Chapter – 11, Gravitation

Gravitational Force : The force of attraction between two point masses is called as Gravitational force.

Gravity : The gravitational force due to the the gravitational field of the earth by which any object is attracted towards the center of the earth, is called as gravity.

Newton’s law of gravitation :

Statement : ” The force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.”

Mathematical form :

     Gravitational Attraction between two point                                      masses

According to Newton’s gravitational law,

∝ Mm

∝ 1 / r2

on combing the above two equations we get

∝ Mm / r2

F = GMm / r2

Where  ‘ M ‘ and ‘m ‘ are the masses of point masses

‘ r ‘  is the distance between the point masses.

G is the proportionality constant and it is known as gravitational constant.

Gravitational constant ( G ) :

It is also called as universal constant as it’s value is independent of the nature of surrounding of the bodies that is whether the body is celestial or terrestrial, It’s value is independent of mass of the body as well as the distance between them, It is also independent of the position of the bodies and strength of gravitational field.

Value of G :

( 1 ) G = 6.67 * 10-11 N m2 .kg-2

( 2 ) G = 6.67 * 10-8 dyn cm2 g-2

Dimension Of G :

[ G ] = [ F ] [ r2 ] / [ M ] [ m ]

[ G ] = [ ML2T-2 ] * [ L2 ] / [ M ] [ M ]

[ G ] = M-1L3T-2

Dimensional Unit Of G :

( 1 )  In SI – System ⇒ kg-1 . m3 . s-2

( 2 ) In CGS – System ⇒ g-1 . cm3 . s-2

Expression Of G :

According to Newton’s law of Gravitation

F = GMm / r2

G = Fr2 /Mm

Vector form of Newton’s Law Of Gravitation :

Let ‘ M ‘ and ‘ m ‘ are the point masses having position vectors r2 and r1 respectively, In the adjacent figure, applying vector law of addition, we get

Vector form of Newton’s law of Gravitation

r1 + r = r2

r = r1r2

r . r = ( r1r1 ) . ( r1r1 )

r2 = Ι r1r2 Ι2

Now, from Newton’s Law of gravitation,we have

F = GMm Î / r2

F = GMm r/ r3

This the vector form of Newton’s law of gravitation.

Here, the bold letters represent the vector quantities and Î represent the unit vector given by

Î = r / r

Properties of Newton’s Law of Gravitation :

1 ) It obeys Newton’s  3rd law of motion.

2) It acts over the long range of distance.

3 ) It is true only for point masses

4 ) The gravitational force is central and conservative force in nature.

Acceleration due to gravity :

The acceleration produced in a freely falling body under the action of gravitational field is called as acceleration due to gravity.

value of g :

( 1 ) g = 9.8 m / s2

( 2 ) g = 980 cm / s2

Relation between ‘ g ‘ and ‘G ‘ :

From Newton’s Law Of Gravitation :

F = GMm / r2

F = gravitational force

G = gravitational constant

‘ M ‘ and ‘ m ‘ are the point masses

r = distance between the point masses

If a body is freely falls then it is under the action of gravitational field and it’s weight is equivalent to the gravitational force.

∴ mg = GMm/ r2

g = GM / r2

Factors influencing the value of acceleration due to gravity :

1 ) Altitude

2 ) depth

3 ) Rotation of the Earth

4 ) Shape of the Earth

Now, let us discuss these factors in detail except hose which are not included in our syllabus.

1 ) Variation in the value of ‘ g ‘ due to height or altitude :

 

Let us consider that earth is a homogeneous sphere of radius ‘ R ‘ and mass ‘ M ‘.

Acceleration due to gravity on the earth’s surface is given by

g = GM / R2 ——————————– ( 1 )

Let a body is raised to a height of ‘ h ‘. The acceleration due to gravity at a height ‘ h ‘ above the earth’s surface is given by

g’ = GM / ( R + h )2 ———————————— ( 2 )

( 1 ) ÷ ( 2 ), we get

g / g’ = { GM / R2 } / {  GM / ( R + h )2 }

g / g’ =   { ( R + h ) / R }2

g / g’ = ( 1 + h / R ) 2

g’ = g ( 1 + h / R ) -2

Applying Binomial theorem of negative index and neglecting higher ordered termed, we get

g’ = g ( 1 – 2h / R ) 

2 ) Variation in the value of ” g ‘ due to depth :

