An exploratory rocket of mass `m` is an orbit about the sun at radius of `R_(ES)//10` (one tenth of the radius of the earth's orbit about the sun). To exit this orbit, it fires its engine over a short period of time. This quickly doubles the velocity of the rocket while halfing its mass the burn, what is the kinetic energy of the rocket? take mass of sun as `M_(S)`

Options

A. `10GM_(S)m//R_(ES)`
B. `5GM_(S)m//R_(ES)`
C. `20GM_(S)m//R_(ES)`
D. `GM_(S)m//2R_(ES)`

Answer

A

Explanation

`KE=(1)/(2)mv^(2)=(GM_(s)m)/(2(R_(ES)//10))`
Final `KE = (1)/(2)((m)/(2))(2v)^(2)=mv^(2)=(10GMsm)/(R_(ES))`