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My answer is coming 4 . but answer given is 2. Plz explain!
Q. Find the ratio of the kinetic energy required to be given to the satellite to escape earth's gravitational field to the KE required to be given so that the satellite moves in a circular orbit just above earth's atmosphere. 

Dear Student
E s c a p e space v e l o c i t y space v subscript e equals square root of fraction numerator 2 G M over denominator R end fraction end root O r b i t a l space v e l o c i t y space v subscript o equals square root of fraction numerator G M over denominator R end fraction end root K. E. space r e q u i r e d space f o r space s a t e l l i t e space t o space e s c a p e space e a r t h apostrophe s space g r a v i t a t i o n a l space f i e l d K. E. space subscript e equals 1 half m v subscript e superscript 2 K. E. space r e q u i r e d space f o r space s a t e l l i t e space t o space m o v e space i n space c i r c u a l r space o r b i t space j u s t space a b o v e space e a r t h apostrophe s space s u r f a c e K. E. subscript o space equals 1 half m v subscript o superscript 2 fraction numerator K. E. space subscript e over denominator K. E. subscript o end fraction equals fraction numerator v subscript e superscript 2 over denominator v subscript o superscript 2 end fraction equals 2
Regards

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The energy needed to raise the satellite to a height 'h' is
Δv=vA−vB
=−GMmR+h−(−GMmR)
=GMmhR(R+h)
=GMmhR2(1+hR)GMR2=g
=mgh(1+hR)=E1
E2= energy of the satellite
=12mv02(v0−orbital velocity )
r- distance of satellite from center of earth
=12m(GMr)
=12m(GMR+h)
=12m(GMR2)R(1+hR)
E2=mgR2(1+hR)
∴E1E2
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