GRAVITATION

MADE BY-
HIMANSHU
CLASS 11TH A
 Definition for Gravitation
 Acceleration Due to Gravity
 Variation Of “G” With Respect to Height And Depth
 Escape Velocity
 Orbital Velocity
 Gravitational Potential
 Time period of a Satellite
 Height of Satellite
 Binding Energy
 Various Types of Satellite
 Kepler’s Law of Planetary motion
• Gravitation is an Natural Phenomenon by physical
bodies attract each other.

• Newton focused on his attention on the motion of

the moon about the earth.

• He declared that the laws of nature that operating

between earth and the apple is same as the laws
operating between earth and moon.

• He calculated the acceleration due to gravitation

(on earth) to be 9.8m/s2 .

1
• Newton predicted that a ∝
𝑟2
• Also,

𝑚
• F ∝
𝑟2

• By , The third law of motion, the force on a body

due to the earth must be equal to the force on the
earth due to the body. Therefore, this force should
also be proportional to the mass of the earth. Thus ,
the Force between the earth and a body is

𝑀𝑚
• F ∝
𝑟2

𝐺𝑀𝑚
• F=
𝑟2

• G is called the universal constant of gravitation and

its value is found to be 6.67x10-11 N-m2 /kg2 .
 The rotation of the earth around the sun or that of the
moon around earth is explained on the basis of this law.
 The tide are formed in ocean due to the gravitational
force of attraction between earth and the moon.
 The value of g can be used to predict the orbits and time
period of an satellite.

Attraction Force between two bodies

 The Gravitation force between two masses is independent
of the intervening medium.
 The mutual gravitational forces between two bodies are
equal and opposite i.e. Gravitational Forces obey
Newton’s third law of motion.
 The gravitational force is an conservative force.
 The law of gravitation holds only for point masses.
 The gravitational force between two point masses is a
central force. Its magnitude depends only on r and has
no angular dependence .
 The Gravitational force between two bodies is
independent of the presence of other bodies.
 Consider an Elevator freely falling from North pole to
South pole through the center of earth. Describe its
motion??
 For answer Press Enter ..
 Suppose a mass ‘m’ is situated outside the earth at a
distance ‘r’ from it centre. The gravitational force on the
mass is
𝑚𝑣
 F=
𝑟2
 F=mg.
𝐺𝑀𝑚
 mg=
𝑟2
𝐺𝑀
 g= 2
𝑟
 By the above equation we can say that Acceleration due
to gravity is independent of mass.
 Determine the mass of earth using the knowledge of G.
G=6.6x1011 and g=9.8 m/s2 and r = 6.36x106 .
 For answer press enter…
𝐺𝑀
 As we know g= 2 .
𝑟
 m= 𝑔𝑟2
𝐺
.
9.8 ∗(6.96∗106)2
 m=
𝐺
 m=5.97x1024
 Do you know ?? Cavendish was the first person the
calculate the mass of earth because he was the first
person to calculate value of G.
 Consider a Body placed at the point ‘p’
above the earth surface at a distance of
‘h’. The g will be .
𝐺𝑀
 g= at earth surface. …(i).
𝑟2
𝐺𝑀
 gh = 2 at point m ….(ii).
𝑟+ℎ
 Dividing (ii) by (i). We get
𝐺𝑀
2
𝑔ℎ 𝑟+ℎ
 = 𝐺𝑀
ℎ 2
𝑟
𝑟2

𝑔ℎ
= 2…..(iii)
ℎ 𝑟+ℎ

𝑔ℎ
ℎ
= 𝑟2
ℎ
𝑟2(1+ 𝑟 )2
ℎ

𝑔ℎ
= (1 + )-2
ℎ 𝑟
 𝑔ℎ 1−2ℎ
 = …(iv)
𝑔 𝑟
 Both the eq. (iii and iv) show that the ‘g’ decreases as the
height increases.
 Eq. iii must be used in the numerical when ‘h’ is
comparable to ‘r’.
 Eq. iv must be used in the numerical when h<<r.

 Variation of g with respect to the

altitude.
 Consider a Body placed at the point ‘d’
below the earth surface at a distance of
‘d’. The g will be .
𝐺𝑀
 g= 2
𝑟
 M = Volume x Density = 4𝜋𝑟3𝜌.
4
𝐺 3𝜋𝑟3𝜌
 g=
𝑟2
4
 g = 𝐺 𝜋𝑟𝜌. At earth surface ...(i)
3
4
 gd= 𝐺 𝜋(𝑟 − 𝑑)𝜌. At the point d …(ii)
3
 Dividing (ii) by (i).
 gd/g = G4/3π(r-d)ρ / G4/3πrρ
 gd/g = g(1-d/r).
 Clearly the “g” increases with increase in depth.
 The Variation of g with respect to Depth and Height
(graph).
 The Variation of g with respect to height will never be
zero. It will increase after certain height.
 The Variation of g with respect to depth will be zero
when the body is placed at the center of earth.
 Earth is not perfectly sphere. It is
flattened at the poles and bulges
out at equator.
𝐺𝑚
g= 2
𝑅

1
g∝ Type equation here..
𝑅2

 The radius at Equator is greater than the radius at pole by

21 km.

