Two satellites of same mass are launched in circular orbits at height of 2R and 3R respectively. The ration of kinetic energies is

Options

A. `2:3`
B. `1:1`
C. `3:2`
D. `1:3`

Answer

C

Explanation

Kinetic energy is given by
`KE_(1)=(1)/(2)mv^(2)=(1)/(2)=(1)/(2)m[(GM)/(2R)]....(i)`
`KE_(2)=(1)/(2)m[(GM)/(3R)]....(ii)`
On dividing Eq.(i) by Eq.(ii) , we get
`(KE_1)/(KE_2)=(3)/(2)`