and 4M have a centre to
centre separation 6R. The two sphere are held Practice Questions:
fixed. A projectile of mass m is projected from 1.[NCERT] In Kepler’s law of period :
the surface of the sphere of mass M directly
T2 = k R3, the constant k = 10-13 s2m-3.
towards the centre of the second sphere. Obtain
Express the constant k is days and kilometres.
an expression for the minimum velocity v of the The moon is at a distance of 3.84 X 10 5 km
projectile so that it reaches the surface of the
from earth. Obtain its time-period of revolution
3𝐺𝑀
second sphere. {Ans. v = √ } in days.
5𝑅
{Ans. 1.33 X 10-14 d2 km-3, 27.3 days}
4. The escape velocity of a projectile on the 2. Calculate the period of revolution of Neptune
earth’s surface is 11.2 kms-1. A body is
around the sun, given that diameter of its orbit
projected out with thrice this sped. What is the is 30 times the diameter of earth’s orbit around
speed of the body for away from the earth?
the sun, both orbits being assumed to be
Ignore the presence of the sun and other circular. {Ans. 164.3 years}
planets. {Ans. 31.68 kms-1}
3. [NCERT] The planet Mars has two moons,
** Kepler’s Laws of Planetary Motion:
Phobos and Delmos. (i) Phobos has an orbital
1.First Law (Law of Orbit): Each planet radius of 9.4 X 103 km. Calculate the mass of
Mars. (ii) Assume that Earth and Mars move in
revolves around the sun in an elliptical orbit
with the sun situated at one of the two foci. circular orbits around the Sun, with the Martian
orbit being 1.52 times the orbital radius of the
Earth. What is the length of the Martian year in
days? {Ans. 6.48 X 1023 kg, 684 days}

4. A geostationary satellite is orbiting the earth

at a height 6R above the surface, where R is
the radius of earth. Find the time period of
another satellite at a height of 2.5R from the
surface of earth in hours.
{Ans. 6√2 h}
8 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376
COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
5. The radius of a planet is double that of the
earth but their average densities are the same.
If the escape velocities at the planet and at the
earth are VP and VE respectively, then prove
that VP = 2 VE

Satellite: A satellite is a body revolves on

its own around a much larger body in a stable
orbit.
Expression for orbital velocity:
Natural satellite: A satellite created by
nature is called natural satellite. E.g. Moon is Let M = mass of eart; R = radius of earth;
the satellite of Earth and Earth is a satellite if m = mass of satellite; v0 = orbital velocity of
sun etc. the satellite; h = height of the satellite above
the surface of earth; R + h = orbital radius of
Artificial satellite: A man made satellite is
the satellite.
called artificial satellite. Russian were the first
to put an artificial satellite, SPUTNIK-1. India In equilibrium
entered space age on April 19, 1975 by
putting in orbit its first satellite Aryabhatta 𝑚 𝑣02
=
𝐺𝑀𝑚
from Russian soil. 𝑅+ℎ (𝑅+ℎ)2

𝐺𝑀
Principle of launching satellite: Let us throw Or 𝑣02 = 𝑅+ℎ
a body horizontally from the top of the tower
with different velocities. When the velocity is 𝐺𝑀
Or v0 = √ ....(i)
low, the body describes a parabolic path under 𝑅+ℎ
the effect of gravity and hits the earth’s surface
at A. With somewhat larger velocity, its path is If g is the acceleration doe to gravity, then
still parabolic but hits the surface at B covering 𝐺𝑀
a larger horizontal range. As we go on g= or G M = g R2
𝑅2
increasing the velocity of horizontal projection,
the body will hit the ground at a point farther hence v0 = √
𝑔 𝑅2
=R√
𝑔
......(ii)
and farther from the foot of the tower. At a 𝑅+ℎ 𝑅+ℎ

certain horizontal velocity, the body will not hit

the earth, but will always be in a state of free When the satellite revolves close to the earth,
fall under gravity and attempt to fall to the h=0 and the orbital velocity will become
earth but missing all the time. Then the body
will follow a stable circulating path around the V0 = √𝑔 𝑅
earth and will become a satellite of the earth.
This horizontal velocity is called orbital velocity. As g = 9.8 ms-2 and R = 6.4 X106 m, so

V0 = √9.8 𝑋 6.4 𝑋 106 = 7.92 X 103 ms-1.

