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 an expression for h in terms of m, M. R, T and constants.

Answer 1

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be (  R+ h )

orbital velocity  V₀ =  $\sqrt{\frac{GM}{( R+h)} }$

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x $\sqrt{\frac{( R+h)}{GM } }$

$\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ = R +h

h = $\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ - R.

Answer 2

Answer:

The expression is:

$h=\sqrt[3]{\frac{GMT^{2} }{4\pi ^{2} } } -R$

Explanation:

please, the solution is in the attached Word file

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