A rocket is launched vertically from the surface of the earth with an initial velocity v. How far above the surface of earth will it go? Neglect the air resistance. (where R is the radius of the earth )

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 earth ,
Total energy = Kinetic energy + Potential energy
`=1/2mv^2-(GmM)/R`
At the highest point, v=0,
Potential energy = `-(GmM)/((R+h))`
where h is the maximum height.
According to the law of conservation of mechanical energy,
we get
`1/2mv^2-(GmM)/R=-(GmM)/(R+h),1/2v^2=(GMh)/((R)(R+h))`
`1/2v^2=(gRh)/(R+h) " " (because g=(GM)/g^2)`
`(R+h)/h=(2gR)/v^2, h=R((2gR)/v^2-1)^(-1)`