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

Options

A. `T^(-1)`
B. `T^(-2//3)`
C. `T^(-2)`
D. `T^(-1//3)`

Answer

B

Explanation

`KE =1/2 mv_0^2`
By Kepler.s third law, `T^2 prop r^3 " or " T^2 = Kr^3` ....(i)
As `v_0 = sqrt((GM)/( r))` where, r = radius of orbit
` therefore KE = 1/2 m (GM)/( r) = (GMm)/(2) 1/r = (GMm)/(2) ((K)/(T^2) )^(1//3) `
or `KE prop T^(-2//3)`