**This article is formulated according to the 8th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.**

Observation of stars, planets, and their motion has been the subject of attention in many countries since the earliest of times. According to this theory, the sun, the moon, and all planets were in uniform motion in circles called epicycles with the motionless earth at the center.

However, a more elegant model in which the sun was the center around which the planet revolved was mentioned by Aryabhata in the 5 century AD in his treatise. In the 15 century, Nicolas Copernicus proposed a definitive model, the heliocentric theory, according to which the earth and all other planets move in a circular orbit around the sun.

In the 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,

∆𝐴 = (1/2)(𝑟⃗ × 𝑣⃗∆𝑡)

∆𝐴/∆𝑡 = (1/2)(𝑟⃗ × 𝑣⃗)

∆𝐴/∆𝑡 = (1/2)(𝑟⃗ × (𝑃⃗/𝑚))

∆𝐴/∆𝑡 = (1/2𝑚)(𝑟⃗ × 𝑝⃗)

∆𝐴/∆𝑡 = (1/2𝑚)𝐿⃗ − − − −> (1)

Now we have, 𝑑𝐿⃗/𝑑𝑡 = 𝜏⃗

But 𝜏⃗ = 𝑟⃗ × 𝐹⃗ where 𝐹⃗ is Gravitation Force

But the force 𝐹⃗ is central force, i e. 𝐹⃗ is along 𝑟⃗ ∴ 𝑟⃗ × 𝐹⃗ = 𝑟𝐹 sin 0 = 0

Then 𝑑𝐿⃗/𝑑𝑡 = 0

𝐿⃗ = constant

Using the above statement in equation (1)

∆𝑨/∆𝒕 = 𝐂𝐨𝐧𝐬𝐭𝐚𝐧𝐭

**Note:** The law of areas is the consequence of the conservation of angular momentum.

**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 the semi-major axis, then 𝑇^{2} ∝ 𝑎^{3}.

## 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 the 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 that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

**Explanation:** If 𝑚_{1} and 𝑚_{2} are the masses of two bodies respectively and are separated by a distance 𝑟 then,

|𝐹⃗| ∝ 𝑚_{1}𝑚_{2}/𝑟^{2}

|𝑭⃗| = 𝑮𝒎_{𝟏}𝒎_{𝟐}/𝒓^{𝟐}

where 𝐺 is the universal gravitational constant.

## Vector form

The force 𝐹⃗ is acting on a point mass m_{2} due to another point mass m_{1}, and the force is directed towards point mass m_{1}. This is given by,

𝐹⃗_{21} = [𝐺𝑚_{1}𝑚_{2}/ |𝑟⃗|^{2}] (−𝑟̂)

𝑭⃗_{𝟐𝟏} = −𝑮𝒎_{𝟏}𝒎_{𝟐}/|𝒓⃗|^{𝟐} (𝒓̂) where 𝑟̂is the unit vector from 𝑚_{1} to 𝑚_{2} and

𝑟⃗ = 𝑟⃗_{2} − 𝑟⃗_{1}

**Note:** The gravitational force on point mass m_{1} due to point mass m_{2} has the same magnitude as the force on point mass m_{2} but in 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_{1} is,

𝐹_{1 }=[𝐺𝑚_{2}𝑚_{1}/𝑟_{21}^{2}] 𝑟̂_{21} + [𝐺𝑚_{2}𝑚_{1}/𝑟_{31}^{2}] 𝑟̂_{31} + [𝐺𝑚_{2}𝑚_{1}/𝑟_{41}^{2}] 𝑟̂_{41}

In case of gravitational force on a particle from a real (extended) object, we will divide the extended object into deferential parts each of mass dm and each producing deferential force 𝑑𝐹 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 mass of the object is concentrated at the object’s center.