Let us consider the Earth is a homogeneous sphere of radius ‘ R ‘ and mass ‘ M ‘ . The acceleration due to gravity on the Earth’s surface is given by

g = GM / R

We know that Mass = density * volume  = ρ * 4πR3 / 3

Now

g = GM / R

putting the value of M we get

g =  4πR3ρG / 3R2

g = 4πRρG / 3

The acceleration due to gravity at a depth ‘ d ‘ below the Earth’s surface is given by

g’ =  4π( R – d ) ρG / 3

dividing both sides by ‘ g ‘ we get

g’ / g = ( R – d ) / R

g’ = g ( 1 – d / R ) 

3 ) Variation in the value of acceleration due to gravity due to shape of the Earth : 

The shape of the Earth is not exactly like homogeneous sphere but is oblate spheroid that is radius of the Earth at the pole is less than that of equator.

R = radius at the equator

R = radius at the pole

Re > Rp

ge = GM / R2e

gp = GM / R2p

On comparing the above values we get

gp < ge

Hence, acceleration due to gravity at the pole is greater than that of equator.

Kepler’s laws of Planetary Motion :

1 ) Kepler’s first law of planetary motion : “Each planer revolves around the sun in an elliptical orbit with the sun at one focus of the elliptical path.”

2 ) Kepler’s second law of planetary motion : ” Position vector of the planet from the sun sweeps out equal areas in equal time.”

OR

” The areal velocity of the planet around the sun is always constant ”

3 ) Kepler’s 3rd law of planetary motion : ” ” The square of time period of any planet about the sun is directly proportional to the cube semi-major axis of elliptical path.”

Mathematical form of Law of periods or Kepler’s 3rd law of planetary motion :

Let

T = time period of revolution of a planet around the sun

r = semi – major axis of the elliptical path

According to Kepler’s 3rd law of planetary motion

∝ r2

Gravitational Field : The space around the mass over which it can exert gravitational force on other masses , called as gravitational field.

F = GMm / r2

putting r = ∝ ( i.e infinity ) 

F = 0

That is, gravitational force becomes zero at infinite distance .

Gravitational field intensity ( I ) : The gravitational field intensity at a point in the gravitational field may be defined as gravitational force experienced by unit mass placed at that point. It is the force per unit mass.

From Newton’s law of gravitation,

F = GMm / r2

for unit mass, m = 1 unit

I = GM / r2

Free Acceleration ( a ) = If a test mass is free to move, it will move towards the body of greater mass with a particular acceleration. This acceleration is termed as Free Acceleration.

I = F / m

I = ma / m

I = a 

Hence, the field strength is equal to free acceleration of an object.

In case of Earth,

a = g = acceleration due to gravity

∴ I = g

Gravitational Potential : Gravitational potential at a point in the gravitational field may be defined as the work done to bring a unit mass from infinity to that point. 

                     Gravitational Potential

Expression of Gravitational Potential :

Let there is an isolated body of mass M .

Let a unit mass is placed at a point A at a distance of x from the center of the body

From Newton’s law of gravitation,

F = GM / x2

Now, let us displace the unit mass through a small distance of ‘ dx’

work done = F . dx = ( GM / r) . dx

Total work done by gravitational field in bringing the unit mass from infinity to that point at a distance of ‘ r ‘ from the isolated body.

w = r ( GM / r) . dx = GM ∫r r-2  . dx = { GM / ( -2 + 1 )} [ r-2+1 ]r

w = – GM / r

This amount of work done is stored as gravitational potential.

Gravitational Potential ( V ) = – GM / r

Here, negative sign indicates that the work is done by gravitational field of the isolated body

Gravitational Potential difference : 

Gravitational potential difference  in the gravitational field may be defined as the work done to bring a unit mass from one point to another point.

Expression of Gravitational Potential Difference :  

Gravitational Potential at point A ( VA ) = – GM / rA

Gravitational Potential at point B ( VB ) = – GM / rB

where G = Gravitational Constant

M = mass of an isolated body

rA = distance of point A from the isolated body

rB = distance of point B from the isolated body

Gravitational Potential Difference in bringing a unit mass from point B to Point A in the Gravitational Field of an isolated body is given by

ΔV = VA – VB = ( – GM / rA ) – (- GM / rB ) = GM / { ( 1 / rB ) – ( 1 / rA ) }

ΔV = GM / { ( 1 / rB ) – ( 1 / rA ) }

Gravitational Potential Energy :

Gravitational potential energy at a point in the gravitational field may be defined as the work done to bring an object from infinity to that point. 

