GRAVITATION

CHAPTER – 8 GRAVITATION

Observation of stars, planets and their motion has been the subject of attention in many
countries since the earliest of times.
The earliest recorded model for planetary motions proposed by Ptolemy about 2000 years
ago was a geocentric model. According to this theory, the sun, the moon and all planets, were in a
uniform motion in circles called epicycles with the motionless earth at the centre. However a more
elegant model in which the sun was the centre around which the planet revolved was mentioned
by Aryabhatta in 5 century AD in his treatise.
In 15 century Nicolas Copernicus proposed a definitive model, the helio-centric theory,
according to which the earth and all other planets move in a circular orbit around the sun. In 16
century Johannes Kepler analyzed the data collected by Tycho Brahe and put forth his discoveries
in the form of three laws known as Kepler’s laws.

Kepler’s laws of planetary motion.

1. Law of orbits: All planets move in elliptical orbits with the sun situated
at one of the foci of the ellipse.

Explanation: 𝑆′ and 𝑆 are the foci, 𝑎 is semi major axis and 𝑏 is the semi
minor axis.

2. Law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of
time.

Explanation: Let the sun be at one of the foci of the ellipse.

Let the position and momentum of the planet be 𝑟⃗ and 𝑝⃗ respectively.
Then the area sweep out by the planet of mass m in time ∆t is,
1
∆𝐴 = (𝑟⃗ × 𝑣⃗ ∆𝑡)
2
∆𝐴 1
= (𝑟⃗ × 𝑣⃗)
∆𝑡 2
∆𝐴 1 𝑃⃗
)
∆𝑡 = 2 (𝑟⃗ × 𝑚
∆𝐴 1
= (𝑟⃗ × 𝑝⃗ )
∆𝑡 2𝑚
∆𝐴 1
= 𝐿⃗⃗ − − − − (1)
∆𝑡 2𝑚
𝑑𝐿⃗⃗
Now we have, = 𝜏⃗
𝑑𝑡
But 𝜏⃗ = 𝑟⃗ × 𝐹⃗ where 𝐹⃗ is Gravitation Force
But the force 𝐹⃗ is central force, i e. 𝐹⃗ is along
𝑟⃗
∴ 𝑟⃗ × 𝐹⃗ = 𝑟⃗𝐹⃗ sin 0 = 0
𝑑𝐿⃗⃗
Then =0
𝑑𝑡
𝐿⃗⃗ = constant
Using the above statement in equation (1)
∆𝑨
= 𝐂𝐨𝐧𝐬𝐭𝐚𝐧𝐭
∆𝒕
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3. Law of periods: The Square of the time period of revolution of a planet is proportional to the
cube of the semi major axis of the ellipse traced out by the planet.

Explanation: If 𝑇 is the time period and 𝑎 is semei major axis, then 𝑇2 ∝ 𝑎3.

Gravitational force: Gravitational force is the force of attraction between the two bodies due to
their masses. It is one of the basic forces of nature and is always attractive.

Gravitation: The tendency of bodies to move toward each other is called gravitation.

Gravity: The attractive force between earth and any other body is called gravity.

Newton’s Universal law of Gravitation: Everybody in the universe attracts every other body with
a force which is directly proportional to the product of their masses and inversely proportional to
the square of the distance between them.

Explanation: If 𝑚1 and 𝑚2 are the masses of two bodies respectively and are separated by a
distance 𝑟⃗ then,
𝑚1𝑚2
|𝐹⃗ | ∝
𝑟⃗2
𝒎𝟏𝒎𝟐
|𝑭| = 𝑮
𝒓𝟐
where 𝐺 is universal gravitational constant.

