# In a satellite if the time of revolution is T, then kinetic energy is proportional to

1T

1T2

1T3

T−2/3

## The correct Answer is:D

## v=√GMr⇒v2∝1r

Since, r3∝T2

r∝T2/3,v2∝T−2/3,∴KE∝T−23.

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