For-a-geosynchronous-satellite-of-mass-m-moving-in-a-circular-orbit-around-the-earth-at-a-constant-speed-v-and-an-altitude-h-above-the-earth-surface-Show-the-velocity-v-GM-e-R-e-h-1-2-I

Question Number 25684 by NECx last updated on 13/Dec/17

F o r a g e o s y n c h r o n o u s s a t e l l i t e o f

m a s s m m o v i n g i n a c i r c u l a r

o r b i t a r o u n d t h e e a r t h a t a c o n s t a n t

s p e e d v a n d a n a l t i t u d e h a b o v e

t h e e a r t h s u r f a c e . S h o w t h e

v e l o c i t y v = {(\frac{{G M}_{e}}{R_{e} + h})}^{\frac{1}{2}} .

I f t h e s a t e l l i t e a b o v e i s s y n c h r o n o u s

h o w f a s t i s i t m o v i n g t h r o u g h

s p a c e, t a k i n g t h e p e r i o d t o b e 24 h r s

a n d M_{e} = m a s s o f t h e s a t e l l i t e a n d

i s e q u a l t o 5.98 \times 10^{24} k g

Answered by jota@ last updated on 13/Dec/17

a_{n} = (1 / m) \frac{{G M}_{e} m}{{(R_{e} + h)}^{2}} = \frac{v^{2}}{R_{e} + h}

v = \sqrt{\frac{{G M}_{e}}{R_{e} + h} .}

Commented by NECx last updated on 13/Dec/17

t h a n k s b o s s

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