Vector form: The force 𝐹⃗ is acting on a point mass m2 due to another point
mass m1, and the force is directed towards point mass m1. This is given by,
𝐺 𝑚1𝑚2
𝐹⃗ 21 = (−𝑟⃗)
|𝑟⃗ |2
𝑮 𝒎𝟏𝒎𝟐
𝑭𝟐𝟏 = − (𝒓)
where 𝑟⃗ is the unit vector from 𝑚1to 𝑚2 and 𝑟⃗ = 𝑟⃗ 2 − 𝑟⃗ 1

Note: The gravitational force on point mass m1 due to point mass m2 has the same magnitude as
the force on point mass m2 but the opposite direction.
i.e 𝐹⃗ 12 = −𝐹⃗ 21

Gravitational force due to multiple point masses:

If we have a collection of point masses the force on any one of
them is the vector sum of the gravitational force exerted by the
other point masses.
The total force on m is, 𝐹⃗ 𝐺 𝑚2𝑚1 𝑟⃗ + 𝐺 𝑚3𝑚1 𝑟⃗ + 𝐺 𝑚4𝑚1 𝑟⃗
1 1
2 21 2 31 2 41
𝑟⃗21 𝑟⃗31 𝑟⃗41
In case of gravitational force on a particle from a real (extended) object, we will divide the
extended object in to deferential parts each of mass dm and each producing deferential force 𝑑𝐹⃗ on

the particle, in this limit sum becomes integral and 𝐹⃗ 1 = ∫ 𝑑𝐹⃗ .

Note: If the extended object is a uniform sphere or a spherical shell we can avoid the integration by
assuming that the objects mass in concentrated at the objects centre.
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Determination of Gravitational constant:

In 1798 Henry Cavendish determined the value of G. The experimental arrangement is as shown.

The Bar AB has two small lead sphere attached at it ends. The bar is suspended from a rigid
support by a fine wire. Two large lead spheres S1 and S2 are brought close to the small ones but on
opposite sides. The big sphere attracts the nearby small ones by equal and opposite forces. There is
no net force on the bar but only torque which is equal to the F times the length of the bar and F is
the force of attraction between a big and its neighboring small sphere. Due to this torque the
suspended wire gets twisted such that the restoring torque of the wire equals to the gravitational
torque.

If 𝜃 is the angle of twist, then restoring torque is equal to 𝜏⃗𝜃.

𝐺𝑀𝑚
∴ 𝐿⃗⃗ = 𝜏⃗𝜃
𝑑2
where 𝑀 → mass of big sphere, 𝑚 → mass of small spheres, 𝐿⃗⃗ → length of the bar AB.
𝜏⃗𝜃𝑑2
𝐺=
𝑀𝑚𝐿⃗⃗
The measurement of G has been refined and the currently accepted value is,
𝐺 = 6.67 × 10−11𝑁𝑚2𝐾𝑔−2

Acceleration due to gravity: The acceleration experienced by a body due to gravitational force of
the earth is known as acceleration due to gravity.

Expression for Acceleration due to Gravity:

Let us assume that the earth is uniform sphere of mass ME.
Consider a body of mass m lying on the surface of the earth.
The magnitude of gravitational force acting on the body is given by,
𝐺𝑀𝐸𝑚
𝐹⃗ =
𝑅𝐸2
The acceleration experienced by the body of mass m due to gravity is given by
𝐹⃗ 𝐺𝑀𝐸𝑚
𝑔= = 2
𝑚 𝑅𝐸𝑚
𝑮𝑴
𝒈 = 𝟐𝑬
𝑹𝑬

Dependence of Acceleration due to gravity: The above equation suggests that g depends on
(i) mass of the earth and (ii) Radius of the earth

