When a body id lifted from surface of earth height equal to radius of earth, then the change in its potential energy is

Options

A. mgR
B. `(3)/(2) mg R`
C. `(mgR)/(2)`
D. `(mgR)/(4)`

Answer

C

Explanation

The gravitational potential energy of mass m in the gravitational field of mass M at a distance R from centre of earth, is
`U_(i)= -(GMm)/(R )`
If mass m is raised to a height R from earth's surface then final potential energy of mass m, is
`U_(f)=-(GMm)/(R+R)= -(GMm)/(2R)`
Hence, change in potential energy
`Delta U=U_(f)-U_(i)`
`=-(GMm)/(2R)-(-(GMm)/(R ))`
`=-(GMm)/(2R)+(GMm)/(R)`
`=(GMm)/(2R)`
Also, we known that
`g=(GM)/(R^(2))`
`rArr GM=gR^(2)`
Hence, `Delta U=(gR^(2)m)/(2R)=(mgR)/(2)`