7 Gravitation

GRAVITATION 1.

KEPLER’S LAWS Kepler’s First Law (Law of Orbits): Each planet moves in an elliptical orbit with the Sun at one focus. Kepler’s Second Law (Law of Areas): The speed of planet varies in such a way that the radius vector drawn from the Sun to a planet sweeps out equal areas in equal times.  Thus the law states that the aerial velocity of the planet is constant.  The orbit of a planet around the sun is shown in fig. the areas; A , A and A are swept by the radius vector in equal 1 2 3 times. So, according to Kepler ’s second law, A1 = A2 = A3  Also, the planet covers unequal distances S , and S in S 1 2 3 equal times due to the variable speed of the planet.  Maximum distance is covered in a given time when planet is closet to the Sun.  When the planet is closest from the sun, its velocity and the kinetic energy of the planet is maximum.  When the planet is farthest from the Sun, its velocity and the kinetic energy is minimum.  However, the total energy of the planet remains constant. Kepler’s Third Law (Law of Periods)  The square of the period of revolution of a planet around the Sun is proportional to the cube of the semi-major axis of its elliptical orbit.  AB is the major axis and CD is the minor axis. AO or OB is called semi-major axis.  Let, T = Period of revolution of planet around Sun. R = length of semi- major axis  According to Kepler ’s third law, T

 T 

2

R

3

2

or T  KR

3

Where K is a constant for all planets. Let and T be the periods of any two planets around the Sun. 1

2

Let R1 and R2 be the lengths of their respective semi – major axes 2

3

R 1 1 Then,T  3 T2 R2  Since different planets are at different distances from the Sun, therefore, their time periods are different. Proof of Kepler’s Third Law  Consider the motion of a planet around the Sun.  Let m and M represent the masses of the planet and Sun respectively.  Let us assume that the planet moves around the Sun in a circular orbit. Let r be the radius of the orbit.  The gravitational force between the planet and the Sun provides the necessary centripetal force to the planet. 

 

GMm 2  mrω 2 r  2π  GM 2 r2 r 

2

T

2

2

4π r 3 

T 

 4π  3 2

GM Or

2

T r

 3

r T   GM   

2.

UNIVERSAL LAW OF GRAVITATION Every particle of matter in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. 

The force of attraction between any two particles in the universe is known as force of gravitation.



The force of gravitational attraction between the two bodies acts along the line joining their centres. This force is mutual.

 m 



Or

Consider two bodies of masses and m with their centres separately by a distance r. 1 2 Let F be the force (in magnitude) of gravitational attraction between the two bodies. According to Newton’s law of gravitation, (ii) 1 Fα 2 (i) F α m1m2 r Combining both the factors, we get F α

FG

m 1m 2 2 r

m1m2 2 r

… (i)

Where G is a constant of proportionality known as gravitational constant 

It has the same value everywhere in the universe. So, it is a universal constant of gravitation.



In SI, the value of G is (6.67 × 10–11 Nm2 kg–2).

 

If m1 = m 2 = 1 unit and r = 1 unit, then from equation (1), F = G. So, the universal gravitational constant (G) is numerically equal to the force of attraction between two bodies, each of unit mass, separated by unit distance.



In SI, the gravitational constant is numerically equal to the force of attraction between two bodies, each of mass

one kilogram, separated by a distance of one metre. Special Features of Gravitational Force 1.

The gravitational forces between two bodies constitute an action and reaction pair, i.e., the force is equal in magnitude but opposite in direction.

2.

The gravitational force between two bodies does not depend upon the nature of the intervening medium.

3.

The gravitational force between two bodies does not depend up the presence or absence of other bodies.

4.

The gravitational force is extremely small in the case of light bodies. However, it becomes appreciable in the case of massive bodies. As an illustration, although the distance between the Sun and Earth is yet the corresponding gravitational force is of the order of 1027 N.

Vector Form of Newtons’s Law of Gravitation 

Consider two particles A and of masses m and m respectively. 1 2 Let, r12 = displacement vector from A to B, r21 = displacement vector from B to A.

F21 = gravitational force exerted on B by A.

F12 = gravitational force exerted on A by B. 