Note: The value of g is independent of mass of the body

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Acceleration due to gravity below and above the surface of earth
(i) Acceleration due to gravity above the surface of earth:
𝐺𝑀𝐸
The acceleration due to gravity on the surface of the earth is given by, 𝑔 =
𝑅 2𝐸
Now consider a body at a height h above the surface of earth.
The acceleration due to gravity at height h is given by,
𝐺𝑀𝐸
𝑔ℎ =
(𝑅𝐸 + ℎ)2
𝑔ℎ 𝑔ℎ 𝐺𝑀𝐸 𝑅𝐸2
by taking , we have, = ×
𝑔 𝑔 (𝑅𝐸 + ℎ)2 𝐺𝑀𝐸
𝑔ℎ 𝑅𝐸2
=
𝑔 (𝑅𝐸 + ℎ)2
𝑔ℎ = 𝑅𝐸2
𝑔 2 ℎ 2
𝑅𝐸 (1 + 𝑅 )
𝐸
𝑔ℎ = 1
𝑔 ℎ 2
(1 + 𝑅 )
𝐸
𝒈
𝒈𝒉 = 𝒉 𝟐
(𝟏 + )
𝑹𝑬
This shows that the acceleration due to gravity decreases as we go away from the surface of earth.
ℎ −2
𝐅𝐮𝐫𝐭𝐡𝐞𝐫: 𝑔ℎ = 𝑔 (1 + )
𝑅𝐸
−2 2ℎ
ℎ
Using binomial theorem, (1 + ) ≈ (1 − )
𝑅𝐸 𝑅𝐸
ℎ
As ℎ ≪ 𝑅, higher powers of can be neglected.
𝑅
𝟐𝒉
𝒉
𝑹𝑬

(ii) Acceleration due to gravity below the surface of the earth:

𝐺𝑀𝐸
The value of 𝑔 on the surface of the earth is given by, 𝑔 =
𝑅𝐸2
Now 𝑀𝐸 = volume × density
4
𝑀 = 𝜋𝑅3𝜌
𝐸 𝐸
3
4
𝐺 ( 𝜋𝑅𝐸3 𝜌)
Now, 𝑔= 3
𝑅𝐸3
4
𝑔 = 𝜋𝐺𝑅𝐸𝜌
3
When the body of mass 𝑚 is taken to a depth d, the mass of the earth
of radius (𝑅𝐸 − 𝑑) will only be effective for the gravitational pull.
The outward shell will have no resultant effect on the mass of the body.
The acceleration due to gravity on the surface of the earth of radius (𝑅𝐸 − 𝑑) is given by,
4
𝑔𝑑 = 𝜋𝐺(𝑅𝐸 − 𝑑)𝜌
43
𝑔 𝜋𝐺(𝑅 − 𝑑)𝜌
𝑑 3 𝐸
By taking =
𝑔 4
3 𝜋𝐺 𝑅𝐸𝜌

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𝑔𝑑 𝑅𝐸 − 𝑑
=
𝑔 𝑅𝐸
𝑑
𝑅 (1 − )
𝐸 𝑑
𝑔𝑑 𝑅𝐸
= = (1 − )
𝑔 𝑅𝐸 𝑅𝐸
𝒅
𝒅
𝑹𝑬

𝑑
As we go down below earth’s surface the value of g decreases by a factor (1 − )
𝑅𝐸

𝑅
Note: When 𝑑 = 𝑅, 𝑔𝑑 = 𝑔 (1 − ) = 0 , At the centre of the earth the value of g is zero. The value
𝑅
of g is more on the surface of the earth.

Gravitational potential energy: Gravitational potential energy of a body at a point is defined as

the work done in displacing the body from infinity to that point in the gravitational field.
Potential energy of the body arising due to gravitational force is called gravitational potential
energy.

Expression for Gravitational potential energy:

