Question

A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 35880Km above the earth surface. Find the period of revolution of the satellite. (Take the mass of earth =5.98×10²⁴kg, the radius of the earth 6370km and G=6.6×10–¹¹Nm²/kg2)​

Answer 1

Period is 86811.5 seconds.

Step-by-step explanation:

`{ \boxed{ \bf{T {}^(2) = (\frac{4 {\pi}^(2) }{GM}) {r}^(3) }}}`

`{ \tt{T {}^(2) = \frac{4 {(3.14)}^(2) }{(6.6 * {10}^( - 11) ) * (5.98 * {10}^(24) )} * {((35880* {10}^(3)) } + (6370 * {10}^(3) )) {}^(3) }} \\ \\ { \tt{T {}^(2) = 7.54 * {10}^(9) }} \\ { \tt{T = \sqrt{7.54 * {10}^(9) } }} \\ { \tt{T = 86811.5 \: seconds}}`