In vector notation, Newton’s law of gravitation is written as follows:  mm F12  – G 1 2 2 r21ˆ r21

… (1)

Where rˆ21 is a unit vector pointing from B to A. The negative sign indicates that the direction of is opposite to F12 that of rˆ21 . The negative sign also shows that the gravitational force is attractive in nature.  F

 m m Similarly,  – G 1 2 rˆ 21 12 r122 Where rˆ12 is a unit vector pointing from A to B. 2

Also, r21 = r12

2

  F  Gm 1m rˆ 2 21 r21

2 21

Equating (1) and (2), F12  – F21 3.

ACCELERATION DUE TO GRAVITY OF THE EARTH Let us consider a body of mass m lying on the surface of the Earth of mass M and radius R. Let g be the value of acceleration due to gravity on the free surface of Earth. g F GMm  GM   2   2  .......(1) Then,    R  m R m Variation of g with Altitude (Height)  Suppose the body is taken to a height ‘h’ above the surface of Earth where the value of acceleration due to gravity is g . h GM g  h  R  h2 … (2)  Then, where  R  h is the distance between the centres of body and Earth. 

Dividing (2) by (1), we get 2 gh GM R2 gh R    2 Or 2 g  R  h GM g     R 1  h R 



Let us now derive an expression for hg when h << R. 2 g R 1 2 R g   



h

–2

h   g  

h  2 R 1     R  

h

or 2 h 2 R 1  R  

g2 

  1

h R 

 1   R 



Since h < < R, therefore, h/R is very small as compared to 1.

 Expanding the right hand side of the above equation by Binomial theorem and neglecting the square an d higher powers of

h , we get R

Or

gh 1– 2h g R   2h g h  g1 –   R

Or

gh  g –



gh – g  g

2h g R – 2h

g

or



g – h

R

2h

g R



Here,  g – g h  gives the decrease in the value of g.



Since the value of g at a given place on the Earth is constant and R is also constant Therefore g – g h  h .



Thus, the value of acceleration due to gravity decreases with increase in height above the surface of Earth. Note 1: The fractional decrease in the value of g is given by

Note 2: The percentage decrease in the value of g is

g  g h  g



2h . g

2h 100 . g

Loss of Weight at Height h(<
mg – mg h

2 mgh R

Loss in weight =

2 mgh R

Variation of g with Depth 

Assume the Earth to be a homogeneous sphere (having uniform density) of radius R and mass M.



Let p be the mean density of Earth.



Let a body be lying on the surface of Earth where the value of acceleration due to gravity is ‘g’.



Or 

Or

GM Then g  2 R 4 3 G pR g r 3 2 R g

4 p R rG 3

… (1)

 Let the body be now taken to a depth d below the free surface of Earth where the value of acceleration due to gravity isd. g  Here, the force of gravity acting on the body is only due to the inner solid sphere of radius (R – d) g  d

Or

GM  d 2

R –

where M  is the mass of the inner solid sphere of radius (R – d). g   4 R – d 3r G p d

3

 R – d 2 Or

gd 



4 p G R – d r 3

 get

Dividing (2) by (1), we

Or

gd  1 – g or

O

r

d R

 …(2) 4

p G  R – d r  3 4 g p GR r 3

gd



R –d R

 d  gd  g 1  R  –

 d  or g – g d g d –1  g g R 



d

R

 g – gd  gives the decrease in the value of g.



Here



Since g is constant at a given place of the Earth and R is also a constant,



g – gd  d



Thus the value of acceleration due to gravity decreases with the increase of depth. Note1: The fractional decrease in the value of g is given by Note2: The percentage decrease in the value of g is

d

g – g g

 100

d

d  R

R

 Weight of a body at the centre of Earth.

g

At a depth d below the free surface of Earth, d

g



d 

1 –  R 

At the centre of Earth, d = R  R  gd  g1 – R  0 

If ‘m’ is the mass of a body lying at the centre of earth, then its weight = mg =0 d

Comparison of Height and Depth For The Same Change in ‘g’ 

As we have seen above, the value of g decreases as we go above the surface of Earth or when we go below the surface of Earth.