Consider a body of mass m placed at a distance x
from the earth of mass ME.
The gravitational force of attraction between the
body and earth is given by ,
𝐺𝑀𝐸𝑚
𝐹⃗ =
𝑥2
Now let the body of mass m be displaced from point C to B through a distance dx towards the
earth, then Work done, 𝑑𝑊 = 𝐹⃗𝑑𝑥
𝐺𝑀𝐸𝑚
𝑑𝑊 = 𝑑𝑥
𝑥2
The total work done in displacing the body of mass m from infinity to a distance r towards the
earth can be calculated by integrating the above equation between the limits 𝑥 = ∞ to 𝑥 = 𝑟⃗.
𝑥=𝑟⃗ 𝐺𝑀 𝑚
𝐸
∫ 𝑑𝑊 = ∫ 2 𝑑𝑥
𝑥=∞ 𝑥
𝑟⃗ 1
𝑊 = 𝐺𝑀𝐸𝑚 ∫ 2 𝑑𝑥
∞𝑥
𝑟⃗ 𝑟⃗
𝑥−1
𝑊 = 𝐺𝑀𝐸𝑚 ∫ 𝑥−2𝑑𝑥 = 𝐺𝑀𝐸𝑚 [ ]
∞ −1 ∞
𝑊 = − 𝐺𝑀𝐸𝑚 ( − ) 1 1
𝑟⃗ ∞
𝐺𝑀𝐸𝑚
𝑊=−
𝑟⃗
The work done is equal to the gravitation potential energy of the body and it is represented by V.
𝑮𝑴𝑬𝒎
𝑽=−
𝒓

Gravitation potential: The gravitational potential due to the gravitational force of the earth is
defined as the potential energy of a particle of unit mass at that point.
𝐺𝑀𝐸
𝑉=− ×1 (∵ 𝑚 = 1𝑘𝑔)
𝑟⃗

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𝑮𝑴𝑬
𝑽=−
𝒓
The unit of gravitational potential is J kg-1 and dimensional formula is [M0L2T-2]

Escape speed: The minimum initial speed required for an object to escape from the earth’s
gravitational field (to reach infinity) is called escape speed.

Expression for Escape speed: Consider an object of mass m is thrown upward so that it can reach
infinity, then the speed there was vf.
The energy of an object is the sum of Potential Energy and Kinetic energy.
At infinity, 𝐸 = 1 𝑚𝑣⃗2 . . . . . … … … … … . . . . (1) (Potential energy = 0 at infinity)
∞ 𝑓
2
Initially if the object was thrown with a speed vi from point at a distance (𝑅𝐸 + ℎ) from the centre
of the earth, the energy is given by,
1 𝐺𝑚𝑀𝐸..........................
𝐸 = 𝑚𝑣⃗2 − (2)
𝑖 2 𝑖 (𝑅𝐸 + ℎ)
By the principle of conservation of energy, equation (1) and (2) are equal.
1 𝐺𝑀𝐸𝑚 1
𝑚𝑣⃗2 − = 𝑚𝑣⃗2
𝑖
2 (𝑅𝐸 + ℎ) 2 𝑓

The RHS of the above equation is positive quantity with a minimum value zero, hence so must be
the LHS.
1 𝐺𝑀𝐸𝑚
𝑚𝑣⃗2 − ≥0
𝑖
2 (𝑅𝐸 + ℎ)
1 𝐺𝑀𝐸𝑚
𝑚(𝑣⃗ )2 − =0
2 𝑖 𝑚𝑖𝑛 (𝑅𝐸 + ℎ)
1 𝐺𝑀𝐸𝑚
𝑚(𝑣⃗2) =
2 𝑖 𝑚𝑖𝑛 (𝑅 + ℎ)
1 𝐺𝑀𝐸𝐸
(𝑣⃗2) =
2 𝑖 𝑚𝑖𝑛 𝑅𝐸 + ℎ
(𝑣⃗2) = 2𝐺𝑀𝐸
𝑖 𝑚𝑖𝑛
𝑅𝐸 + ℎ

(𝑣⃗ ) = √ 2𝐺𝑀𝐸
𝑖 𝑚𝑖𝑛
𝑅𝐸 + ℎ
If the object is thrown from the surface of the earth then ℎ = 0
2𝐺𝑀𝐸
(𝑣⃗𝑖)𝑚𝑖𝑛 = √
𝑅𝐸
𝐺𝑀𝐸 𝐺𝑀𝐸
But = 𝑔, then = 𝑔𝑅
𝑅𝐸2 𝐸
𝑅𝐸
(𝒗𝒊)𝒎𝒊𝒏 = √𝟐𝒈𝑹𝑬

Note: (1) Using the value of g and 𝑅𝐸, (𝑣⃗𝑖)𝑚𝑖𝑛 = 11.2 𝑘𝑚/𝑠 for the earth.
(2) (𝑣⃗𝑖)𝑚𝑖𝑛 for moon is 2.3 km/s, This is why moon has no atmosphere.