Therefore Ge > Gp.

 The variation of g between the poles and the equator is

about 0.5%.
Gravitational field
 Gravitational field . Two bodies attract each other by the gravitational force
even if they are not in direct contact. This interaction is called action at a
distance. It can best explained in terms of concept of field. According to the
field concept.
 Every mass modifies the space around it . This modified space is called gravitational field
 When any other mass is placed in this field , it feels a gravitational force of attraction due
to its interaction with the gravitational field

 The space surrounding a material body within which its gravitational force of
attraction can be experiences a force of attraction towards the centre of
earth
Gravitational Potential Energy
 Amount of work done in bringing a body from infinity to the given point in
the gravitational field of the other.
 Expression for Gravitational potential energy.
 F=
𝐺𝑀𝑚 A
𝑥2
dx
 Work down in bringing the body to point
B from Point A. B
 W = Fdx P
 W=
𝐺𝑀𝑚
dx x
𝑥2
𝑟 𝐺𝑀𝑚
 𝑤 = ∞ 𝑥2
dx
𝑟 −2
 W = GMm ∞
x r
𝑟 −1
 W = -GMm ∞
x R
O
 1 1
 W = -GMm [𝑅-∞]
𝐺𝑀𝑚
 U=- 𝑟

 Gravitational Potential
 Amount of work done in bringing a body of unit mass from infinity to the given
point
𝐺𝑀
 Gravitational Potential = -
𝑟
 If we throw a ball into air , it rises to a certain height and
falls back. If we throw it with a greater velocity , it will rise
higher before falling down. If we throw with sufficient
velocity , it will never come back . i.e. It will escape from
the gravitational pull of the earth.
 The minimum velocity required to do so is called escape
velocity.
 Consider the earth to be a sphere of mass M and radius R
with centre O.
 Using total energy concept we can derive the equation for escape
velocity.
 At any point the Total energy must be zero.
 We know that Kinetic energy = ½ 𝑚𝑣𝑒2
−𝐺𝑀𝑚
 Potential energy =
𝑅
 K.E+P.E = 0
−𝐺𝑀𝑚
 ½ 𝑚𝑣𝑒2 + =0
𝑅
2𝐺𝑀
 ve2 =
𝑅
2𝐺𝑀
 Ve =
𝑅
 Multiply and divide by R
 Ve = (2gR)
 Can you tell why moon has no atmosphere ??
 For answer press enter
 Due to the small value of g. The escape velocity in the moon is 2.38 km/s .
The air molecules have thermal velocity is greater than the escape velocity
and therefore air molecules escape.
 Satellite

Artificial Natural

Geostationary Polar
• Satellite is an body which continuously revolves on it own
around and a much larger body in a stable orbit.
• Natural satellites : A satellite created by nature is called
natural satellite . example : moon.
• Artificial satellite : A man made satellite is called an
artificial satellite. Example Chandrayaan .
• World’s Frist satellite was SPUTNIK-1.
• Principle for launching a satellite : Consider a high tower
with its top projecting outside the earth’s atmosphere.
• Lets throw a body horizontally from the top of the tower
with different velocities.
• As we increase the velocity of horizontal projection , the
body will hit the ground at point farther and farther from
the foot of the tower.
• At certain velocity the body will not hit the ground , but
always be in a state of free fall under the influence of the
gravity.
• Then the body will follow a stable circular orbit . And that
body is called satellite.
Click on the video
• Orbital velocity is the velocity required to put
the satellite into its orbit around earth
 Force of gravity on satellite
𝐺𝑀𝑚
 F=
(𝑅+ℎ)2
 Centripetal Force required by the satellite to keep its orbit
𝑚𝑣𝑜2
 F=
(𝑅+ℎ)2 Vo
 In equilibrium , the centripetal force
is just provided by the gravitational
pull of the earth. R R+h
𝑚𝑣𝑜2
= (𝑅+ℎ)2
𝐺𝑀𝑚 h

(𝑅+ℎ)
𝐺𝑀
 Vo2 =
(𝑅+ℎ)
𝐺𝑀 𝐺𝑀
 Vo = simplified eq is
(𝑅+ℎ) 𝑅
• It is the time takes by a satellite to complete one
revolution around the earth. It is given by
• T = circumference of the orbit / orbital velocity
2𝜋(𝑅+ℎ)
• T=
𝐺𝑀
𝑅+ℎ
3
𝑅+ℎ
• T=2𝜋
𝐺𝑚