## 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 spheres attached at its ends. The bar is suspended from rigid support by a fine wire. Two large lead spheres S_{1} and S_{2} 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 F times the length of the bar and F is the force of attraction between a big and it’s neighboring small sphere. Due to this torque, the suspended wire gets twisted such that the restoring torque of the wire equals the gravitational torque. If 𝜃 is the angle of twist, then restoring torque is equal to 𝜏_{𝜃}. ∴ [𝐺𝑀𝑚/𝑑^{2}]𝐿 = 𝜏𝜃 where 𝑀 → mass of big sphere, 𝑚 → mass of small spheres, 𝐿 → length of the bar AB.

𝐺 = 𝜏_{𝜃}𝑑^{2}/𝑀𝑚𝐿

The measurement of G has been refined and the currently accepted value is,

𝐺 = 6.67 × 10^{−11}𝑁𝑚^{2}𝐾𝑔^{−2}

## Acceleration due to gravity

The acceleration experienced by a body due to the 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 a uniform sphere of mass M_{E}. 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,

𝐹 = 𝐺𝑀_{𝐸}𝑚/𝑅_{𝐸}^{2}

The acceleration experienced by the body of mass m due to gravity is given by 𝑔 = 𝐹/𝑚 = 𝐺𝑀_{𝐸}𝑚/𝑅_{𝐸}^{2}

𝑚𝒈 = 𝑮𝑴_{𝑬}/𝑹_{𝑬}^{𝟐}

## 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 the mass of the body.

## Acceleration is due to gravity below and above the surface of the earth.

**(i) Acceleration due to gravity above the surface of the earth**

The acceleration due to gravity on the surface of the earth is given by,

𝑔 = 𝐺𝑀_{𝐸}/𝑅_{𝐸}^{2}

Now consider a body at a height h above the surface of the earth. The acceleration due to gravity at height h is given by,

𝑔_{ℎ} = 𝐺𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)^{2}

by taking 𝑔_{ℎ}/𝑔, we have,

𝑔_{ℎ}/𝑔 = (𝐺𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)^{2})× (𝑅_{𝐸}^{2}/𝐺𝑀_{𝐸})

𝑔_{ℎ}/𝑔 = 𝑅_{𝐸}^{2}(𝑅_{𝐸} + ℎ)^{2}

𝑔_{ℎ}/𝑔 = 𝑅_{𝐸}^{2}/𝑅_{𝐸}^{2}(1 + (ℎ/𝑅_{𝐸}))^{2}

𝑔_{ℎ}/𝑔 = 1/(1 + (ℎ/𝑅_{𝐸}))^{2}

𝒈_{𝒉} = 𝒈(𝟏 + (𝒉/𝑹_{𝑬}))^{𝟐}^{}

This shows that the acceleration due to gravity decreases as we go away from the surface of earth.

𝐅𝐮𝐫𝐭𝐡𝐞𝐫: 𝑔_{ℎ} = 𝑔 (1 + (ℎ/𝑅_{𝐸}))^{−2}

Using binomial theorem, (1 + ℎ_{𝑅𝐸})^{−2} ≈ (1 – (2ℎ/𝑅_{𝐸}))

As ℎ ≪ 𝑅, higher powers of ℎ/𝑅 can be neglected.

𝒈_{𝒉} = 𝒈(𝟏 – (𝟐𝒉/𝑹_{𝑬}))

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

The value of 𝑔 on the surface of the earth is given by,

𝑔 = 𝐺𝑀_{𝐸}/𝑅_{𝐸}^{2}

Now 𝑀_{𝐸} = volume × density

𝑀_{𝐸} = (4/3)𝜋𝑅_{𝐸}^{3}𝜌

Now, 𝑔 = 𝐺((4/3)𝜋𝑅_{𝐸}^{3}𝜌)/𝑅_{𝐸}^{2}

𝑔 = (4/3)𝜋𝐺𝑅_{𝐸}𝜌

When the body of mass 𝑚 is taken to a depth d, the mass of the earth of radius (𝑅_{𝐸} − 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 (𝑅_{𝐸} − 𝑑) is given by, 𝑔_{𝑑} = (4/3)𝜋𝐺(𝑅_{𝐸} − 𝑑)𝜌.

