Prof. Sachin M.

Shiral
(M.Sc. B.Ed.)

SUCCESS CLASSES, BARSHI

Gravitation
 Gravitation: Attraction between any two matter particles due to their masses
is called as gravitation.
 Gravitational force: Force of attraction between any two objects exerted by
them on each other due to their masses is called as gravitational force.
 Characteristics of gravitational force:
1) It is weakest force
2) It is long range force
3) It acts along the line joining centres of two bodies
4) It obeys inverse square law of distance
5) It is attractive force
 Gravitational force exerted by earth on any body is directed towards the center
of the earth so the object falls vertically downwards on the surface of the earth
if no other force acts on it
Note: Force is necessary to change the speed as well as the direction of motion
of an object
 Newton’s laws of motion
𝑰𝒔𝒕 law: Every object in a state of uniform motion or rest will remain in that
state unless an external force acts on it. It is also called as law of inertia.
𝑰𝑰𝒏𝒅 law: Newton’s second law states that force is equal to the change in
momentum per change in time.
For constant mass m, force = mass × acceleration
𝒊. 𝒆. 𝐹 = 𝑚𝑎
𝒓𝒅
𝑰𝑰𝑰 law: Newton’s 3rd law states that for every action there is an equal and
opposite reaction
 Circular motion: Motion of particle along the circumference of circle is called
as circular motion.
e.g. Motion of a stone tied to one end of a string when sting is rotated by taking
its other end in our hand is circular motion.
 Centripetal force: The force acting on a object moving along a circle and it is
directed towards the centre of the circle is called as centripetal force.
i.e. centre seeking force is centripetal force.
Centripetal force is necessary for circular motion of object.
In above example tension created in the string by holding its one end in hand
is centripetal force.
If we cut off this force the stone will fly off tangentially.
Thus for any circular motion centripetal force is necessary.

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 Prof. Sachin M. Shiral
(M.Sc. B.Ed.)

SUCCESS CLASSES, BARSHI

We know that moon revolves around the earth in circular orbit. For its circular
motion necessary centripetal force is provided by gravitational force exerted by
earth on the moon.
 Kepler’s laws of planetary motion
 Kepler’s 𝑰𝒔𝒕 law:
The orbit of a planet is an ellipse with the sun is at one of the foci

 Kepler’s second law: The line joining the planet and the sun sweeps equal
areas in equal intervals of time
Suppose as shown in fig arc AB and arc CD are distance covered by planet in
equal interval of time then by Kepler’s second law
Area 𝐴𝑆𝐵 = Area 𝐶𝑆𝐷
 Kepler’s third law: The square of period of revolution of planet around the sun
is directly proportional to the cube of mean distance of planet from the sun.
i.e. 𝑇 2 ∝ 𝑟 3
where 𝑇 = period of revolution of planet and 𝑟 = Average distance of the planet
from the sun
𝑇2
∴ = constant
𝑟3
 Newton’s universal law of gravitation:
Statement: Every object in the universe attracts every other object with a force
which is directly proportional to the product of the masses of two objects and is
inversely proportional to the square of the distance between them.

Consider two objects having masses 𝑚1 and 𝑚2 and are separated by distance
d as shown in above fig.
If F is gravitational force between them, then by newton’s law of gravitation
𝐹 ∝ 𝑚1 𝑚2

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 Prof. Sachin M. Shiral
(M.Sc. B.Ed.)

SUCCESS CLASSES, BARSHI

1
And 𝐹 ∝
𝑑2
𝑚1 𝑚2
∴ 𝐹∝
𝑑2
𝐺 𝑚1 𝑚2
∴ 𝐹= ....... (1) where G is
𝑑2

Constant of proportionality called as universal gravitational constant.

 Definition of G: According newton’s law,
𝐺 𝑚1 𝑚2
𝐹=
𝑑2
𝐹 𝑑2
∴ 𝐺=
𝑚1 𝑚2
Putting 𝑚1 = 𝑚2 = 1 𝑎𝑛𝑑 𝑑 = 1 in above equation
∴ 𝐺=𝐹
Thus value of G is the gravitational force acting between two unit masses kept
at unit distance away from each other
In S. I. system 𝐺 = 6.67 × 10−11 𝑁 𝑚2 𝑘𝑔2
 Centre of mass of body: The point at which entire mass of body is supposed
to be concentrated is called as centre of mass of the body.
Centre of mass may be inside or outside the body
 Proof of inverse square law from Kepler’s 𝟑𝒓𝒅 law
1) Consider a planet of mass m is orbiting around the sun in orbit of radius r with
speed v.
2) The necessary centripetal force for circular motion of the planet is given by
𝑚𝑣 2
𝐹= ................ (1)
𝑟
3) Let T is the period of revolution of planet.
Now, distance covered in one revolution = 2𝜋𝑟
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑
Now, speed =
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
2𝜋𝑟
∴𝑣= ............. (2)
𝑇
4) Using equation (2) in equation (1)
2𝜋𝑟 2
𝑚
𝑇
∴𝐹=
𝑟
𝑚 4𝜋 2 𝑟 2
∴𝐹=
𝑇2 𝑟
𝑟3 1
∴ 𝐹 = 4𝜋 2 𝑚 .............. (3)
𝑇2 𝑟2

5) Now, in R. H. S. of equation (3) 4𝜋 2 𝑚 is constant

Also according to Kepler’s 3𝑟𝑑 law
∴ 𝑇2 ∝ 𝑟3
𝑇2
∴ = constant
𝑟3
𝑟3
∴ = constant
𝑇2
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 Prof. Sachin M. Shiral
(M.Sc. B.Ed.)

