Newtonion Gravitation

Gravitation is defined as the force that is produce between two bodies by virtue of their masses.

Acceleration due to gravity :

Earth attracts all bodies towards its centre. The acceleration produced in the body due to attraction force of earth on it is called the acceleration due to gravity. It is denoted by ‘g’.

Newton’s Law of Gravitation :

According to this law any two particles in the universe attract each other, of attraction (F) between two point two masses ($m_1 \, \text{and} \, m_2$) being directly proportional to the product of two masses and inversely proportional to the square of distance (r) between them i.e.

$F \propto \dfrac{m_1 m_2}{r^2}$                    ……………………………………………(1)

where G is a constant of proportionality and is called universal gravitational constant and its value is

G = $6.67 \times 10^{- 11} N m^2 / kg^2$   everywhere in the universe.

Remark :

The gravitational force between two particles is independent of the presence of other bodies or the properties of the intervening medium.

Relation between g and G :

The force of attraction exerted by earth on a body is called the force of gravity. If M is mass of earth and r the distance of body of mass m from the centre-of the earth, then force of gravity

$F = \dfrac{G M m}{r^2}$    ………………………………………………(1)

According to Newton’s law of motion the effect of force on a body is to produce an acceleration in it. Accordingly the force of gravity produces acceleration in a freely falling body called the acceleration due to gravity, denoted by g.

According to Newton’s II law, the force of gravity

F = mg                                       ……………………………………………………………..(2)

Comparing (1) and (2), we get

$g = \dfrac{GM}{r^2}$                              …………………………..(3)

If a body is at or near the surface of earth; then r = R, the radius of earth, therefore

$g = \dfrac{GM}{R^2}$            ……………………………………………….(4)

This is the relation between g and G.

Inertial and Gravitational Mass:

The inertia is the inherent property of the body which opposes any change in its state of rest or uniform motion. The inertia is related to mass. Greater is the mass, greater is said to be the inertia. When we apply external force, the body is accelerated; the acceleration produced in a body is directly proportional to the applied force. This constant of proportionality appearing in the relation $F \propto a$ ; is called the inertial mass. In brief the inertial mass is the inertia offered by body against the inertial (applied) force.

Every freely falling body, irrespective of inertial mass has the same acceleration, called acceleration due to gravity. This acceleration is due to gravitational force of earth on the body. The inertia offered by the body against the gravitational force is called the gravitational mass i.e. $F_g = m_g$ where
$m_g$ is gravitational mass.

The inertial and gravitational masses, are although defined in different ways ; but they are proportional to each other and for convenience this constant of proportionality is taken to be 1 and hence the gravitational and inertial masses are considered to be equivalent.

Variation of g :

The value of g varies from place to place due to following factors:

(i) Shape of earth :

The earth is not perfectly spherical but is slightly ellipsoidal. It is bulging at the equator and flattened at the poles. Its equatorial radius is about 21 km more than the polar radius.

As g $\propto ( 1/R^2 )$ the value of g increases from equator to poles i.e. it is maximum at the poles and minimum at the equator.

(ii) Effect of earth’s rotation :

The earth rotates about its axis with angular velocity $\omega$

Effect's of earth rotation on g

The necessary centripetal force $F_c$ is provided by a component of gravitational attraction force between the body and the earth, so that the effective weight of body is reduced and hence effective g is reduced and at latitude $\lambda$ it is given by

$F' = F - F_c \, \cos \lambda$

But

$F_c = m r {\omega}^2 \, \, \text{and} \, \, r = R \cos \lambda \\ \therefore F' = F - m r {\omega}^2 cos \lambda \\ = F - m ( R \cos \lambda ) {\omega}^2 \cos \lambda$

i.e

$mg' = mg - mR {\omega}^2 {\cos}^2 \lambda \\ \text{or} \, g' = g - R {\omega}^2 {\cos}^2 \lambda$  ………………………………………………….(1)

This equation shows that the value of g decreases due to earth’s rotation.

At equator $\lambda = 0 \, \, \, \, \, \therefore g' = g - R {\omega}^2$

At poles $\lambda = 90^o \, \, \, \, \, \therefore g' = g$

Therefore we note that the decrease in g is maximum at equator and zero at poles.