By taking 𝑔_{𝑑}/𝑔 =[(4/3)𝜋𝐺(𝑅_{𝐸} − 𝑑)/(4/3)𝜋𝐺𝑅_{𝐸}𝜌]

𝑔_{𝑑}/𝑔 = (𝑅_{𝐸} – 𝑑)/𝑅_{𝐸}

𝑔_{𝑑}/𝑔 = 𝑅_{𝐸}(1 – (𝑑/𝑅_{𝐸}))/𝑅_{𝐸} = (1 – 𝑑/𝑅_{𝐸})

𝒈_{𝒅} = 𝒈(𝟏 – 𝒅/𝑹_{𝑬})

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

**Note:** When 𝑑 = 𝑅, 𝑔 = 𝑔 (1 – 𝑅/𝑅) = 0, At the center of the earth the value of g is zero. The value of g is more on the surface of the earth.

## Gravitational potential energy

The 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. The 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 M_{E}. The gravitational force of attraction between the body and the earth is given by,

𝐹 = 𝐺𝑀_{𝐸}𝑚/𝑥^{2}

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 = 𝐹𝑑𝑥

𝑑𝑊 = 𝐺𝑀_{𝐸}𝑚/𝑥^{2}𝑑𝑥

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 = 𝑟.

The work done is equal to the gravitation potential energy of the body and it is represented by V.

𝑽 = −𝑮𝑴_{𝑬}𝒎/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.

𝑉 = −(𝐺𝑀_{𝐸}/𝑟) × 1 (∵ 𝑚 = 1𝑘𝑔)

𝑽 = −𝑮𝑴_{𝑬}/𝒓

The unit of gravitational potential is Jkg^{-1} and dimensional formula is [M^{0}L^{2}T^{-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 v_{f}. The energy of an object is the sum of Potential Energy and Kinetic energy. At infinity, 𝐸_{∞} = (1/2)𝑚𝑣_{𝑓}^{2} . . . . . … … … … … . . . .> (1) (Potential energy = 0 at infinity).

Initially if the object was thrown with a speed v_{i }from point at a distance (𝑅_{𝐸} + ℎ) from the centre of the earth, the energy is given by, 𝐸_{𝑖} = (1/2)𝑚𝑣_{𝑖}^{2} – (𝐺𝑚𝑀_{𝐸})/(𝑅_{𝐸} + ℎ) . . . . . . . . . .> (2)

By the principle of conservation of energy, equation (1) and (2) are equal.

(1/2)𝑚𝑣_{𝑖}^{2} − 𝐺𝑀_{𝐸}𝑚/(𝑅_{𝐸} + ℎ) = (1/2)𝑚𝑣_{𝑓}^{2}

The RHS of the above equation is positive quantity with a minimum value zero, hence so must be the LHS.

(1/2)𝑚𝑣_{𝑖}^{2} − 𝐺𝑀_{𝐸}𝑚/(𝑅_{𝐸} + ℎ) ≥ 0

(1/2)𝑚(𝑣_{𝑖}^{2})_{𝑚𝑖𝑛}_{ }− 𝐺𝑀_{𝐸}𝑚/(𝑅_{𝐸} + ℎ) = 0

(1/2)𝑚(𝑣_{𝑖}^{2})_{𝑚𝑖𝑛} = 𝐺𝑀_{𝐸}𝑚/(𝑅_{𝐸} + ℎ)

(1/2)(𝑣_{𝑖}^{2})_{𝑚𝑖𝑛} = 𝐺𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)

(𝑣_{𝑖}^{2})_{𝑚𝑖𝑛} = 2𝐺𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)

(𝑣_{𝑖})_{𝑚𝑖𝑛} = √[2𝐺𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)]

