The geostationary satellite is revolving round the earth of mean radius R . If g is the acceleration due to gravity at earth's surface, ω is the angular speed of rotation of earth about its own a

Question

The geostationary satellite is revolving round the earth of mean radius R . If g is the acceleration due to gravity at earth's surface, ω is the angular speed of rotation of earth about its own axis and r is the radius of orbit of a satellite, then the correct relation between r , ω , g and R is

TestHub · Accepted Answer

For a geostationary satellite, the angular speed of revolution is equal to the angular speed of rotation of the earth about its own axis, $w$ .

The necessary centripetal force for the circular orbit is provided by the gravitational force of the earth.

$\frac{GM m}{r ^{2}} = m r w^{2}$

$\Rightarrow GM = r^{3} w^{2}$

The acceleration due to gravity at the surface of the earth is given by:

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

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

Equating the two expressions for $GM$ :

$g R^{2} = r^{3} w^{2}$

$\Rightarrow R^{2} = \frac{r ^{3} w ^{2}}{g}$

Answer: $R^{2} = \frac{r ^{3} w ^{2}}{g}$

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