If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass m raised from surface of the earth to a height equal to radius R of the earth is - [M = mass of earth]

Options

A. 2mgR
B. mgR
C. `1/2` mgR
D. `1/4` mgR

Answer

C

Explanation

The potential energy of an object at the surface of the earth
`U_(1)=-(GMm)/(R)" "......(i)`
The potential energy of the object at a height h=R from the surface of the earth
`U_(2)=-(GMm)/(R+h)=-(GMm)/(R+R)" "......(ii)`
Hence, the gain in potential energy of the object
`DeltaU=U_(2)-U_(1)`
`DeltaU=-(GMm)/(R+R)+(GMm)/(R)`
`DeltaU=-(GMm)/(2R)+(GMm)/(R)`
`DeltaU=(1)/(2)(GMm)/(R)`
But we know that `GM=gR^(2)`
Hence, `DeltaU=(1)/(2)(gR^(2)m)/(R)`
or `DeltaU=(1)/(2)gRm`
or `DeltaU=(1)/(2)mgR`