Satellite: Satellites are the celestial objects revolving around the planet.

Earth’s satellites: Earth satellite is an object which revolves around earth.

Earth has only one natural satellite - moon with a time period 27.3 days and its rotational period
about it axis is also same as that of the time period. But earth has so many artificial satellites which
have practical use in fields like telecommunication, geophysics, meteorology, cartography etc.

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Orbital velocity: The velocity required to put a satellite into its orbit around the earth is called
orbital velocity.

Expression for Orbital speed: Consider a satellite of mass m and speed 𝑣⃗0 in a circular orbit at a
distance (𝑅𝐸 + ℎ) from the centre of the earth.
𝑚𝑣⃗02
The centripetal force required for this obit is, 𝐹⃗𝑐 =
(𝑅𝐸 + ℎ)
𝐺𝑚𝑀𝐸
The centripetal force is provided by the gravitational force, 𝐹⃗ =
(𝑅𝐸 + ℎ)2
𝑚𝑣⃗02 𝐺𝑚𝑀𝐸
Equating the equations, we get =
(𝑅𝐸 + ℎ) (𝑅𝐸 + ℎ)2
𝐺𝑀𝐸
𝑣⃗20 =
𝑅𝐸 + ℎ
𝑮𝑴𝑬
𝑹𝑬 + 𝒉

𝐺𝑀𝐸
For ℎ = 0 (since ℎ << 𝑅𝐸 ), we have 𝑣⃗0 = √
𝑅𝐸
𝐺𝑀𝐸
(∵ 𝑔 = )
𝑅𝐸2

Time period of a satellite: In every orbit the satellite travels a distance 2𝜋(𝑅𝐸 + ℎ) with speed 𝑣⃗0,
then its time period is,
2𝜋(𝑅𝐸 + ℎ) 2𝜋(𝑅𝐸 + ℎ)
𝑇= =
𝑣⃗0 𝐺𝑀𝐸
(√ )
(𝑅𝐸 + ℎ)
1
2𝜋(𝑅𝐸 + ℎ)(𝑅𝐸 + ℎ)2
𝑇=
√𝐺𝑀𝐸
3
2𝜋(𝑅𝐸 + ℎ)2
𝑇=
√𝐺𝑀𝐸
𝟑
𝟐𝝅(𝑹𝑬 + 𝒉)𝟐
𝑻=

This is the expression for time period of a satellite.

Note: Further, squaring on both sides

4𝜋2(𝑅𝐸 + ℎ)3
𝑇 =
2
𝐺𝑀𝐸
4𝜋2
𝑇2 = ( ) (𝑅𝐸 + ℎ)3
𝐺𝑀𝐸
4𝜋2
𝑻𝟐 = 𝑲(𝑹𝑬 + 𝒉)𝟑 where 𝐾 = , Which is the Kepler’s law of periods
𝐺𝑀𝐸

Total energy of an orbiting satellite: The energy of a satellite in its orbit is the sum of the potential
energy due to the gravitational force of attraction and kinetic energy due to the orbital motion.

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Expression for total energy of the satellite:
We have, 𝐸 = 𝑉 + 𝐾𝐸
𝐺𝑀𝐸𝑚 1
𝐸=− + 𝑚𝑣⃗2
0
(𝑅𝐸 + ℎ) 2
𝐸 = − 𝐺𝑀𝐸𝑚 + 1 𝑚 𝐺𝑀𝐸
(𝑅𝐸 + ℎ) 2 (𝑅𝐸 + ℎ)
𝐸= 𝐺𝑀 𝐸𝑚 1
(−1 + )
(𝑅𝐸 + ℎ) 2
𝑮𝑴𝑬𝒎
𝑬=− The total energy of an orbiting satellite is negative.
𝟐(𝑹𝑬 + 𝒉)

Note: If total energy of an orbiting satellite is equal or greater than zero then the satellite does not
remain in the orbit, it escapes from the earths pull. Negative energy implies that the satellite is
bound to the earth.

