If g be the acceleration due to gravity of the earth's surface, the gain is 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. `1/2mgR`
B. `2mgR`
C. `mgR`
D. `1/4mgR`

Answer

A

Explanation

At a distance `x` from the centre of earth of gravitational force
`F=(GMm)/(x^(2))`
`dW=Fdx=(GMm)/(x^(2))dx`
We get `W=int_(R)^(R+h)(GMm)/(x^(2))dx=Gmm[1/R-1/(R+h)]`
`=gR^(2)m[1/R-1/(R+h)]`
Therefore the gain in potential energy is
`gR^(2)m[1/R-1/(2R)]=(mgR)/2`