Expression of Gravitational Potential Energy :

Let there is an isolated body of mass M .

Let an object of mass ‘ m ‘  is placed at a point A at a distance of x from the center of the body

From Newton’s law of gravitation,

F = GMm / x2

Now, let us displace the unit mass through a small distance of ‘ dx’

work done = F . dx = ( GMm / r) . dx

Total work done by gravitational field in bringing the unit mass from infinity to that point at a distance of ‘ x ‘ from the isolated body.

w = ∫ ( GMm / r) . dx = GMm ∫r r-2  . dx = { GMm / ( -2 + 1 )} [ r-2+1 ]r

w = – GMm / r

This amount of work done is stored as Gravitational Potential Energy given by

U = – GMm / r

Here, negative sign indicates that the work is done by gravitational field of the isolated body.

Escape Velocity : 

The minimum velocity with which a body must be projected up from the surface of a planet to escape from its gravitational pull, is called as Escape Velocity.

Expression for the Escape Velocity for a Planet :

         Expression for the Escape Velocity

Let us assume that the Planet is homogeneous sphere of mass ‘ M ‘ and radius ‘ r ‘ . Let a body of mass ‘ m ‘ is at a height of ‘ h ‘ from the Planet’s surface.

From Newton’s Law of Gravitation, we have

F = GMm / x2

where x = r + h = distance between the body and the center of the planet.

Let the body of mass ‘ m ‘  is displaced through a small distance ‘ dx ‘. In doing so, the work done is given by

dw = F.dx = { GMm / x} .dx

Total work done in moving the body from the surface of the planet to a infinite distance  ( where its Gravitational force is zero ) is given by

w = r  F.dx =∫r{ GMm / x} .dx = GMm ∫r{ x-2  dx } = GMm [ ( -1/∞ ) – ( – 1 / r ) ] = GMm / r 

This amount of work done supplies kinetic energy to the body to escape from gravitational field of the planet.

∴ Kinetic energy = GMm / r

⇒ mve2 / 2  = GMm / r

⇒ ve2=  2GM / r

ve = √ ( 2 GM / r ) 

ve is the Escape velocity of the body to escape from the gravitational field of the planet.

In case of the Earth, the acceleration due to gravity is given by

g = GM / r2

gr = GM / r

Therefore, escape velocity for the earth , is given by

ve=  2 ( GM / r )

⇒ ve = 2gr

ve  = √ ( 2gr )

OR

ve  = √gD

Where D = 2r = Diameter of the earth.

Orbital Velocity :

The velocity which is given to an artificial satellite a few hundred kilometer above the surface of the planet to start revolving round the planet , is called as Orbital Velocity.

Orbital Velocity of an artificial satellite : 

 Expression for the Orbital Velocity

Let

m = mass of the satellite

r = orbital radius of the satellite

R = Radius of the planet

h = height of the satellite

M = mass of the planet

vo = orbital velocity of the satellite

r = R + h = distance between the planet and satellite

From Newton’s law of gravitation, we have

F = GMm / r2

To revolve round the planet in the orbit, these gravitational force supplies centripetal force to the satellite.

F = mvo2 / r

GMm / r2 = mvo2 /r

GM / r = vo2

vo = √ ( GM / r )  

In case of the earth

g’ = acceleration due to gravity at a height of h

g = acceleration due to gravity on the earth’s surface.

Again we can write

F = mg’

GMm / r2 = mg’

GM / r2 = g’ ——————————– ( 1 )

GM = g’r2

Putting the value of the GM in the expression of the orbital velocity we get

vo = √ ( GM / r )

vo = √ ( g’ r2 / r )

vo = √ ( g’r ) ——————————— ( A )

Again we have g = GM / R2 ——————————– ( 2 )

( 1 ) ÷ ( 2 )

g’ / g = R2 / r2

g’ = g ( R / r )2

Putting the value of g’ in the equation  ( A )

vo = √ ( g’r )

vo = √  ( g R2r / r2 )

vo = R√ ( g / r )

vo = R√ { g / ( R + h ) }

If h << R then h may be neglected

vo = R√ { g / ( R + h ) }

vo = R√ { g / R }

vo = √ ( g R ) 

Satellite : 

A body revolving around another larger body is called as Satellite. It has two types

  1. Natural satellite
  2. Artificial satellite

1 ) Natural Satellite :

A celestial body in space that revolving around the larger body, is called as Natural Satellite.