(iii) Effect of Altitude (Height)

The acceleration due to gravity g at the surface of the earth is

$g = \dfrac{GM}{R^2}$                    ………………………………………….(2)

The acceleration due to gravity $g_h$ at height h from earth’s surface is

$g_h = \dfrac{GM}{( R + h )^2}$     …………………………………………..(3)

( since r = R + h )

Dividing (3) by (2) we get

$\dfrac{g_h}{g} = \dfrac{R^2}{( R + h )^2}$

i.e

$g_h = \dfrac{g}{( 1 + \dfrac{h}{R} )^2}$    ………………………………..(4)

This relation holds for any height.

If h<<R , then using bionomial theorem

$g_h = ( 1 + \dfrac{h}{R} )^{-2} g$

Or

$g_h = ( 1 - \dfrac{2h}{R} ) g$            …………………………………………….(5)

From equation (4) and (5) it is obvious that the value of g decreases with increase of height from the surface of the earth.

(iv) Effect of Depth (i.e. below earth’s surface ):

Assuming earth to be a sphere of uniform density $\rho$ ,we have mass of earth , $M' = \dfrac{4}{3} \pi R^3 \rho$. The acceleration due to gravity at earth’s surface

$g = \dfrac{GM}{R^2} = \dfrac{G ( \dfrac{4}{3} \pi R^3 \rho )}{R^2} = \dfrac{4}{3} G \rho R$……………..(6)

If body is taken to a depth x below earth’s surface ( e.g. in a mine ), the body will be attracted only by the mass ( M’) of earth which is enclosed in a sphere of radius ( R –h) given by

Effect of depth on g

$M' = \dfrac{4}{3} \pi ( R - x )^3 \rho$

The value of acceleration due to gravity at depth x is given by

$g_x = \dfrac{GM'}{( R - x )^2} = \dfrac{G \dfrac{4}{3} \pi ( R - x )^3 ) \rho}{( R - x )^2} \\ = \dfrac{4}{3} \pi G \rho ( R - x )$…..(7)

Dividing  (7) by (6) , we get

$\dfrac{g_x}{g} = \dfrac{R - x}{R} = 1 - \dfrac{x}{R} \\ i.e \, \\ g_x = ( 1 - \dfrac{x}{R} ) g$ …………………(8)

Thus the value of g decreases with increase of depth below earth’s surface.
This decrease is due to reduction in mass of attracting sphere of earth.

Gravitational field strength, Potential and Potential energy :

The gravitational field strength at any point in a gravitational force experienced by a unit mass placed at that point provided the unit mass itself does not cause any change in the field. Thus the gravitational field due to a particle of mass M at point distant r from mass M is

$f_g = \dfrac{F}{m} = \dfrac{GM m/r^2}{m} = \dfrac{GM}{r^2}$
The gravitational potential at any point in a gravitational field is defined as the work done is bringing a unit mass from infinity to that point. The gravitational potential due to a point mass M at a point distant r from it is given by

$V = - \dfrac{GM}{r}$

The relation between gravitational field strength and potential is

$f_g = - grad V$

In one dimension $f_g = - \dfrac{dV}{dr}$
The gravitational potential energy of a system of masses is defined as the work done in assembling the system of masses from infinity to its present configuration.

Gravitational potentiall energy

The gravitational potential energy or self energy of the system of two masses m1 and m2 is given by

$U = - \dfrac{G m_1 m_2}{r}$

where r is the distance between the masses.

Gravitational potential energy has two familiar expressions -
U = mgh     (at height h from earth ’s surface near surface of earth, assuming zero P.E at earth’s surface)

And
$U = - \dfrac{G Mm}{r}$  (assuming zero P.E. at infinity where r = distance of body of mass m from centre of earth usually r = R + h )

Planets and Satellites :

A planet is a celestial body revolving around the sun. The nine planets are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto.

The motion of planets is described by well known Kepler’s laws which are :

1. Each planet revolves around the sun in elliptical orbits, with the sun at one focus of the ellipse.

2. The speed of the planet in its orbit varies in such a way that the radius vector joining the planet to sun sweeps equal areas in equal times. Alternatively this law may be stated as
The areal velocity of radius vector joining the planet to sun is constant.

