As we know that the earth’s gravitational force provides the necessary centripetal force to the satellite               

Hence  \frac{{mv_{0}^{2}}}{r}=\frac{{GMm}}{{{{r}^{2}}}} \frac{{mv_{0}^{2}}}{r}=\frac{{GMm}}{{{{r}^{2}}}}  ( where r=R+h))

or,          v_{0}^{2}=\frac{{GM}}{r}c

{{v}_{0}}=\sqrt{{\frac{{GM}}{r}}}                                                              …(1)

But {{v}_{0}}=\sqrt{{\frac{{GM}}{r}}}

{{v}_{0}}=\sqrt{{\frac{{GM}}{R+h}}} …(2)

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

Then,    m{{g}_{h}}=\frac{{GMm}}{{{{{(R+h)}}^{2}}}}

or,          GM={{g}_{h}}{{(R+h)}^{2}}

From equation (2), {{v}_{0}}=\sqrt{{\frac{{{{g}_{h}}{{{(R+h)}}^{2}}}}{{R+h}}}}

or,          \displaystyle {{v}_{0}}=\sqrt{{{{g}_{h}}(R+h)}}    …(3)

If g be the value of acceleration due to gravity on the surface of Earth, then g=\frac{{GM}}{{{{R}^{2}}}}

Also,      {{g}_{h}}=\frac{{GM}}{{{{{(R+h)}}^{2}}}}

\frac{{{{g}_{h}}}}{g}=\frac{{GM}}{{{{{(R+h)}}^{2}}}}\times \frac{{{{R}^{2}}}}{{GM}}=\frac{{{{R}^{2}}}}{{{{{(R+h)}}^{2}}}}

or            {{g}_{h}}=\frac{{g{{R}^{2}}}}{{{{{(R+h)}}^{2}}}}

from equation (3), {{v}_{0}}=\sqrt{{\frac{{g{{R}^{2}}}}{{{{{(R+h)}}^{2}}}}(R+h)}}

or            {{v}_{0}}=R\sqrt{{\frac{g}{{R+h}}}}

Equation (1), (2), (3) and (4) give different expressions for the orbital velocity of a satellite.

Special Case. If the satellite is close to Earth, say at a height of 100-200 kg, then h may be regarded as zero and g_h may be considered to be equal to g.

In this case, from equation (3), {{v}_{0}}=\sqrt{{gR}}

Substituting values,

{{v}_{0}}=\sqrt{{9.81\times 6400\times 1000}}\ \text{m}{{\text{s}}^{{-1}}}

=\sqrt{{62784000}}\ \text{m}{{\text{s}}^{{-1}}}=7923.6\text{m}{{\text{s}}^{{-1}}}=7.9\text{km}\,{{\text{s}}^{{-1}}}

Note: For an orbit close to the surface of the Earth, escape velocity, {{v}_{e}}=\sqrt{{2gR}}=\sqrt{2}orbit velocity v₀.

Important

(i)  Orbital velocity is independent of the mass of the orbiting body and is always along the tangent to the orbital.

(ii)    Close to the surface of the earth, r = R as h = 0.

{{V}_{0}}=\sqrt{{\frac{{GM}}{R}}}=\sqrt{{gR}}=\sqrt{{10\times 6.4\times {{{10}}^{6}}}}\approx \ 8\ \text{km/s}

(iii)          Close to the surface of the planet

{{V}_{0}}=\sqrt{{\frac{{GM}}{R}}}=\frac{{{{V}_{e}}}}{{\sqrt{2}}}

i.e.,        {{V}_{e}}=\sqrt{2}{{V}_{0}}

Time period of a satellite

As          Time=\frac{{Dis\tan ce}}{{Speed}}                                                                           

therefore              T=\frac{{\text{circumference}\ \text{of}\ \text{circular}\ \text{orbit}}}{{\text{orbital}\ \text{velocity}}}

or,          T=\frac{{2\pi r}}{{{{v}_{0}}}}                                       

or,          T=\frac{{2\pi \,(R+h)}}{{{{v}_{0}}}} [\because \ \ r=R+h]

or,          T=2\pi (R+h)\sqrt{{\frac{{R+h}}{{GM}}}}               \left[ {\because \ \ {{v}_{0}}=\sqrt{{\frac{{GM}}{{R+h}}}}} \right]

or,          T=2\pi \sqrt{{\frac{{{{{(R+h)}}^{2}}}}{{GM}}}}                    

Also,      T=2\pi \sqrt{{\frac{{{{{(R+h)}}^{2}}(R+h)}}{{GM}}}}

or,          T=2\pi \sqrt{{\frac{{R+h}}{{{{g}_{h}}}}}} \left[ {\because \ \ gh=\frac{{GM}}{{{{{(R+h)}}^{2}}}}} \right]                                            

But GM=g{{R}^{2}}

From equation, T=2\pi \sqrt{{\frac{{{{{(R+h)}}^{3}}}}{{g{{R}^{2}}}}}}          ————– (A)               

Equations give different expressions for the time period of a satellite.

Special Case. If the satellite is orbiting very close to the Earth, then

T=2\pi \sqrt{{\frac{{{{R}^{3}}}}{{GM}}}} and T=2\pi \sqrt{{\frac{R}{g}}}.