Geostationary satellites: Satellites in a circular orbits around the earth in the equatorial plane with
time period T = 24 hours are called Geostationary satellites and the orbit is called Geo-synchronous
orbit.
For geostationary orbit,
1. The time period of the satellite is equal to the rotational period of the earth.
2. The height form the equatorial plane must be about 35800km (nearly equal to 36000 km)
3. Direction of rotation of the satellite must be same as that of the earth.

Use of Geostationary Satellites: They are used for telecommunication purpose.

Note: The INSAT groups of satellites are Geo-stationary satellites.

Polar satellites: The low altitude satellites which go around the poles of the earth in a north south
direction are called polar satellites and the orbit in called polar orbits. The time period of a polar
satellite is about 100 minutes and hence it crosses any altitude many times a day.

Uses of polar satellites:

1. They are used for remote sensing. The IRS group of satellites are Remote sensing Satellites.
2. They are used for environmental studies, and also in the field of meterology.
3. They are used for natural resource survey
4. They are used for forest, waste land, drought assessment etc.

Weightlessness: When there is no normal reaction or upward force on the object from any surface,
then the weight of the object will become zero, this particular situation of the object is termed as
weight less ness.
When an object is in free fall, it is weightless.
While a man or an object accelerating downwards, if the lift is cutoff, feels weightless.
In a satellite revolving round the earth, gravitational force of the earth provides necessary
centripetal force to the satellite and this force is opposite to the force exerted by satellite on the
man, thus the person inside a satellite feels weightlessness.

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Suggested Questions.
One mark.
1. How does acceleration due to gravity vary with altitude?
2. What is escape velocity?
3. How does the escape speed of a particle depend on the mass of the earth?
4. What is the time period of revolution of geostationary satellite? or
What is the period of geostationary satellite?
5. What is geo-stationary satellite?
6. How does the speed of the earth changes when it is nearer to the sun?
7. Write the SI unit of 𝐺.

Two marks.
1. Write the relation between 𝑔 and 𝐺 and explain the terms.
2. State and explain Newton’s law of gravitation.
or
State Newton’s law of gravitation and write its mathematical form.
3. Mention any two applications of satellite.

Three marks.
1. State Kepler’s laws of planetary motion.
2. Derive an expression for orbital velocity of a satellite.
3. Derive the relation between 𝑔 and 𝐺. or Arrive at the relation 𝑔 =
𝐺𝑀𝐸
𝑅2𝐸

4. State and explain Newton’s universal law of gravitation and express its equation in vector
form.
5. What are geo-stationary satellites? Mention its time period of revolution.

Five marks.
1. Derive an expression for acceleration due to gravity above the surface of earth.
2. State Kepler’s laws of planetary motion. Draw diagram to explain any two of them.
or
State and explain Kepler’s law of planetary motion.
3. Define orbital velocity and escape velocity. Write the expression for them. How are they
related?
4. State and explain Newton’s universal law of gravitation.
5. Obtain the expression for acceleration due to gravity with depth of the earth.
or
Derive the expression for the variation of gravity 𝑔 with depth.

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Problem (1) Suppose there existed a planet that went around the sun twice as fast as the earth.
What would its orbital size be as composed to that of the earth?