If the object is thrown from the surface of the earth then

ℎ = 0

(𝑣_{𝑖})_{𝑚𝑖𝑛} = √(2𝐺𝑀_{𝐸}/𝑅_{𝐸})

But 𝐺𝑀_{𝐸}/𝑅_{𝐸}^{2} = 𝑔, then 𝐺𝑀_{𝐸}/𝑅_{𝐸} = 𝑔𝑅_{𝐸}

(𝒗_{𝒊})_{𝒎𝒊𝒏} = √(𝟐𝒈𝑹_{𝑬})

**Note:**

(1) Using the value of g and 𝑅_{𝐸}, (𝑣_{𝑖})_{𝑚𝑖𝑛} = 11.2 𝑘𝑚/𝑠 for the earth.

(2) (𝑣_{𝑖})_{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 the earth. Earth has only one natural satellite – the moon with a time period of 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.

## 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 𝑣_{0} in a circular orbit at a distance (𝑅_{𝐸} + ℎ) from the centre of the earth. The centripetal force required for this obit is, 𝐹_{𝑐} = 𝑚𝑣_{0}^{2}/(𝑅_{𝐸} + ℎ). The centripetal force is provided by the gravitational force, 𝐹 = 𝐺𝑚𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)^{2}

Equating the equations, we get 𝑚𝑣_{0}^{2}/(𝑅_{𝐸} + ℎ) = 𝐺𝑚𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)^{2}

𝑣_{0}^{2} = 𝐺𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)

𝒗_{𝟎} = √(𝑮𝑴_{𝑬}/(𝑹_{𝑬} + 𝒉))

For ℎ = 0 (since ℎ << 𝑅𝐸 ), we have 𝑣_{0} = √(𝐺𝑀_{𝐸}/𝑅_{𝐸})

𝒗_{𝟎} = √𝒈𝑹_{𝑬} (∵ 𝑔 = 𝐺𝑀_{𝐸}/𝑅_{𝐸}^{2})

## Time period of a satellite

In every orbit the satellite travels a distance 2𝜋(𝑅_{𝐸} + ℎ) with speed 𝑣_{0}, then its time period is,

𝑇 = 2𝜋(𝑅_{𝐸} + ℎ)/𝑣_{0} = 2𝜋(𝑅_{𝐸} + ℎ)/(√(𝐺𝑀_{𝐸}/(𝑅_{𝐸} + ℎ)))

𝑇 = [2𝜋(𝑅_{𝐸} + ℎ)(𝑅_{𝐸} + ℎ)^{1/2}]/√(𝐺𝑀_{𝐸})

𝑇 = 2𝜋(𝑅_{𝐸} + ℎ)^{3/2}/√(𝐺𝑀_{𝐸})

𝑻 = 𝟐𝝅(𝑹_{𝑬} + 𝒉)^{𝟑}^{/}^{𝟐}/√(𝑮𝑴_{𝑬})

This is the expression for time period of a satellite.

**Note:** Further, squaring on both sides

𝑇^{2} = 4𝜋^{2}(𝑅_{𝐸} + ℎ)^{3}/𝐺𝑀_{𝐸}

𝑇^{2} = (4𝜋^{2}/𝐺𝑀_{𝐸})(𝑅_{𝐸} + ℎ)^{3}

𝑻^{𝟐} = 𝑲(𝑹_{𝑬} + 𝒉)^{𝟑}

where 𝐾 = 4𝜋^{2}/𝐺𝑀_{𝐸}, Which is the Kepler’s law of periods

## 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.

**Expression for total energy of the satellite**

We have, 𝐸 = 𝑉 + 𝐾𝐸

𝐸 = − 𝐺𝑀_{𝐸}𝑚/(𝑅_{𝐸} + ℎ) + (1/2)𝑚𝑣_{0}^{2}

𝐸 = − 𝐺𝑀_{𝐸}𝑚/(𝑅_{𝐸} + ℎ) + (1/2)𝑚(𝐺𝑀_{𝐸}/(𝑅_{𝐸} + ℎ))

𝐸 = 𝐺𝑀_{𝐸}𝑚/(𝑅_{𝐸} + ℎ) (−1 + (1/2))

𝑬 = − 𝑮𝑴_{𝑬}𝒎/𝟐(𝑹_{𝑬} + 𝒉)

The total energy of an orbiting satellite is negative.