This can be taken to mean that value of g is maximum on the surface of Earth.  and    Now g h  g 2h gd  g 1 d    R   1– –   R  gh  g d , then 1 – 2h

 When

R

1–

d R

2h d – R R



–

Or

2h  d

Or

d  2h



4.

Thus, the value of acceleration due to gravity at a height h is same as the value of acceleration due to gravity at a depth d (= 2h). But this is true only if h is very small. GRAVITATIONAL POTENTIAL ENERGY The force of gravity is a conservative force and we can calculate the potential energy of a body arising out of this force, called the gravitational potential energy. 

 this

Suppose, a body of mass ‘ m’ is placed at P in the gravitational field of a body of mass M. Let r be the distance of P from the centre O of the body of mass M. In order to determine the gravitational potential energy of system, let us calculate the work done in moving mass ‘ m‘ from infinity to P.

 When the mass ‘ m ’ is at A, the gravitational force of attraction on it due to mass M is given by F  GMm . 2 x 

When the mass m moves from A to B, i.e. through distance dx, then work done will be: GM m dW  Fdx  dx 2 x

 If W is the total work done by the gravitational field when a body of mass m is moved from infinity to a distance r from O, then

W  GMm 

Or Or

r

GM m x

2

dx 

r

1



x



2

dx

r r 1  x –2  –2  W  GMm  x dx  GM m   – 2  1 GMm  W– r

r

 1 1 1  – GM m   – GMm  –    x    r   

This work done is equal to the gravitational potential energy U of mass m.  Gravitational potential energy, U  – G

M m r

GMm  GM  = – m r r   Gravitational potential energy = gravitational potential × mass  The gravitational potential energy of mass m in the gravitational field of another mass is negative because work is done by gravitational field and not against it in bringing the mass m from infinity to the point under consideration.  The gravitational potential energy of mass m at a height h above the surface of the Earth is given by Now, U = –



U–

 U–

R

The distance between the mass m and the centre of the Earth is (R + h).] GMm [  R+h GMm  h 1+ R   R 

GMm  =– 1 

–1

h  R 

h   1, so expanding the right hand side of the above equation by Binomial Theorem and neglecting R

 Since

squares and higher powers of

h , we get R

 GMm  h U– 1–  R  R  GMmh Or U  –  2 R GMm R But 

GM 2  g (acceleration due to gravity) R

U–

GMm = gravitational potential energy of mass m at the surface of the Earth. R According to convention, the gravitational potential energy at the surface of the Earth is taken to be zero. U = mgh Note: The surface of Earth is an equipotential surface because the gravitational potential at all points on the surface of Earth is the same. But –

 

GMm  mgh R

5.

ESCAPE SPEED  Escape velocity is defined as the least velocity with which a body must be thrown vertically upwards in order that it may just escape the gravitational pull of the Earth.  When we throw a body vertically upwards with certain velocity, the body returns to the Earth’s surface after some time.  However, when the body is thrown with a velocity equal to the escape velocity, the body overcomes the Earth’s gravitational pull and also the resistance of the Earth’s atmosphere. This body never returns to the surface of the Earth again. Expression for Escape Velocity  Consider a body of mass m lying at a distance x from the centre of the Earth and let M be the mass of the Earth.  According to Newton’s law of gravitation, the gravitational force F of attraction between the body and the Earth is given by FG M m x

2

 Let dW be the work done is raising a body through a small distance dx, then the total work W done in raising the body from the surface of the Earth to infinity is given by  GMm dx W  

dW





x2

R

Here, R is the radius of the Earth. GMm  1  2 Or W = dx x

R



 –2



W  GMm x R



x  Or W = GMm   –1 –1





 

Let

R

Or

2

 x –2  1    dx  GMm    – 2  1 R

 1 1 1  GMm  – GMm  = – GMm  –   R R  x      R 

 1    0 

 

2 ve be the escape velocity, then kinetic energy imparted to the body =1 mv e

Or

1 2 mvm  GM 

 GMm is a constant quantity 

R

e

2GM ve  R But g 

Or

ve 

Or

ve 

GM R

2

2

ve 

2GM R …(i)