3. The square of period of revolution of planet around the sun is proportional to the cube of average distance of planet round the sun.
More correctly we may say that the square of period of revolution of a planet around the sun is proportional to the cube of the semi-major axis of its orbit.

A satellite is secondary heavenly body which revolves round a planet in its definite orbit. The satellites may be natural and artificial. Moon is a natural satellite of earth, while Aryabhatta, Rohini, Insat 2-A, 2-D are artificial satellites of earth. The artificial satellites are put in orbits round the earth by multi-stage rocket which carries the satellite to required vertical height and then gives it appropriate horizontal velocity required for a stable orbit around the earth.

Motion of Satellites:

Let a satellite of mass m revolve around the earth in circular orbit of radius.The gravitational pull between the satellite and earth

= $\dfrac{G Mm}{r^2}$
provides the necessary centripetal force i.e.

$\dfrac{m {v_0}^2}{r} = \dfrac{G Mm}{r^2}$

This gives orbital speed

$v_0 = \sqrt{ \dfrac{G M}{r}}$

If h is the height of satellite from earth’s surface and R is the radius of earth, then r = R + h.

$v_0 = \sqrt{\dfrac{GM}{R + h}}$   ……………………………….(2a)

Since acceleration due to gravity at earth’s surface

$g = \sqrt{\dfrac{GM}{R}^2} \\ \, or \, GM = R^2 g \\ v_0 = \sqrt{R^2 g}{R + h}$………..(2b)

Equation (2) shows that the orbital velocity satellite depends only on the height for a given plane If T is the period of revolution, then

$T = \dfrac{2 \pi r}{v_0} = \dfrac{2 \pi r}{\sqrt{( G M/r )}} = 2 \pi \sqrt{\dfrac{r^3}{GM}} \dots (3a) \, \\ \, = 2 \pi \sqrt{\dfrac{( R + h )^3}{R^2 g}}$…………………..(3b)
From (3a), we have $T^2 \propto r^3$, which is Kepler III law.

Remark :

If a satellite is close to earth’s surface such that h << R, then orbital speed

$v_o = \sqrt{\dfrac{GM}{R}} = \sqrt{\dfrac{R^2 g}{R}} = \sqrt{R g}$

and periodic time,

$T = 2 \pi \sqrt{\dfrac{R^3}{GM}} = 2 \pi \sqrt{\dfrac{R^3}{R^2 g}} = 2 \pi \sqrt{\dfrac{R}{g}}$

For earth ,

R = 6400 km = $6.4 \times 10^6$ m

g = $9.8 m/s^2$

orbital speed ,

$v_0 = \sqrt{( 6.4 \times 10^6 \times 9.8 )} \, \\ \, = 8 \times 10^3 m/s = 8 km/s$

and period of revolution,

$T = 2 \pi \sqrt{\dfrac{6.4 \times 10^6}{9.8}} = 5075 sec = 84.6 min$

Energy of a satellite :

The kinetic energy,

$E_k = \dfrac{1}{2} mv^2 = \dfrac{G Mm}{2r}$

The potential energy,

$U = - \dfrac{G Mm}{r}$

So,

Total mechanical energy

$E_t = E_k + U = - \dfrac{G Mm}{2r}$

This is negative, indicating that the satellite is bound.

Binding energy :

The minimum energy required to free a satellite from its gravitational attraction is called binding energy i.e.

Binding energy , $B = -E_t = \dfrac{G Mm}{2r}$

Geostationary or Communication Satellite:

A communication satellite is an artificial satellite of earth and appears stationary to any observer on the surface of earth. Such a satellite is called a geostatic satellite or a geo-synchronous satellite. Insat 2-A, 2-D are satellites of India. For such a satellite:

(1) The orbit of the satellite must be circular and in the equatorial plane of the earth.

(2) The angular velocity of the satellite must be in the same direction as the angular velocity of rotation of earth.

(3) The period of revolution of the satellite must be equal to the period of rotation of earth about its axis.
i.e. 24 hours = 24 X 60 X 60 = 86400 sec.
If calculations are made from the relation

$T = 2 \pi \sqrt{\dfrac{r^3}{GM}} = 2 \pi \sqrt{\dfrac{r^3}{R^2 g}}$

the radius of circular orbit for such a satellite is 42000 km.

