If g is the acceleration due to gravity on earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

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) = (GMn)/(R+h) = -(GMm)/(R+R) …(ii)`
Hence, the gain in potential energy of the object
`triangleU = U_(2) -U_(1)`
`rArr triangleU = -(GMm)/(R+R) + (GMn)/(R )`
`rArr triangleU = -(GMm)/(2R) + (GMm)/(R )` ltbr. `rArr triangleU = (1)/(2)(GMm)/(R )`
BUt we know that, `GM = gR^(2)`
Hence, `triangleU = (1)/(2) (gR^(2)m)/(R )`
`rArr triangleU = (1)/(2)gRm rArr triangleU = (1)/(2)mgR`