A body of mass 'm' taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of the body will be?

Let the mass of the body be 'm'.
Let the radius of the earth be 'R'.
The potential energy of a body of mass 'm' at a distance 'r' from the center of the earth is given by:
PE = -GMm/r
where G is the gravitational constant, M is the mass of the earth.
At the earth's surface (r = R), the potential energy is:
PE₁ = -GMm/R
At a height h = 2R above the earth's surface, the distance from the center of the earth is r = R + h = R + 2R = 3R.
The potential energy at this height is:
PE₂ = -GMm/3R
The change in potential energy is:
ΔPE = PE₂ - PE₁ = (-GMm/3R) - (-GMm/R) = GMm/R - GMm/3R = (2/3)(GMm/R)
Since GM/R² = g (acceleration due to gravity at the earth's surface),
GM/R = gR
Therefore,
ΔPE = (2/3)mgR
Thus, the change in potential energy is (2/3)mgR or 23mgR.