A body of mass `m` rises to a height `h=R//5` from the earth's surface where `R` is earth's radius. If `g` is acceleration due to gravity at the earth's surface, the increase in potential energy is

Options

A. mgh
B. `4/5` mgh
C. `5/6` mgh
D. `6/7` mgh

Answer

C

Explanation

`{:(DeltaU =U_(F) -U_(i)),(DeltaU=U_(n )-U _(h)):}|{:(because U_(h) =(-GMm )/((R+h))),(U_(h) =(-GMm)/((R+R//5))=-(5GMm)/(6R)):}`
`=(-(5)/(6) (GMm)/(R))-((-GMm)/(R))|because U_(s) =-(GMm)/(R)`
`=(GMm)/(6R) (because GM =gR^(2))`
`=(mgR)/(6) =5/6 mgh " "(because R= 5h)`