 

Example :

  • Moon is the natural satellite of Earth.
  • Phobos and Deimos are two natural satellites of Mars.
  • Jupiter has 67 natural satellites.

2 ) Artificial Satellite : 

A satellite which is made by people and launched by rocket, is called as Artificial satellite.

Examples:

  • Sputnik I is the first artificial satellite launched by Soviet Union on 4 October 1957.
  • Aryabhata is the first Indian satellite built by ISRO and launched by Soviet Union on 19 April 1975.
  • Rohini is the first satellite to be placed in orbit by Indian – made launched vehicle, SLV-3 in 1980
  • The Polar Satellite Launch Vehicle ( PSLV – C42 ) of ISRO succesfully launched two satellite on 16 September 2018 ( Sunday ) from Satish Dhawan Space Center ( SDSC ) SHAR, Sriharikota.
    • NovaSAR
    • S1-4
  • INSAT  – 1A ( Indian National Satellite ). It is the first multipurpose satellite.
  • APPLE ( Arian Passenger Pay Load Experiment ). It is a type of a communication satellite.

At present, there are 1,886 artificial satellite orbiting the Earth’s Orbit.

                Indian’s first satellite

               Rohini Satellite RS – 1

          Sputnik I, World’s first satellite

Types Of Artificial Satellite : 

  1. Navigation satellite
  2. Communication satellite
  3. Weather Satellite
  4. Earth Observation Satellite
  5. Astronomical Satellite

1 ) Navigation Satellite :

A satellite that provides autonomous Geo – Spatial Positioning or Global Positioning system, is termed as Navigation Satellite. There are seven Navigation Satellites in India

Example :

  • IRNSS ( Indian Regional Navigation Satellite System )
  • GPS ( Global Positioning system )

     Navigation Satellite

2 ) Communication Satellite :

The satellite which is used for communication purpose, is called as Communication Satellite. It creates a communication channel between a source transmitter and a receiver at any location on the Earth. It communicates by using radio waves to send signals to the antennas on the Earth. Then the antennas capture the signals and work with the signals.

3 ) Weather Satellite : 

The satellite used for monitoring the weather and the climate on the Earth is known as Weather satellite.

Examples :

  1. METEOSAT 9 (MSG 2) 
  2. NOAA 18
  3. MTSAT-1R
  4. FENGYUN 2C

                     weather satellite

4 ) Earth Observation Satellite :

The satellite used for the observation of different position of the Earth and taking images of the Earth is termed as Earth Observation Satellite.

Examples :

  1. Geostationary Operational Environmental Satellites
  2. Polar Operational Environmental Satellites
  3. Joint Polar Satellites

5 ) Astronomical Satellite : 

The satellite which provide the information about space is termed as Astronomical Satellite. Actually it is big telescope floating in space.

                  Astronomical Satellite

Examples :

  1. Galileo ( the first probe thrust into orbit around Jupiter in 1950 )
  2. Voyager
  3. Cassini
  4. Huygens
  5. Pioneer

Geostationary Satellite :

It is an earth – orbiting satellite rotating in the direction as same as the direction of the earth’s rotation. It is placed at a height of 35, 800 km ( approximately ).

                 Geostationary Satellite

Geostationary Stationary orbit : The orbit which is synchronous with the equator of the earth, is called as Geostationary Satellite.

Geostationary orbit

Properties of Geostationary satellite : 

  • It is an artificial satellite of the Earth.
  • Its orbit is synchronous with the equator of the earth.
  • Its height is about 35,800 km.
  • Its direction of rotation is as same as the Earth.
  • It is a communication satellite.
  • Its angular speed is as same as that of the Earth.
  • The plane of orbit  of the satellite is perpendicular or normal to the axis of rotation of the Earth.
  • Its orbital velocity is nearly about 3.1 km /sec.
  • It appears to be stationary to an observer on the Earth.

Uses Of Geostationary Satellite : 

  1. It is used for communication purpose.
  2. It is used as Broad – cast Satellite.

Question – Answer Zone 

1 ) Define gravitational force.

Ans :  The force of attraction between two point masses is called as Gravitational force.

