The kinetic energy of a revolving satellite m a s s m at a height equal to thrice the radius of the earth R is

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The kinetic energy of a revolving satellite

(m a s s m)

at a height equal to thrice the radius of the earth

(R)

is

PhysicsGravitationMHT CETMHT CET 2019 (02 May Shift 1)

Options:

A $\frac{m g R}{8}$
B $\frac{m g R}{16}$
C $\frac{m g R}{2}$
D $\frac{m g R}{4}$

Solution:

1767 Upvotes Verified Answer

The correct answer is:

\frac{m g R}{8}

K \cdot E . = \frac{G M m}{2 (R + h)}

K \cdot E . = \frac{G M m}{2 (R + 3 R)} = \frac{G M m}{8 R} \times \frac{R}{R} = \frac{1 m .}{8} (\frac{G M m}{R^{2}}) \cdot R

K \cdot E . = \frac{m g R}{8} (∵ \frac{G M}{R^{2}} = g)

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