The kinetic energy of a satellite in its orbit around the earth is...

The kinetic energy of a satellite in its orbit around the earth is E. What should be the kinetic energy of the satellite so as to enable it to escape from the gravitational pull of the earth?

Options

A $4 E$
B2 E
C $\sqrt{2} \mathrm{E}$
DE

Correct answer

B. 2 E

Step-by-step solution

We know that

v_e=\sqrt{2} v_0

, where

\mathrm{v}_0

is orbital velocity. K.E. in the orbit,

E=\frac{1}{2} M v_0{ }^2

K.E. to escape

E=\frac{1}{2} M v_e^2=\frac{1}{2} M\left(2 v_0{ }^2\right)=\frac{1}{2} M v_0^2 \times 2=2 E

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