1.

An exploratory rocket of mass m is in orbit about the sun at a 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 reducing its mass by half (due to fuel consumption). Immediately after the burn, what is the kinetic energy of the rocket?

Answer»

`10GM_(S)m//R_(ES)`
`5GM_(S)m//R_(ES)`
`20GM_(S)m//R_(ES)`
`GM_(S)m//2R_(ES)`

Solution :`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))`


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