2 ) What do you mean by gravity ?

Ans : The gravitational force due to the the gravitational field of the earth by which any object is attracted towards the center of the earth, is called as gravity.

3 ) Define acceleration due to gravity ?

Ans : The acceleration produced in a freely falling body under the action of gravitational field is called as acceleration due to gravity.

4 ) What is the value of G in SI – system ?

Ans : G = 6.67 * 10-11 N m2 .kg-2

5 ) What is the value of G in CGS – system ?

Ans : G = 6.67 * 10-8 dyn cm2 g-2

6) Why Gravitational Constant is called as Universal Constant ?

Ans : It is also called as universal constant as it’s value is independent of the nature of surrounding of the bodies that is whether the body is celestial or terrestrial, It’s value is independent of mass of the body as well as the distance between them, It is also independent of the position of the bodies and strength of gravitational field.

7 ) Write down the dimension of G if velocity ( V ), mass ( M ), and time ( T ) are taken to be fundamental quantities.

Ans : According to Newton’s law of gravitation

F = GMm / r2

∵ F = ma

G = Fr / Mm = mar / Mm

Where a = acceleration of the body = v / t

G = ar / M = ( v / t ) ( vt ) / M

G = v2 / M =

[ G ] = [ V2 ] / [ M ] = M-1V2T0

[ G ] = M-1V2T0

8 ) State Newton’s law of Gravitation.

Ans :  The force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

9 ) Write down the vector form of Newton’s Law of Gravitation.

Ans :

Let ‘ M ‘ and ‘ m ‘ are the point masses having position vectors r2 and r1 respectively, In the adjacent figure, applying vector law of addition, we get

Vector form of Newton’s law of Gravitation

r1 + r = r2

r = r1r2

r . r = ( r1r1 ) . ( r1r1 )

r2 = Ι r1r2 Ι2

Now, from Newton’s Law of gravitation,we have

F = GMm Î / r2

F = GMm r/ r3

This the vector form of Newton’s law of gravitation.

10 ) Write down the mathematical form of Newton’s Law of Gravitation and mention each term.

Ans :

F = GMm / r2

Where  ‘ M ‘ and ‘m ‘ are the masses of point masses

‘ r ‘  is the distance between the point masses.

G is the proportionality constant and it is known as gravitational constant.

11 ) What is the value of acceleration due to gravity on the Earth’s surface?

Ans : 9.8 m/sec2

12 ) What is the value of acceleration due to gravity at the center of the earth ?

Ans : Zero

13 ) Why gravitational force is conservative in nature ?

Ans : The gravitational force is conservative in nature because work done by the gravitational field is independent of the path followed and it depends only on the initial and final position of the body.

14 ) Compare the value of acceleration due to gravity at equatorial and polar region.

Ans :The shape of the Earth is not exactly like homogeneous sphere but is oblate spheroid that is radius of the Earth at the pole is less than that of equator.

R = radius at the equator

R = radius at the pole

Re > Rp

ge = GM / R2e

gp = GM / R2p

On comparing the above values we get

gp < ge

Hence, acceleration due to gravity at the pole is greater than that of equator.

15 ) State Kepler’s Law of Planetary Motion.

Ans :

1 ) Kepler’s first law of planetary motion : “Each planer revolves around the sun in an elliptical orbit with the sun at one focus of the elliptical path.”

2 ) Kepler’s second law of planetary motion : ” Position vector of the planet from the sun sweeps out equal areas in equal time.”

OR

” The areal velocity of the planet around the sun is always constant ”

3 ) Kepler’s 3rd law of planetary motion : ”  The square of time period of any planet about the sun is directly proportional to the cube semi-major axis of elliptical path.”

16 ) Write down the Kepler’s Law of Orbit.

Ans : Each planer revolves around the sun in an elliptical orbit with the sun at one focus of the elliptical path.

17 )  Write down the Kepler’s Law of Area.

Ans : Position vector of the planet from the sun sweeps out equal areas in equal time.

18 )  Write down the Kepler’s Law of Period.

Ans : The square of time period of any planet about the sun is directly proportional to the cube semi-major axis of elliptical path.

19 )  Write down the mathematical form of  Kepler’s Law of Period.

Ans :

Let

T = time period of revolution of a planet around the sun

r = semi – major axis of the elliptical path

According to Kepler’s 3rd law of planetary motion

∝ r2

20 ) What do you mean by natural year of planet ?