SUCCESS CLASSES, BARSHI

4𝜋 2 𝑚 𝑟 3
Thus = constant
𝑇2

6) Equation (3) can be written as

1
F = (constant)
𝑟2
1
∴𝐹∝
𝑟2
7) Thus necessary centrepetal force for circular motion of planet around the sun
is inversely proportional to square of distance of the planet from the sun.
8) But this centrepetal force is provided by gravitational force of attraction exerted
by the sun on the planet.
9) Hence gravitational force obeys inverse square law of distance
 Tides:
1) The level of water in the sea changes because of the gravitational force exerted
by the moon.
2) Water directly under the moon gets pulled towards the moon and the level of
water there goes up causing high tide at that place.
3) At two places on the earth at 90° from the place of high tide, the level of water
is minimum and causes low tide at those places.
4) Low and high tides occur at a place twice a day at regular intervals.
 Earth’s gravitational force:-
1) Earth exerts gravitational force of attraction on any object near to it.
2) Due to this object falls vertically downwards.
3) When we throw a stone vertically upwards its velocity decreases due to
gravitational force exerted by earth on it and finally at some height velocity
becomes zero, pull continues and the stone starts moving vertically
downwards.
4) The gravitational force due to the earth also acts on moon and artificial
satellites due to which they revolve around the earth.
 Acceleration due to gravity:
1) Gravitational force of attraction exerted by earth on any body is called as
gravity or weight of that body.
2) According to newton’s second law of motion a force acting on a body results in
its acceleration.
3) Thus gravitational force exerted by earth on the body produces acceleration in
the body this acceleration is called as acceleration due to gravity.
4) Acceleration due to gravity is denoted by g and it is directed towards the centre
of the earth.
 Value of g on the surface of the earth
1) Let 𝑀 = mass of the earth
𝑚 = mass of the body
𝑟 = distance of the body from centre of the earth

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 Prof. Sachin M. Shiral
(M.Sc. B.Ed.)

SUCCESS CLASSES, BARSHI

𝐺 = universal gravitational constant
𝑔 = acceleration due to gravity at a point where the body is
2) According to newton’s law of gravitation, gravitational force exerted by earth on
the body is given by
𝐺𝑀𝑚
𝐹= ............... (1)
𝑟2

3) Also, Force = mass × acceleration

∴ 𝐹 = 𝑚𝑔 .............. (2)
4) ∴ by equation (1) and (2)
𝐺𝑀𝑚
∴ 𝑚𝑔 =
𝑟2
𝐺𝑀
∴𝑔= ........... (3)
𝑟2

5) If the object is situated on the surface of the earth 𝑟 = 𝑅

∴ equation (3) becomes
𝐺𝑀
𝑔= ............... (4)
𝑅2

6) Equation (4) shows that acceleration due to gravity depends upon mass (M)
and radius (R) of earth and it does not depend upon mass of the object
i.e. acceleration due to gravity at a given point on the earth is same for all
objects.
7) Putting 𝐺 = 6.67 × 10−11 𝑁 𝑚2 𝑘𝑔2
𝑀 = 6 × 1024 𝑘𝑔
𝑅 = 6.4 × 106 𝑚
In equation (4) we get 𝑔 = 9.77 𝑚 𝑠 2
 Variation in value of g:
I. Variation in value of g due to shape of the earth:
1) Earth is not spherical it is flatten at poles and bulge at equator.
2) Hence polar radius of earth is smaller than that of equatorial radius.
𝐺𝑀
3) We have 𝑔 =
𝑅2
1
∴𝑔∝ ...... (∵ M is constant)
𝑅2

Hence as we move from equator to pole radius decreases and acceleration due
to gravity increases.
4) On the surface of the earth g is minimum at equator and its value there is
9.78 𝑚 𝑠 2 and maximum at pole and its value at pole is 9.832 𝑚 𝑠 2
II. Variation in g due to Height:
𝐺𝑀
1) We know that 𝑔 =
𝑟2

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 Prof. Sachin M. Shiral
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SUCCESS CLASSES, BARSHI

1
∴𝑔∝ ......... (∵ M is constant) ... (1)
𝑟2
2) As height from surface of earth increases r increases and by (1) acceleration
due to gravity decreases.
III. Variation in g due to depth
1) We know that
𝐺𝑀
𝑔=
𝑟2
𝑀
∴𝑔∝ ......... (∵ M is not constant) ... (1)
𝑟2

2) The value of g is maximum on the surface of the earth.

