First slide

Motion of planets and satellites

Question

Imagine a geo-stationery satellite of earth which is used as an inter continental telecast station. At what height will it have to be established

Moderate

A

$10^{3} m$
B

$6.4 \times 10^{3} m$
C

$35.94 \times 10^{3} m$
D

$i n f i n i t y$

Solution

$T = 2 π \sqrt{\frac{{(R + h)}^{3}}{G M}} \Rightarrow {(\frac{T}{2 π})}^{2} = \frac{{(R + h)}^{3}}{g R^{2}} \Rightarrow (\frac{T^{2}}{4 π^{2}}) g R^{2} = {(R + h)}^{3} \Rightarrow h = {(\frac{T^{2} g R^{2}}{4 π^{2}})}^{\frac{1}{3}} - R \Rightarrow h = {(\frac{{(86400)}^{2} (9.8) {(6400 \times 1000)}^{2}}{4 π^{2}})}^{\frac{1}{3}} - 6400 \times 1000 = 35.94 \times 10^{3} m (∵ R a d i u s o f e a r t h = R = 6400 K m, T i m e p e r i o d o f g e o s t a t i o n a r y s a t e l l i t e = 24 h = 24 \times 3600 = 86400 s a n d g = 10 m s^{- 2})$

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