= 7.92 ≈ 8 kms-1.

Relation between orbital velocity and

escape velocity:

Escape velocity of a body from earth’s surface

is Ve = √2 𝑔 𝑅

The orbital velocity of a satellite revolving close

to earth’s surface V0 = √𝑔 𝑅
Orbital velocity: Orbital velocity is the
velocity required to put the satellite its orbit 𝑣𝑒 √2 𝑔 𝑅
around the earth. 𝑣𝑜
= = √2 or 𝑣𝑒 = √2 𝑣𝑜
√𝑔 𝑅

9 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376

COSMOS ACADEMY Prepared By: Arvind Kashyap (Kashyap Sir)
Gravitation
Hence the escape velocity of a body from the Angular momentum:
earth’s surface is √2 times its velocity in a
circular orbit just above the earth’s surface. The angular momentum of a satellite of mass m
moving with velocity v0 in an orbit of radius r =
Time period of satellite: (R+h) is given by

It is the time taken by a satellite to complete 𝐺𝑀

L = m 𝑣0 r = m√ . r = √𝐺 𝑀 𝑚2 𝑟
one revolution around the earth. 𝑟

T=
𝐶𝑖𝑟𝑐𝑢𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑟𝑏𝑖𝑡
=
2 𝜋 (𝑅+ℎ) Practice Questions:
𝑂𝑟𝑏𝑖𝑡𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣0

1.Two particles of equal mass m go round a

As the orbital velocity, v0 = √𝑅+ℎ
𝐺𝑀
circle of radius R under the action of their
mutual gravitational attraction. What is the
1 𝐺𝑚
2 𝜋 (𝑅+ℎ) (𝑅+ℎ)3 speed of each particle? {Ans. v =2 √ }
T= 𝐺𝑀
= 2π √ 𝐺𝑀
....(i) 𝑅
√
𝑅+ℎ
2. The radius of a planet is R. A satellite
But g=
𝐺𝑀
or GM=gR 2 revolves round it in a circle of radius r. The time
𝑅2
period of revolution is T. Determine the
(𝑅+ℎ)3 acceleration due to gravity of the planet at its
T = 2π √ .....(ii) 4 𝜋2 𝑟3
g R2 surface. {Ans. g = . }
𝑇2 𝑅2

If the earth is a sphere of mean density ρ, then

its mass would be
4
M = Volume X density = 𝜋 𝑅3 𝜌
3

(𝑅+ℎ)3 3 𝜋 (𝑅+ℎ)3
T = 2π √ 4 = √
𝐺. 𝜋 𝑅3 𝜌 𝐺 𝑅3 𝜌
3

When the satellite revolves close to the earth,

h=0 and the time period will be

(𝑅)3 (𝑅)3 𝑅 3𝜋
T = 2π √ 𝐺 𝑀 = 2π √g R2 = 2π √𝑔 = √𝐺 𝜌

Putting g = 9.8 ms-2 and R = 6.4 X 106 m, we

6.4 𝑋 106
get T = 2π √ 9.8
= 5078 s = 84.6 min

Height of satellite above the earth’s

surface:

Squaring both side of equation (ii)

4 𝜋2 (𝑅+ℎ)3
T2 = g R2

𝑇2 𝑅2 𝑔
Or (𝑅 + ℎ)3 =
4 𝜋2

1/3
𝑇2 𝑅2 𝑔
Or R+h=[ 4 𝜋2
]

Height of satellite
1/3
𝑇2 𝑅2 𝑔
h=[ ] –R
4 𝜋2

10 | SH-299, Shashtri Nagar, Ghaziabad. Phone: 09350764376