Question

A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which the Earth completes one revolution around its own axis. The satellite has mass m and the Earth has mass M and radius R. In order to be geosynchronous, the satellite must be at a certain height h above the Earth’s surface. Derive equatio of h in terms of M, R, T.

Answer 1

Answer:

   h = r- R   with   r = ∛ (G M T² / 4π²)

Explanation:

As the satellite interacts with the Earth, let's use the law of universal gravitation

            F = G m M / r²

Let's write Newton's second law

           F = m a

           a = v2 / r

   

          G m M / r² = m v² / r

          G M / r = v²

          r = G M / v²

Let's work on satellite speed, counting speed, so we use uniform speed expressions

          v = d / t

Let's look for the distance it travels in a complete orbit and its time is equal to one (1) day since the satellite is geosynchronous

           d = 2π r

           t = T = 1 day (24h / 1day) (3600s / 1h) = 86400 s

           v = 2 π R / T

Let's calculate

           r = G M (T / 2π r)²

           r = G M T² / 4π² r²

           r³ = G M T² / 4π²

           r = ∛ (G M T² / 4π²)

This distance (r) is measured from the center of the earth, let's get the height (h) from the planet's surface

          r = R + h

          h = r- R