3) As depth from surface increases r decreases and along with this the part of the
earth which contributes towards the gravitational force decreases
i.e. M also decreases.
4) Thus due to combined effect of change in r and M the value of g decreases.
5) At the centre of the earth value of g is zero
 Difference between mass and weight
Mass Weight
i) Amount of matter contained in i) Gravitational force exerted by
an object is called as mass of earth on the object is called as
that object weight of that object
ii) Mass remains same everywhere ii) Weight of body changes from
place to place since g changes
iii) Its S. I. unit is kg iii) Its S. I. unit is newton (N)
iv) It is scalar quantity iv) It is vector quantity
v) Mass can never be zero v) Weight of object becomes zero at
centre of the earth
 Free fall:
1) Motion of object under the influence of the force of gravity alone is called as
free fall.
2) A stone released from particular height its motion is free fall if resistive
forces due to medium are neglected.
For its motion the velocity on reaching the earth and the time taken for it
can be calculated by sing kinematical equation and for this initial velocity
𝑢 = 0 and g is +𝑣𝑒.
3) For calculating the motion of an object thrown upward final velocity 𝑣 = 0
and g is −𝑣𝑒
4) Motion of moon and satellites of earth are also free fall.
Note: Kinematical equations are
1
𝑣 = 𝑢 + 𝑎 𝑡, 𝑠 = 𝑢𝑡 + 𝑎𝑡 2 , 𝑣 2 = 𝑢 2 + 2𝑎𝑠
2
For body falling freely equations becomes

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 Prof. Sachin M. Shiral
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SUCCESS CLASSES, BARSHI

1
𝑣 = 𝑔𝑡, 𝑠 = 𝑔𝑡 2 , 𝑣 2 = 2𝑔𝑠
2
 Gravitational potential energy:
1) The energy stored in an object because of its position or state is called as
potential energy.
2) When height h is small compared to the radius of earth (R) we can assume g
to be constant and potential energy of an object of mass m is given by
𝑃. 𝐸 = 𝑚𝑔ℎ ........ (1)
3) But for larger values of h, g is not constant hence above formula (1) is not
applicable
Hence potential energy of object of mass m is given by
𝐺𝑀𝑚
𝑃. 𝐸 = − ....... (2) where M = mass of the earth
𝑅+ℎ
4) From above equation (2) it is clear that when object is at infinite height from
surface of the earth value of its potential energy is zero there. And for
smaller heights the potential energy is less than zero i.e. it is negative
 Escape velocity:
1) The minimum velocity with which a body should be projected from the
surface of a planet or moon, so that it escapes from the gravitational
influence of the planet or moon is called as escape velocity.
 Expression for escape velocity from earth
2) When object is thrown vertically upwards with initial velocity equals escape
velocity it escape from earths gravitational influence.
3) That is object reaches at infinite height where force exerted by earth on it
becomes zero and this happens at infinite height this means that for the
object to be free from the gravity of the earth it has to reach infinite distance
from the earth and will come to rest there.
4) On the surface of the earth
1
𝐾. 𝐸. = 𝑚 𝑣𝑒 2 where 𝑚 = mass of object 𝑣𝑒 = escape velocity
2
𝐺𝑀𝑚
𝑃. 𝐸. = −
𝑅
1 𝐺𝑀𝑚
Total energy, 𝑇. 𝐸 = 𝑚𝑣𝑒 2 − ........... (1)
2 𝑅

5) When object is at infinite distance from surface of earth it comes to rest

there
∴ 𝐾. 𝐸. = 0
𝐺𝑀𝑚
𝑎𝑛𝑑 𝑃. 𝐸. = − =0
∞

𝑇. 𝐸. = 𝐾. 𝐸. +𝑃. 𝐸
∴ 𝑇. 𝐸. = 0 + 0
∴ 𝑇. 𝐸. = 0 .................(2)
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 Prof. Sachin M. Shiral
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SUCCESS CLASSES, BARSHI

6) But according to law of conservation of energy
Total energy at infinite distance = total energy on surface of the earth
1 𝐺𝑀𝑚
∴ 0 = 𝑚𝑣𝑒 2 −
2 𝑅
1 𝐺𝑀𝑚
∴ 𝑚𝑣𝑒 2 =
2 𝑅
2𝐺𝑀
∴ 𝑣𝑒 2 =
𝑅

2𝐺𝑀
∴ 𝑣𝑒 = .............. (3) is expression for escape velocity
𝑅

7) Equation (3) shows that escape velocity depends upon mass and radius of
planet and is independent of mass of the object
𝐺𝑀
8) Also, =𝑔
𝑅2

∴ 𝐺𝑀 = 𝑔𝑅2 ............... (4)

Using equation (4) in equation (3)
2𝑔𝑅 2
𝑣𝑒 =
𝑅

∴ 𝑣𝑒 = 2𝑔𝑅 ................ (4)

9) Escape velocity from earth is obtained by
Putting 𝑅 = 6.4 × 106 𝑚 in
𝑔 = 9.8 𝑚 𝑠 2 and equation (4)
∴ 𝑣𝑒 = 11.2 𝑘𝑚/𝑠

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