Height of satellite

From equation (A)       {{T}^{2}}=4{{\pi }^{2}}\frac{{{{{(R+h)}}^{3}}}}{{g{{R}^{2}}}} or {{(R+h)}^{3}}=\frac{{g{{R}^{2}}{{T}^{2}}}}{{4{{\pi }^{2}}}}

or            R+h={{\left[ {\frac{{g{{R}^{2}}{{T}^{2}}}}{{4{{\pi }^{2}}}}} \right]}^{{1/3}}}

or            h={{\left[ {\frac{{g{{R}^{2}}{{T}^{2}}}}{{4{{\pi }^{2}}}}} \right]}^{{1/3}}}-R

Angular momentum of a satellite (L)

In case of satellite motion, angular momentum will be given by

L=mvr=mr\ \sqrt{{\frac{{GM}}{r}}}

L={{\left( {{{m}^{2}}GMr} \right)}^{{1/2}}}                                          

important

In case of satellite motion, the net force on the satellite is centripetal force. The torque of this force about the center of the orbit is zero. Hence, angular momentum of the satellite is conserved, i.e., L = constant.

Energy of a satellite

The P.E. of a satellite is

U =mV =-\frac{{GMm}}{r} \,\,\left[ {\because \ V=-\frac{{GM}}{r}} \right]

The kinetic energy of the satellite is

K=\frac{1}{2}mv_{0}^{2}\ \ \ =\frac{{GMm}}{{2r}} \ \ \ \ \left[ {\because \ {{v}_{0}}=\sqrt{{\frac{{GM}}{r}}}} \right]

Total mechanical energy of the satellite = -\frac{{GMm}}{r}+\frac{{GMm}}{{2r}}=-\frac{{GMm}}{{2r}}

Important

(i)            We have,  \frac{K}{E}=-1  i.e.,  K = –E

Also,      \frac{U}{E}=2

U = 2E

therefor   E = -K = U/2

(ii)     Total energy of a satellite in its orbit is negative. Negative energy means that the satellite is bound to the central body by an attractive force and energy must be supplied to remove it from the orbit to infinity.

Binding energy of the satellite

The energy required to remove the satellite from its orbit to infinity is called binding energy of the satellite, i.e.,

Binding energy = – E =\frac{{GMm}}{{2r}}

Weightlessness in a satellite

In the frame of reference of a satellite the astronauts experience the following forces

1.

The gravitational force of the planet around which the satellite is moving = \frac{{GM{{m}_{a}}}}{{{{r}^{2}}}} and it acts along the radius toward the center of the orbit.

2.

The centrifugal force due to circular motion of the satellite = \frac{{{{m}_{a}}v_{o}^{2}}}{r} where v_o is the orbital velocity and is given by {{\nu }_{o}}=\sqrt{{\frac{{GM}}{r}}}, this force acts radially outward i.e. away from the center of the orbit.

Hence substituting v_o we get  the centrifugal force = \displaystyle \frac{{{{m}_{a}}v_{o}^{2}}}{r}=\frac{{GM{{m}_{a}}}}{{{{r}^{2}}}} which is exactly equal to the

gravitational inward force therefore the net force acting on the astronauts becomes zero hence they feel weightlessness.

Uses of artificial satellites

(i)   To study various phenomena in the outer regions of the Earth’s atmosphere.

(ii)   To study various phenomena connected with sea.

(iii)   For communication purposes.

(iv)  For weather forecasting.

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 synchronized with the angular speed of Earth about its axis. When such a satellite is used for communication purposes, it is also known as communication satellite.

In order that a satellite may appear stationary to an observe on the Earth, the following conditions must be satisfied.

(i)   It should be at a height of nearly 36000 km above the equator.

(ii)   It should revolve in an orbit which is concentric and coplanar with the equatorial plane. So, the plane of the orbit of the satellite in normal to the axis of rotation of the Earth.

(iii)   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 around the Earth from west to east as does the Earth. Its orbit velocity is nearly 3.1 km s^–1.

(iv)  The orbital period of the satellite, i.e., the time taken by the satellite to complete one revolution around the Earth is the same as that of the Earth about its own axis i.e., 24 hours.

Illustration

Calculate the orbital velocity of a satellite revolving at a height h above the earth’s surface if h = R. Also calculate the time-period of this satellite. (g = 9. m/s²,
R = 6400 km)

Solution: 

For the orbital velocity in a circular orbit, we have

v=\sqrt{{\frac{{GM}}{r}}}=\sqrt{{\frac{{GM}}{{R+h}}}} (Qr = R + h)

v=\sqrt{{\frac{{g{{R}^{2}}}}{{2R}}}}=\sqrt{{\frac{{gR}}{2}}}  (QGM = gR² and h = R)   Þ v=\sqrt{{\frac{{9.8\,\,\times \,6400\times {{{10}}^{3}}}}{2}}} = 5.6 km/s

Time period = T=\frac{{2\pi r}}{v}=\frac{{2\pi \,(2R)}}{{4\sqrt{2}\times {{{10}}^{3}}}}

T=\frac{{4\pi \times 6400\times {{{10}}^{3}}}}{{4\sqrt{2}\times {{{10}}^{3}}}} =3.95\ \text{hrs}

Bound and Unbound Trajectories

Suppose that a cannonball is fired from the peak of very tall tower with speed v. Let us consider the shapes of the paths for various values of v.

If the ball is located at a distance r from the center of the earth, the speed that is required to escape from the point is given by

v_esc= \sqrt{{\frac{{2GM}}{r}}}

For a circular orbit of radius r the orbital speed is    v_o= \sqrt{{\frac{{GM}}{r}}}

(a)  When v <v_o, the orbit is elliptical, with the peak as the apogee (the farthest point from earth). If v is too small, the projectile will strike the earth.

When v = v_o, the orbit is circular.

When v_esc>v >v_o, the orbit is again an ellipse, but now the peak is the perigee (the closest point to earth).

(b)  When v = v_esc= \sqrt{2}v_o, the path is parabolic and is not a closed orbit. The object is not bound.

(c)  When v >v_esc, the path is hyperbola and not closed. The object is not bound.