If a body of mass m has to be taken from the surface to the earth to a height h=R, then the amount of energy required is (R= radius of the earth)

Options

A. mgR
B. `(mgR)/(3)`
C. `(mgR)/(2)`
D. `(mgR)/(12)`

Answer

C

Explanation

We know that,
Gravitational potential energy
`U =- ( GM _(e) m )/(R) " "...(i)`
and gravitational kinetic energy
`K = (1)/(2). ( GM _(e) m )/(R) " "...(ii)`
`therefore ` Total energy of a body is
`E = U + K`
`=- ( GM _(e) m )/(R) + ( GM _(e) m )/( 2 R)`
`=- ( 2 GM _(e) m + GM _(e) m )/(2R) =- (GM _(e) m )/( 2R)`
But, acceleration due to gravity (g) in terms of graviational constant (G) is
`g = (GM _(e))/( R ^(2)) " "...(iii)`
`therefore - (GM _(e))/( R ^(2)) xx R ^(2) xx (m)/( 2R) `[From Eq. (iii)]
`= g xx R ^(2) xx (m)/( 2R)`
(Concelation of negative sign, because energy cen never be negative)
`= (gmR)/( 2) = ( mgR)/(2)`