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Q.

An artificial satellite is revolving around the earth in a circular orbit. Its velocity is one third of the escape velocity. Its height from the earth’s surface is (in km)

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a

3200

b

1600

c

22400

d

12800

answer is A.

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Detailed Solution

$v = \frac{v_{e}}{3} \sqrt{\frac{G M}{R + h}} = \frac{1}{3} \sqrt{\frac{2 G M}{R}} S q u a r i n g o n b o t h s i d e s w e g e t \Rightarrow \frac{G M}{R + h} = \frac{1}{9} \frac{2 G M}{R} \Rightarrow 9 R = 2 R + 2 h \Rightarrow 7 R = 2 h \Rightarrow h = \frac{7 R}{2} = \frac{7 \times 6400 k m}{2} = 22400 k m$

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