Or GM  g R

2

2

Or

ve 

 

2gR

…(ii)

g  2R  gD

…(iii)

Here, D is the diameter of the Earth. 2G 4 3  p R r From equation (1), ve R 3 

(where r is the mean density of Earth)

 ve  R

    

8p Gr 3

…(iv)

Equation (i), (ii), (iii) and (iv) give different expressions for the escape velocity of a body. If a body is to be projected from the surface of any planet other than Earth, then the escape velocity will be different as this escape velocity depends upon the mass and radius of the planet. For the moon, the escape velocity is only 2.4 km s–1. This is a very low velocity. Such a small velocity is easily acquired by gas molecules. This explains as to why gases have escaped the surface of the moon. Moon has practically no atmosphere. On the other hand, the escape velocity is very large in the Sun. So, sun holds its gaseous envelope very firmly.

6. EARTH SATELLITES A satellite is a body which is continuously revolving around a bigger body. Satellite may be regarded as a ‘secondary body’.  The centripetal force required by a satellite to move in a circular orbit is provided by the gravitational force of attraction between the satellite and the body around which it revolves. 

Planets can be said to be the natural satellites of the sun.



Moon is a natural satellite of the Earth which revolves around the Earth in a nearly circular orbit of radius 5 3.85  10 km and completes one revolution is 27.3 days.

Motion of A Satellite (i)

All satellites today get into orbit by riding on a rocket Or by riding in the cargo bay of the space shuttle.

(ii) A rocket is aimed straight up at first which gets the rocket through the thickest part of the atmosphere most quickly. This minimises fuel consumption. (iii) Once the rocket reaches extremely thin air (at 193 km), the rocket’s navigational system fires small rockets. This puts the launch vehicle into horizontal position. (iv) The rockets are now again fired to ensure separation between the launch vehicle and the satellite. Thus, the satellite is put on the path described in the flight plan. Note: (a) We consider the gravitational force only between the satellite and the Earth. The distribution effect of the gravitational force of other bodies is ignored. (b) The centre of mass of the Earth’s satellite system is at the centre of the Earth. Orbital Velocity  Orbital velocity (v0 ) is the velocity which is given to an artificial Earth’s satellite a few hundred kilometers above the Earth’s surface so that it may start revolving round the Earth. 

Expression for orbital velocity Let

m = mass of satellite, r = radius of the circular orbit of satellite h = height of the satellite above the surface of Earth, R = radius of Earth, v0 = orbital velocity, M = mass of the Earth

2 m v 0 required by the satellite to keep moving in a circular orbit is produced by the r

 The centripetal force

Mm between the satellite and the Earth. 2 r

gravitational force G mv r

2 0

GM m  2 r



Then,

Or

v0 

Or

v0  GM r

… (i) … (ii)



rRh GM v0  Rh



Let g h be the value of acceleration due to gravity at a height h above the free surface of the Earth .

2

GM r

But

mg h 

 Then,

Or GM  g h



GMm R  h 2

Rh 2

gh

From equation (ii), v0  Or v0  

gh

Or

2

Rh

Rh

…(iii)

 GM If g be the value of acceleration due to gravity on the surface of the Earth, then g  2 R GM Also, gh   R  GM

gh



 R h

 g GM

gh 

h2

 R  h 2 gR

R



2

R 

2

 R  h 2

2

 R  h 2

From equation (iii), Or v0  R

v0 

gR

2

2  R  h  R  h 

g Rh

 Special Case: If the satellite is close to the Earth say at a height of 100 – 200 km, then h may be regarded as zero and g h may be considered to be equal to g 

v0 

gR

Substituting values, v0  9.81  6400  1000 m s –1  

62784000 m s

For an orbit close to the surface of the Earth, escape velocity, v e =

1

 7923. 6 m s

2gR =

1

 7.9 km s

–1

.. (iv)

2 vo

Time Period of a Satellite (T)  The period of revolution of a satellite is the time taken by the satellite to complete one revolution round the Earth. circumference of circular orbit orbital velocity



T

O

2 T  rv 0 2 R  h T

r

Or

…(v)

 r  R  h 





v0 Or

Or

 R  h

…(vii)

 R  h  R  h



gh

… (viii)

Or

 

 g

  GM h  R  h 2 

2

From equation (vii),

Rh 3

T  2

gR 

R h

T  2

GM

But GM  gR

GM  Rh

3

2



   v0 

GM

Also, T  2

 

T  2   R  h  R h GM T  2

…(vi)

… (ix)

2

Equations (v), (vi), (vii), (viii) and (ix) give different expressions for the time period of a satellite.



