Question

You would like to put a GEO satellite for QU_SAT in orbit. Its weight is 3000 Kg. What should be its

exact altitude for being stationary with respect to Qatar. Show the steps of your calculations.

Answer 1

A satellite at altitude h from earth surface is at distance R+h from center of earth where R is radius of earth.

According to Newton's law of universal gravitation, the force of attraction between a satellite with mass m at altitude h is:

$F = G\frac{Mm}{(R+h)^2}$

Where G = gravitational constant = 6.674x10^-11 m³/(kgs²⁾
            R = radius of earth = 6 378 km
            M = mass of earth = 5.972 × 10²⁴ kg
            m = mass of satellite = 3000 kg, as given in question

A geostationary satellite orbits round earth in 24 hours, thus effectively being stationary w.r.t a certain place on earth surface.

=> The satellite moves through an angle of 2 $\pi$ rad (assuming circular orbit for simplicity of calculation) in 24x3600 seconds

=> angular velocity of satellite:

$\omega = \frac{2\pi}{24\times 3600}$

The centripetal force required to keep the satellite in orbit is given by the formula

$F = m(R+h)\omega^2$

When a satellite is in orbit the gravitational force must equal the centripetal force, i.e.

$G\frac{Mm}{(R+h)^2} = m(R+h)\omega^2$

$\Rightarrow R+h = \left(\frac{GM}{\omega^2} \right )^{\frac{1}{3}}$
Putting the values of G, M and $\omega$ in above equation,

R+h $\approx$ 422.4x10⁵ m = 42240 km

=> h = (42240 - 6378) km = 35862 km [Ans]

We find that altitude of geostationary satellite is independent of it's mass.