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Question

Physics Question on Gravitation

A satellite is revolving in a circular orbit at a height ' hh' from the earth's surface (radius of earth R;h<<RR ; 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 : (Neglect the effect of atmosphere.)

A

2gR\sqrt{2\,gR}

B

gR\sqrt{g\,R}

C

gR/2\sqrt{g\,R / 2 }

D

gR(21)\sqrt{gR} ( \sqrt{2} - 1 )

Answer

gR(21)\sqrt{gR} ( \sqrt{2} - 1 )

Explanation

Solution

Orbital velocity v=GMR+h=GMRv = \sqrt{\frac{GM}{R + h}} = \sqrt{\frac{GM}{R}} as h < < R
Velocity required to escape
12mv2=GMmR+h;v=2GMR+h=2GMR(h<<R)\frac{1}{2} mv'^{2} = \frac{GMm}{R+h} ; v' = \sqrt{\frac{2GM}{R+h}} = \sqrt{\frac{2GM}{R}} \left(h < < R\right)
\therefore Increase in velocity
vv=2GMRGMR=2gRgR=gR(21)v' - v = \sqrt{\frac{2GM}{R}} - \sqrt{\frac{GM}{R}} = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR} \left( \sqrt{2} - 1 \right)