**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 meteorology.

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 cut off, 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.

## Important Questions for Exams.

**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 = 𝐺𝑀_{𝐸}/𝑅_{𝐸}^{2}

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.

## Problems

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, 𝑇^{2}/𝑅^{3} = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑇_{1}^{2}/𝑅_{1}^{3} = 𝑇_{2}^{2}/𝑅_{2}^{3}

𝑅_{2}^{3} = (𝑇_{2}^{2} × 𝑅_{1}^{3})/𝑇_{1}^{2}

𝑅_{2}^{3} = [(2𝜋/𝜔_{2})^{2} × 𝑅_{1}^{3}]/(2𝜋/𝜔_{1})^{2} (𝑇 = 2𝜋/𝜔)

𝑅_{2}^{3} = [(2𝜋/𝜔^{2})^{2} × 𝑅_{1}^{3} × 𝜔_{1}^{2}]/(2𝜋)^{2}

𝑅_{2}^{3} = [𝜔_{1}^{2}/𝜔_{2}^{2}] . 𝑅_{1}^{3}

𝑅_{2} = (𝜔_{1}/𝜔_{2})^{2/3}𝑅_{1}

Where 𝑅_{2} → Radius of the orbit of planet around the sun.

𝑅_{1} → Radius of the orbit of the earth

𝜔_{1}→ angular speed of earth

𝜔_{2} → angular speed of planet

𝑅_{2} = (𝜔_{1}/2𝜔_{2})^{2/3}𝑅_{1}

𝑅_{2} = (1/2)^{2/3}𝑅_{1}

𝑅_{2} = (1/1.49)𝑅_{1} = 0.63𝑅_{1}

Orbital size of the planet is smaller than the earth.

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 × 10^{8} km away from the sun?

Earth year 𝑇_{𝑒} = 1 year

Saturn year 𝑇_{𝑠} = 29.5 year

Radius of the orbit of earth = 𝑅_{𝑒}_{ }= 1.50 × 10^{8} km

From Kepler’s III law, 𝑇_{𝑒}^{2}/𝑅_{𝑒}^{3} = 𝑇_{𝑠}^{2}/𝑅_{𝑠}^{3}

𝑅_{𝑠}^{3} = (𝑇_{𝑠}^{2} × 𝑅_{𝑒}^{3})𝑇_{𝑒}^{2}

𝑅_{𝑠}^{3} = [(29.5)^{2} × (1.50 × 10^{8})^{3}]/(1^{2})

𝑅_{𝑠}^{3} = 2.947 × 10^{27}km

𝑅_{𝑠} = 1.43 × 10^{9}𝑘𝑚 = 1.43 × 10^{12}𝑚

3. Assuming earth to be a sphere of uniform mass density, how much would a body weighs halfway down to the Centre of earth, if it weighed 250N on the surface?

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.

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 × 10^{24}kg; radius of the earth = 6.4 × 10^{6}m; G = 6.67 × 10^{−11}𝑁𝑚^{2}𝑘𝑔^{−2}.

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?

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 × 10^{6}m and g on the earth’s surface = 9.8ms^{-2}.

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}).

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.

10. You are given the following data: g = 9.81 ms^{–2}, R_{E} = 6.37×10^{6} m, the distance to the moon R = 3.84×10^{8} m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth M_{E} in two different ways.

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.

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 = 9.8𝑚𝑠^{−2}

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^{8}km. Calculate that of planet Mars.

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 × 10^{24} 𝑘𝑔, 𝐺 = 6.67 × 10^{−11} 𝑁𝑚^{2}𝑘𝑔^{−2}.