• If the earth is a sphere the density = 𝜌 then mass would

4
be = volume * density = 𝜋𝑅3𝜌
3
3 3
𝑅+ℎ 3𝜋 𝑅+ℎ
• T= 2𝜋 4 =
𝐺 3𝜋𝑅3𝜌 𝐺𝜌𝑅3
 3
𝑅+ℎ
• T=2𝜋 (Gm = gR2)
𝐺𝑚
3
𝑅+ℎ
• T =4𝜋
2 2
𝑔𝑅2
𝑇2𝑅2𝑔
• (R+h)3 =
4𝜋2
3
𝑇2𝑅2𝑔
• R+h = 4𝜋2
3
𝑇2𝑅2𝑔
•H= 4𝜋2 -R
• A satellite which revolves around the earth in tis
equatorial plane with the same angular speed and in the
same direction as the earth rotates about its own axis is
called a geostationary or synchronous satellite.
• It should revolve in an orbit concentric and coplanar
with the equatorial plane of the earth.

• Its sense of rotation should be same as that of the

earth , i.e From west to east.

• Its period of revolution around the earth should be

exactly same as that of the earth about its own axis , i.e
24 hours

• It should revolve at a height of exactly 35930 km.

• In communicating radio,T.V and telephone signals
across the world.
• In studying the upper regions of the atmosphere.
• In Forecasting weather.
• In studying meteorites.
• In studying solar radiation and cosmic rays.
• And used in GPS (Global positioning System).
• A satellite that revolves in a planar orbit is called a polar
satellite. Eg IERS (Indian earth resources satellites)
• Uses of Polar satellite
– Polar satellites are used in weather and environment monitoring.
– Spying
– Study topography of other celestial
bodies
• Consider a satellite of mass m moving around the earth
with velocity in an orbit of a radius r.
• Because of the gravitation pull of the earth, the satellite
has a potential energy which is given by
𝐺𝑀𝑚
• U= -
𝑟
1 𝐺𝑀
• Kinetic energy = m .
2 𝑅
𝐺𝑀𝑚 1 𝐺𝑀
• Total Energy = - + m
𝑟 2 𝑅
𝐺𝑀𝑚
• E=-
2𝑟

• The negative sign shows that the satellite is bound to the

earth.
• The energy required by a satellite to leave its orbit
around the earth and escape to infinity is called binding
energy.
𝐺𝑀𝑚
• Binding energy = +
2𝑟
𝐺𝑀𝑚
• Because the total energy of the satellite is -
2𝑟
in order to escape into infinity , it must be supplied extra
energy so that its energy E becomes zero.
• Law of orbits (first law) : Each planet revolves around the
sun in an elliptical orbit with the sun situated at the one
of the two foci.
• Law of areas (second law) : The radius vector drawn
from the sun to a planet sweeps out equal areas in equal
intervals of time i.e the areal velocity ( area covered per
unit time) of a planet around the sun is constant.
 𝑣→
 Proof : 𝑣→
P’
 𝜏 = 𝑟 → x 𝐹 → = 0 [ opp. Faced] planet
∆𝐴
𝑑𝐿→
 𝜏= = 0.
𝑑𝑡
𝑟→ 𝐹→
 Or 𝐿→ = constant .
→ sun
 ∆𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
→ 1 →
 ∆𝐴 = 𝑟 → x 𝑃→ P (∆𝐴 = Area)
2
𝑃→
 𝑃→ P =∆ 𝑟→ = 𝑣→ ∆t = ∆t
𝑚
 → 1 𝑃→
 ∆𝐴 = 𝑟→ x ∆t
2 𝑚
→
∆𝐴 1 → 𝑃→
 = 𝑟 x
∆t 2 𝑚
→
∆𝐴 1
 = (𝑟 → x 𝑃→ )
∆t 2𝑚
→ →
∆𝐴 𝐿
 =
∆t 2𝑚
→
∆𝐴
 = constant [L and m are constant]
∆t
• Law of periods (third law) : The square of the period of
revolution of a planet around the sun is proportional to
the cube of the semi major axis of the its elliptical orbit.
• Proof : Suppose a planet of mass m moves around the
sun in a circular orbit of radius r with orbital speed v. Let
M be the mass of the sun. The force of gravitation
between the sun and the planet provides the necessary
centripetal force.
𝑚𝑣2 𝐺𝑀𝑚 𝐺𝑀 .
• = or v2 =
𝑟 𝑟2 𝑟
𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 2𝜋𝑟
• v= =
𝑃𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑡
4𝜋2𝑟2 𝐺𝑀
• =
𝑡2 𝑟
 4𝜋2
 𝑡=
2 x r3 = constant r3
𝐺𝑀

 𝑡2 ∝ r3 .
 For better understanding please click the here.
THANKS FOR
WATCHING