A satellite is revolving in a circular orbit at a height lsquo-h rsquo- from the earth rsquo-s surface -radius of earth-R- h lt- lt-R- The minimum increase in its orbital velocity required- so that the satellite could-escape from the earth rsquo-s gravitational field- is close to -

Dear Student Orbital velocity of satellite revolving in circular orbit can be given as vo=GM/R+hEscape #160;velocity #160;required #160;to #160;escape ...

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Question

A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth
R; h <<R). The minimum increase in its orbital velocity required, so that the satellite could
escape from the earth’s gravitational field, is close to :

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Solution

Dear Student

Orbital velocity of satellite revolving in circular orbit can be given as
$v_{o} = \sqrt{G M / R + h} E s c a p e v e l o c i t y r e q u i r e d t o e s c a p e f r o m e a r t h' s g r a v i t a t i o n f i e l d i s g i v e n a s v_{e} = \sqrt{2 G M / R + h} S a t e l l i t e w i l l e s a c p e f r o m e a r t h' s g r a v i t a t i o n a l f i e l d f r o m h e i g h t h i f i t s o r b i t a l s p e e d i s e q u a l t o e s c a p e v e l o c i t y . L e t x i s t h e i n c r e a s e i n o r b i t a l v e c l o t i y s o t h a t i t e q u a l t o e s c a p e v e l o c i t y . \sqrt{G M / R + h} + x = \sqrt{2 G M / R + h} x = \sqrt{2 G M / R + h} - \sqrt{G M / R + h} x = \sqrt{G M / R + h} (\sqrt{2} - 1) A s h < < R x = \sqrt{G M / R} (\sqrt{2} - 1) = \sqrt{g R} (\sqrt{2} - 1)$
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A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R; h <<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to :

A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth
R; h <<R). The minimum increase in its orbital velocity required, so that the satellite could
escape from the earth’s gravitational field, is close to :