Ans : The period of revolution of a planet around the sun is termed as Natural Year of a planet.

21 ) what is meant by areal velocity ?

Ans : The area covered in unit time is called as areal velocity.

22 ) Write down the nature of the Kepler’s Law of Planetary Motion.

Ans : The Kepler’s Law of Planetary Motion is empirical in nature.

23 ) What do you mean by Empirical Law ?

Ans : A law which is derived from experimental data ( without complete understanding ) rather than from some elegant theory or by logic from first principle is termed as Empirical Law.

Example : Kepler’s Law of Planetary Motion.

24 ) Why weight of a body becomes zero at the center of the Earth?

                                                              Or

         Why does a body lose weight at the center of the Earth ?

Ans : As the value of acceleration due to gravity becomes zero at the center of the earth, weight of a body becomes zero at the center of the earth or a body loses weight at the center of the earth.

25 ) Where does a body weighs more at the surface of the Earth or inside a mine ?

Ans : The value of the acceleration due to gravity decreases with depth so the body weighs more at the surface of the Earth than inside a mine.

26 ) Gravitational force between two bodies is 1 N. If the distance between them is made twice, what will be the force ?

Ans : Gravitational Force ( F ) = 1 N

distance between the two bodies = d

∝ 1 / d 

( F / F’ ) = ( d’ / d )

⇒ F’ = F * ( d / d’ )

⇒ F’ = 1 * ( d / 2d )

⇒ F ‘ = 1 / 2

Hence, the required force will be 0.5 N

27 ) Derive Kepler’s third Law of Planetary Motion by using Newton’s Law of Gravitation.

Ans : From Newton’s Law of Gravitation, we have

F = GMm / r2

where

F = Gravitational Force

G = Gravitational Constant

M = Mass of the sun

m = mass of the planet

r = distance between the planet and the sun.

To revolve around the sun in the orbit, the gravitational force supplies the centripetal force to the planet.

2r = GMm / r2

where ω = the angular speed of the planet = 2π / T

T = time period of revolution

Now, putting the value of ω we get

m ( 2π / T )2 =  GMm / r3

4 π2 / T2 = GM / r3

T2 = ( 4 π2 / GM )r3

( 4 π2 / GM ) = Constant

T2 ∝ r3 

That is the square of the time period of revolution of a planet is directly proportional to the cube of the mean orbital radius of planet. This is nothing but the Kepler’s third law of planetary motion.

 

28 ) Derive Newton’s Law of Gravitation by using Kepler’s Law of Planetary Motion.

29 ) Show that the Kepler’s second Law of Planetary Motion follows the law of conservation of momentum.

30 ) Prove that g = GM / r2 where symbols have usual meaning.

Ans :

From Newton’s Law Of Gravitation :

F = GMm / r2

F = gravitational force

G = gravitational constant

‘ M ‘ and ‘ m ‘ are the point masses

r = distance between the point masses

If a body is freely falls then it is under the action of gravitational field and it’s weight is equivalent to the gravitational force.

∴ mg = GMm/ r2

g = GM / r2

31 ) Distinguish between ‘g’ and ‘ G ‘.

Ans :

32 ) Show that the acceleration of free fall ‘g’ at the earth’s surface and Gravitational constant are related by the expression

g = ( 4πρR / 3 ) G

where g = acceleration due to gravity

G = Gravitational constant

ρ = density of the earth

R = Radius of the earth.

33 ) Show that the acceleration due to gravity decreases with the altitude.

Ans :

 

Let us consider that earth is a homogeneous sphere of radius ‘ R ‘ and mass ‘ M ‘.

Acceleration due to gravity on the earth’s surface is given by

g = GM / R2 ——————————– ( 1 )

Let a body is raised to a height of ‘ h ‘. The acceleration due to gravity at a height ‘ h ‘ above the earth’s surface is given by

g’ = GM / ( R + h )2 ———————————— ( 2 )

( 1 ) ÷ ( 2 ), we get

g / g’ = { GM / R2 } / {  GM / ( R + h )2 }

g / g’ =   { ( R + h ) / R }2

g / g’ = ( 1 + h / R ) 2

g’ = g ( 1 + h / R ) -2

Applying Binomial theorem of negative index and neglecting higher ordered termed, we get

g’ = g ( 1 – 2h / R ) 

34 ) Show that the acceleration due to gravity decreases with the increase of depth.

