# 4.10 Artificial satellites  (Page 2/4)

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We can write the equation of orbital speed in terms of acceleration due to gravity at the surface (g = ${g}_{0}$ ), which is given by :

$g=\frac{GM}{{R}^{2}}$

$⇒GM=g{R}^{2}$

Substituting in the equation of orbital velocity, we have :

$⇒v=\sqrt{\left(\frac{GM}{r}\right)}=\sqrt{\left(\frac{g{R}^{2}}{r}\right)}=\sqrt{\left(\frac{g{R}^{2}}{R+h}\right)}$

where “h” is the vertical height of the satellite above the surface. Rearranging,

$⇒v=R\sqrt{\left(\frac{g}{R+h}\right)}$

## Time period of revolution

Time period of revolution is equal to time taken to travel the perimeter of circular path. The time period of rotation is :

$T=\frac{2\pi r}{v}$

Substituting expression of “v” as obtained earlier,

$T=\frac{2\pi {r}^{\frac{3}{2}}}{\sqrt{\left(GM\right)}}$

This is the expression of time period for a satellite revolving in a circular orbit. Like orbital speed, the time period is also independent of the mass of the satellite. Now, squaring both sides, we have :

$⇒{T}^{2}=\frac{2\pi {r}^{3}}{GM}$

Clearly, square of time period of a satellite is proportional the cube of the linear distance for the circular orbit,

$⇒{T}^{2}\propto {r}^{3}$

## Example

Problem 1: Two satellites revolve around Earth along a coplanar circular orbit in the plane of equator. They move in the same sense of direction and their periods are 6 hrs and 24 hrs respectively. The satellite having period of 6 hrs is at a distance 10000 km from the center of Earth. When the satellites are at the minimum possible separation between each other, find the magnitude of relative velocity between two satellites.

Solution : Let us denote two satellites with subscripts “1” and “2”. Let the satellite designated with “1” is closer to the Earth. The positions of satellites, corresponding to minimum possible separation, are shown in the figure.

The distance between center of Earth and the satellite “1” is 10000 km, but this data is not available for the other satellite. However, we can evaluate other distance, using the fact that square of time period of a satellite is proportional to the cube of the linear distance for the circular orbit.

$\frac{{r}_{2}^{3}}{{r}_{1}^{3}}=\frac{{T}_{2}^{2}}{{T}_{1}^{2}}={\left(\frac{24}{6}\right)}^{2}=16$

$⇒\frac{{r}_{2}}{{r}_{1}}=2$

$⇒{r}_{2}=2{r}_{1}=2x{10}^{4}=2X{10}^{4}\phantom{\rule{1em}{0ex}}km$

We can now determine velocity of each satellite as :

$v=\frac{2\pi r}{T}$

For the first satellite,

$⇒{v}_{1}=\frac{2\pi X10000}{6}=\frac{10000\pi }{3}$

For the second satellite,

$⇒{v}_{2}=\frac{2\pi X20000}{24}=\frac{10000\pi }{6}$

Hence, magnitude of relative velocity is :

$⇒{v}_{1}-{v}_{2}=\frac{\pi }{6}X10000=5238\phantom{\rule{1em}{0ex}}km/hr$

## Energy of “earth-satellite” system

For consideration of energy, Earth can be treated as particle of mass “M”. Thus, potential energy of “Earth – satellite” as two particles system is given by :

$U=-\frac{GMm}{R}$

Since expression of orbital speed of the satellite is known, we can also determine kinetic energy of the satellite as :

$K=\frac{1}{2}m{v}^{2}$

Putting expression of speed, “v”, as determined before,

$⇒K=\frac{GMm}{2r}$

Note that kinetic energy of the satellite is positive, which is consistent with the fact that kinetic energy can not be negative. Now, mechanical energy is algebraic sum of potential and kinetic energy. Hence, mechanical energy of “Earth – satellite” system is :

$⇒E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}$

$⇒E=-\frac{GMm}{2r}$

Here, total mechanical energy of the system is negative. We shall subsequently see that this is characteristic of a system, in which bodies are bounded together by internal force.

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