From the law of periods, Where

𝑇2 𝑅2 → Radius of the orbit of planet around the
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑅23 sun.
𝑇 𝑇2
1
=
2 𝑅1 → Radius of the orbit of the earth
𝑅13 𝑅23
𝜔1 → angular speed of earth
𝑇2 × 𝑅3
𝑅3 = 2 1 𝜔2 → angular speed of planet
2
𝑇12
2𝜋 2 3 2
(𝜔 ) × 𝑅1 2𝜋 𝜔1 3
𝑅23 = 22𝜋 2 (𝑇 = 𝜔 ) 𝑅2 = ( ) 𝑅1
2𝜔2
(𝜔 ) 2
𝑅2 = (1) 𝑅1
1 3
2𝜋 2 3 2
(𝜔 ) × 𝑅1 × 𝜔1 2
𝑅32 = 2 1
(2𝜋)2 𝑅2 = 𝑅 = 0.63𝑅1
𝜔2 1.49 1
𝑅3 = 1 . 𝑅3 Orbital size of the planet is smaller than the
2
𝜔22 1 earth.
2
𝑅 = ( 𝜔1 3 𝑅
2 ) 1
𝜔2

Problem (2) A Saturn year is 29.5 times the earth’s years. How far is the Saturn from the sun if the
earth is 1.50 × 108 km away from the sun?

2 3
Earth year 𝑇𝑒 = 1 year 𝑇𝑠 × 𝑅𝑒
𝑅𝑠3 =
Saturn year 𝑇𝑠 = 29.5 year 𝑇𝑒2
Radius of the orbit of earth = 𝑅𝑒 = 1.50 × 108 (29.5)2 × (1.50 × 108)3
𝑅𝑠3 =
km 12
𝑠 = 2.947 × 10 km
𝑅3
2 2 27
𝑇𝑒 𝑇𝑠
From Kepler’s III law, = 𝑅𝑠 = 1.43 × 109𝑘𝑚 = 1.43 × 1012𝑚
𝑅𝑒 𝑅𝑠3
3

Problem (3) Assuming earth to be a sphere of uniform mass density, how much would a body
weighs half way down to the Centre of earth, if it weighed 250N on the surface?
Problem (4) An aircraft executes horizontal loop of radius of 1km with a steady speed of 900kmph.
Compare the centripetal acceleration with acceleration due to gravity.
Problem (5) A satellite orbits the earth at a height of 400km above the surface. How much energy
must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the
satellite = 200kg; mass of the earth = 6.0 × 1024kg; radius of the earth = 6.4 × 106m; G = 6.67 ×
10−11𝑁𝑚2𝑘𝑔−2.
Problem (6) A body weighs 63N on the surface of the earth. What is the gravitational force on it
due to the earth at a height equal to half the radius of the earth?
Problem (7) Calculate g at the bottom of a mine 10km deep and at an height 20km above the
earth’s surface. Radius of the earth=6.4 × 106m and g on the earth’s surface =9.8ms-2.
Problem (8) The size of a planet is same as that of the earth and its mass is four times that of the
earth. Find the potential energy of the mass 2kg at a height of 2m from the planet. (g on the earth’s
surface =10ms-2)

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 GRAVITATION
Problem (9) The escape speed of a projectile on the earth’s surface is 11.2kms-1. A body is projected
out with thrice this speed. What is the speed of the body far away from the earth? Ignore the
presence of sun and other planets.
Problem (10) You are given the following data: g = 9.81 ms–2, RE = 6.37×106 m, the distance to the
moon R = 3.84×108 m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of
the Earth ME in two different ways.
Problem (11) The mass of the planet is 90 times that of the moon and its radius 3 times that of
moon. Compare the weight of a body on the surface of the moon with its weight on the surface of
the planet.
Problem (12) The mass and diameter of a planet are 3 times those of the earth. what is the
acceleration due to gravity on the surface of the planet? Given 𝑔 = 9.8 𝑚 𝑠−2
Problem (13) The planet mars take 1.88 years to complete one revolution around the sun. The
mean distance of the earth from the sun is 1.5×10 8km. Calculate that of planet Mars.
Problem (14) Calculate the orbital velocity and period of revolution of an artificial satellite of earth
moving at an altitude of 2000 𝑘𝑚 if radius of the earth is 6000 𝑘𝑚, mass of the earth is 6 ×
1024 𝑘𝑔, 𝐺 = 6.67 × 10−11 𝑁𝑚2𝑘𝑔−2.

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