Ans:

Let us consider the Earth is a homogeneous sphere of radius ‘ R ‘ and mass ‘ M ‘ . The acceleration due to gravity on the Earth’s surface is given by

g = GM / R

We know that Mass = density * volume  = ρ * 4πR3 / 3

Now

g = GM / R

putting the value of M we get

g =  4πR3ρG / 3R2

g = 4πRρG / 3

The acceleration due to gravity at a depth ‘ d ‘ below the Earth’s surface is given by

g’ =  4π( R – d ) ρG / 3

dividing both sides by ‘ g ‘ we get

g’ / g = ( R – d ) / R

g’ = g ( 1 – d / R ) 

35 ) Discuss the effect of depth on the value of acceleration due to gravity and what will be its value at the center of the earth?

Let us consider the Earth is a homogeneous sphere of radius ‘ R ‘ and mass ‘ M ‘ . The acceleration due to gravity on the Earth’s surface is given by

g = GM / R

We know that Mass = density * volume  = ρ * 4πR3 / 3

Now

g = GM / R

putting the value of M we get

g =  4πR3ρG / 3R2

g = 4πRρG / 3

The acceleration due to gravity at a depth ‘ d ‘ below the Earth’s surface is given by

g’ =  4π( R – d ) ρG / 3

dividing both sides by ‘ g ‘ we get

g’ / g = ( R – d ) / R

g’ = g ( 1 – d / R ) 

At the center of the Earth d = R

g’ = g ( 1 – 1 )

g’ = 0

At the center of the Earth, the value of the acceleration due to gravity is zero.

36 ) How does the shape of the earth affects the value of acceleration due to gravity ?

Ans :

The shape of the Earth is not exactly like homogeneous sphere but is oblate spheroid that is radius of the Earth at the pole is less than that of equator.

R = radius at the equator

R = radius at the pole

Re > Rp

ge = GM / R2e

gp = GM / R2p

On comparing the above values we get

gp < ge

Hence, acceleration due to gravity at the pole is greater than that of equator.

37 ) What do you mean by Gravitational Field ?

Ans : The space around the mass over which it can exert gravitational force on other masses , called as gravitational field.

38 ) What is meant by Gravitational Field Intensity ?

Ans : The gravitational field intensity at a point in the gravitational field may be defined as gravitational force experienced by unit mass placed at that point. It is the force per unit mass.

39 ) Prove that the gravitational field intensity is equivalent to free acceleration of an object.

Ans :

If a test mass is free to move, it will move towards the body of greater mass with a particular acceleration. This acceleration is termed as Free Acceleration.

I = F / m

I = ma / m

I = a 

Hence, the field strength is equal to free acceleration of an object.

40 ) Define gravitational potential.

Ans : Gravitational potential at a point in the gravitational field may be defined as the work done to bring a unit mass from infinity to that point. 

41 ) Derive the expression for the gravitational potential energy.

Ans :

Let there is an isolated body of mass M .

Let an object of mass ‘ m ‘  is placed at a point A at a distance of x from the center of the body

From Newton’s law of gravitation,

F = GMm / x2

Now, let us displace the unit mass through a small distance of ‘ dx’

work done = F . dx = ( GMm / r) . dx

Total work done by gravitational field in bringing the unit mass from infinity to that point at a distance of ‘ x ‘ from the isolated body.

w = ∫ ( GMm / r) . dx = GMm ∫r r-2  . dx = { GMm / ( -2 + 1 )} [ r-2+1 ]r

w = – GMm / r

This amount of work done is stored as Gravitational Potential Energy given by

U = – GMm / r

Here, negative sign indicates that the work is done by gravitational field of the isolated body.

42 ) Define the gravitational potential energy.

Ans :Gravitational potential energy at a point in the gravitational field may be defined as the work done to bring an object from infinity to that point. 

43 ) What do you mean by Escape velocity ?

Ans : The minimum velocity with which a body must be projected up from the surface of a planet to escape from its gravitational pull, is called as Escape Velocity.

44 ) Derive the expression for the escape velocity of an object to escape from the earth’s gravitational field.

Ans :

         Expression for the Escape Velocity

Let us assume that the Earth is homogeneous sphere of mass ‘ M ‘ and radius ‘ r ‘ . Let a body of mass ‘ m ‘ is at a height of ‘ h ‘ from the Earth’s surface.