Special case: If the satellite is orbiting very close to the Earth, then equations (vii) and (viii) become 3 R R T  2 and T  2 GM g Height of Satellite from Earth’s Surface  2

We know that the time period of satellite is T 

gR satellite above the earth.



Rh 3

Squaring the above equation we get;

T

2

 4

 R h

3

2

gR

Or



2 gR2 T 2 4

R  h  3

2

R h

2

 gR T

2



1/ 3

2

Where R is radius of earth, h is the height of





 gR 2 2 T 4

2



h 

Or

1/ 3

 



2

4 



–R

 

 and

This is the relation which gives the orbital radius of a geostationary satellite. Substituting values of g = 9.8 m s–2, T = 1 day = 86400 s, R = 6.38  106 m, we get 2 2 1/3 1/3 2  86400    42, 237 – 6, 380  35, 857 km  gR T  2 6 h  9.8   6.38  10  – R   2   4 2    4    The geostationary satellite should be taken to a height of 35,857 km above the Earth’s surface at the equator

given the required orbital velocity in the horizontal direction. Uses of Artificial Satellites 1. To study various phenomena in the outer regions of the Earth’s atmosphere. 2. To study various phenomena connected with sea. 3. For communication purposes. 4. For weather forecasting. Energy of An Orbiting Satellite 1 2  The kinetic energy of a satellite moving with orbital velocity v is K E = mv . 2 Therefore, g

GmM E Kg E =

2(R E +h)

 Considering the gravitational potential energy at infinity to be zero, the potential energy at distance (R E +h) from GmM E the centre of the earth is P.E   . (R E +h)  The K.E is positive whereas the P.E is negative. However, in magnitude the K.E. is half the P.E, so that the total E is K.E + P.E  

GmM E 2(R E +h)

 The total energy of circularly orbiting satellite is thus negative, with the potential energy being negative but twice the magnitude of the positive kinetic energy.  When the orbit of a satellite becomes elliptic, both the K.E. and P.E. vary.  The total energy, which remains constant, is negative in the circular orbit case.  If the total energy is positive Or zero, the object escapes to infinity.  Satellites are always at finite distance from the earth and hence their energies cannot be positive or zero. Geostationary or Synchronous Satellite  A geostationary satellite is so named because it appears to be stationary to an observer on the Earth.  This satellite is also named as synchronous satellite because the angular speed of the satellite is synchronised with the angular speed of the Earth about its axis.  When such a satellite is used for communication purposes, it is also known as communication satellite.  Conditions for a satellite to appear stationary to an observer on the Earth: 1. It should be at a height of nearly 36,000 km above the equator. 2. It should revolve in an orbit which is concentric and coplanar with the equatorial plane. So, the plane of the orbit of the satellite is normal to the axis of rotation of the Earth. 3. The sense of rotation of the satellite should be the same as that of the Earth about its own axis. So, the satellite should orbit

4.

around the Earth from west to east as does the Earth. Its orbital velocity is nearly 3.1 km s–1. The orbital period of the satellite, i.e. the time taken by the satellite to complete one revolution around the Earth should be the same as that of the Earth about its own axis. i.e. 24 hours.

Weightlessness     

Weight of an object is the force with which the earth attracts it. When an object is in free fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness. In a satellite around the earth, every part and parcel of the satellite has an acceleration towards the center of the earth which is exactly the value of the earth’s acceleration due to gravity at that position. Thus in a satellite everything inside it is in a state of free fall. This is just as if we were falling towards the earth from a height. Therefore, in a manned satellite, people inside experience no gravity. Gravity for us defines the vertical direction and thus for them there are no horizontal or vertical directions, all directions are the same.