Escape velocity:

The minimum velocity required to just free a body from the binding of the gravitational attraction is called the escape velocity $( v_e )$

$v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2Rg}$ ………………………….(1)

The escape velocity is independent of mass of body projected and the direction of projection.

For earth

$R = 6.4 \times 10^6 m \, \, and g = 9.8 m/s^2 \\ \therefore v_c = 11.3 km/sec$

Relation between orbitial velocity and escape velocity $v_e$ near earth surface is

$V_e = \sqrt{2} v_o$

Weightlessness in Artificial Satellites:

A astronaut in a satellite experiences weightlessness. The reason is that there act two forces on the astronaut.

(i) Gravitational pull $F_g = \dfrac{G M_e m}{r^2}$

Where

$M_e$ = mass of earth, m = mass of astronaut,  r= distance between satellite from earth’s centre

(ii) Centrifugal force :

As astronaut is in rotating frame; the astronaut experiences a centrifuge force whose direction is away from the centre of earth and magnitude given by,

$F_c = ma = \dfrac{m v^2 0}{r}$

where v0 is orbital speed of satellite.

Net force on astronaut:

$F = f_g - f_c = \dfrac{G M_e m}{r^2} - \dfrac{m v_o^2}{r}$

Also the condition of circular motion of satellite is

$\dfrac{G M_e m}{r^2} = \dfrac{m {v_o}^2}{r}$

Using this equation, equation (1) gives

F = 0

i.e. the net force on astronaut is zero the astronaut experiences weightlessness in artificial satellite.

Remark:

The weight of a person/body is zero in artificial satellite, but it is not zero in natural satellite, because natural satellite has its own g. The value of g on moon is one -sixth that on earth.

Absence of atmosphere on moon:

For the presence of atmosphere on a planet/moon the condition is that the root mean square speed of molecule: on planet/moon should be less than the escape velocity i.e. vrms < vescape. This condition is satisfied on earth but not on moon. Since on earth

vrms = 2.5 km/s, ve = 11.2 km/h

Therefore earth has atmosphere.

For moon vrms = 2.5 km/s, ve = 2.35 km/h

i.e. at moon vrms > vhence, moon has no atmosphere.

hence moon has no atmosphere.
Thus a planet will have no atmosphere if root moon square of molecules on its surface is larger than escape velocity on the planet’s surface.

Cosmic velocities:

There are three cosmic velocities :

(i) First cosmic velocity

The minimum horizontal velocity required to launch a satellite in an orbit near earth’s surface is called the first cosmic velocity. Its value is

$v_0 = \sqrt{\dfrac{G m_e}{R_e}} = \sqrt{R_e g} = 8 km/s$

(ii) Second cosmic velocity:

The minimum velocity required to project a body from earth’s surface to escape from earth’s gravitational field is called the second cosmic velocity. Its value is

$v_e = \sqrt{\dfrac{2G m_e}{R_e}} = \sqrt{2 R_e g} = 11.2 km/s$

(iii) Third cosmic velocity:

The minimum velocity required to project a body so as to escape from the solar system is called the third cosmic velocity. Its value from earth’s surface is nearly 17 km/s.

A planet in elliptical orbit:

If a planet is in an elliptical orbit, the following two equations are used. Let a planet of mass m revolve around the sun of mass M in elliptical orbit, with sun at one focus of orbit.

(i) Conservation of energy : Total energy at farthest point A = Total energy at nearest point B

$\dfrac{1}{2} m{v_A}^2 - \dfrac{GMm}{r_1} = \dfrac{1}{2} m{v_B}^2 - \dfrac{GMm}{r^2}$ …….(1)

conservation of energy

(ii) Conservation of angular momentum gives

Angular momentum at A = angular momentum at B

$m v_A r_1 = m v_B r_2$……………………….(2)

Geometry of elliptical orbits :

If a and b are semi-major and semi-minor axis of orbit, then eccentricity of orbit is given by

$e = \sqrt{1 - \dfrac{b^2}{a^2}}$      as b<a , e<1

Also maximum distance of planet from sun = perihelion, r1 = a (l+ e)

Minimum distance of planet from sun

= aphelion , r2 = a (1 – e)

Semi-latus rectum of L = $\dfrac{b^2}{a}$

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