A rocket is fired vertically from the surface of the earth with a speed v. How far from the earth does the rocket go before returning to the earth ? (where `R_(E)` is the radius of the earth and g is acceleration due to gravity )

Options

A. `R((2gR)/(v^(2))-1)^(-1//2)`
B. `R((2gR)/(v^(2))-1)`
C. `R((2gR)/(v^(2))-1)^(-1)`
D. `R((2gR)/(v^(2))-1)^(2)`

Answer

C

Explanation

On the surface of the earth
Total energy `=`Kinetic energy `+` Potential energy
`=1/2mv^(3)(GmM)/R`
At the highest point `v=0` potential energy `=-(GmM)/(R+h)`
where `h` is the maximum height
`1/2mv^(2)-(GmM)/R=-(GmM)/(R+h)`
`1/2v^(2)-(gRh)/R+h` or `(R+h)/h=(2gR)/(v^(2))`
`h=R((2gR)/(v^(2))-1)`