From Newton’s Law of Gravitation, we have

F = GMm / x2

where x = r + h = distance between the body and the center of the Earth.

Let the body of mass ‘ m ‘  is displaced through a small distance ‘ dx ‘. In doing so, the work done is given by

dw = F.dx = { GMm / x} .dx

Total work done in moving the body from the surface of the Earth to a infinite distance  ( where its Gravitational force is zero ) is given by

w = r  F.dx =∫r{ GMm / x} .dx = GMm ∫r{ x-2  dx } = GMm [ ( -1/∞ ) – ( – 1 / r ) ] = GMm / r 

This amount of work done supplies kinetic energy to the body to escape from gravitational field of the Earth.

∴ Kinetic energy = GMm / r

⇒ mve2 / 2  = GMm / r

⇒ ve2=  2GM / r

ve = √ ( 2 GM / r ) ——————- ( 1 ) 

ve is the Escape velocity of the body to escape from the gravitational field of the planet.

We know that the acceleration due to gravity is given by

g = GM / r2

gr = GM / r

Putting these value in equation ( 1 ) we get

ve = √ ( 2 GM / r ) = √ ( 2gr )

ve = √ ( 2gr )

45 ) What is meant by orbital velocity ?

Ans : The velocity which is given to an artificial satellite a few hundred kilometer above the surface of the planet to start revolving round the planet , is called as Orbital Velocity.

46 ) Derive the expression for the orbital velocity of a geostationary satellite.

Ans :

 Expression for the Orbital Velocity

Let

m = mass of the geostationary satellite

r = orbital radius of the geostationary satellite

R = Radius of the Earth

h = height of the geostationary satellite

M = mass of the Earth

vo = orbital velocity of the geostationary satellite

r = R + h = distance between the planet and geostationary satellite

From Newton’s law of gravitation, we have

F = GMm / r2

To revolve round the Earth in the orbit, these gravitational force supplies centripetal force to the geostationary satellite.

F = mvo2 / r

GMm / r2 = mvo2 /r

GM / r = vo2

vo = √ ( GM / r )  ———————– ( 1 )

We know that the acceleration due to gravity is given by

g = GM / r2

gr = GM / r

Putting these value in equation ( 1 ) we get

vo = √ ( GM / r ) = √ ( gr )

vo = √ ( gr )

47 ) Define the term ‘ Satellite ‘.

Ans : A body revolving around another larger body is called as Satellite. It has two types

48 ) Give an example of an artificial satellite.

Ans : IRNSS ( Indian Regional Navigation Satellite System ) is an example of an artificial satellite.

49 ) What do you mean by geostationary satellite ?

Ans : The orbit which is synchronous with the equator of the earth, is called as Geostationary Satellite. It is an artificial satellite of the Earth. Its height is about 35,800 km. Its direction of rotation is as same as the Earth. It is a communication satellite. Its angular speed is as same as that of the Earth. The plane of orbit  of the satellite is perpendicular or normal to the axis of rotation of the Earth. Its orbital velocity is nearly about 3.1 km /sec. It appears to be stationary to an observer on the Earth.

50 ) What are uses of geostationary satellite ?

Ans : The uses of geostationary satellite are given below.

  1. It is used for communication purpose.
  2. It is used as Broad – cast Satellite.

51 ) Why geostationary satellite is so called ?

Ans : Geostationary satellite is so called because it appears stationary to an observer on the Earth.

52 ) Why geostationary satellite is appeared to be stationary to an observer on the earth ?

Ans : Geostationary satellite is appeared to be stationary to an observer on the earth because its orbit of revolution is synchronized with the equator of the Earth.

53 ) Derive the expression for the gravitational potential.

Ans :

Let there is an isolated body of mass M .

Let a unit mass is placed at a point A at a distance of x from the center of the body

From Newton’s law of gravitation,

F = GM / x2

Now, let us displace the unit mass through a small distance of ‘ dx’

work done = F . dx = ( GM / r) . dx

Total work done by gravitational field in bringing the unit mass from infinity to that point at a distance of ‘ r ‘ from the isolated body.

w = r ( GM / r) . dx = GM ∫r r-2  . dx = { GM / ( -2 + 1 )} [ r-2+1 ]r

w = – GM / r

This amount of work done is stored as gravitational potential.

Gravitational Potential ( V ) = – GM / r

 

 


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