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 )`
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)`
hence the gain in potential energy of the object
`triangleU=-(GMm)/(2R)+(GMm)/(R )`
`=1/2 . (GMm)/(R )`
But we know that `GM=gR^(2)`
hence `triangle U =1/2 (gR^(2)m)/(R )=1/2gRm=1/2mgR`