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) . If the satellite above is synchronous how fast is it moving through space,taking the period to be 24hrs and M_e =mass of the satellite and is equal to 5.98×10^(24) kg


Question and Answers Forum

All Questions Topic List
Gravitation Questions
Previous in All Question Next in All Question
Next in Gravitation
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

	$t h a n k s b o s s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum