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. `(3)/(4)mgh`
C. `(4)/(5)mgh`
D. `(5)/(6)mgh`

Answer

D

Explanation

The change in gravitational P.E.,
`Delta V` = P.E. at a height `(R )/(5)`
- P.E. at the surface of the earth
`=-(GMm)/(R+(R )/(5))-(-(GMm)/(R ))`
`=(GMm)/(R )-(5GMm)/(6R)=(1)/(6)(GMm)/(R )`
But `GM=gR^(2)`
`therefore Delta v = (1)/(6).(gR^(2).m)/(R )=(1)/(6)gmR`
But `R=5h`
`therefore Delta v =(1)/(6)gm xx 5h=(5)/(6)(mgh)`