A body of mass \(m\) is taken from the surface of earth to a height equal to twice the radius of earth (\(R_e\)). The increase in potential energy will be _______.
(\(g\) is acceleration due to gravity at the surface of earth)

Initial potential energy of the body at the surface of the earth is given by:

\(U_i = -\dfrac{GMm}{R_e}\)

Final potential energy of the body at a height \(h = 2R_e\) from the surface of the earth is:

\(U_f = -\dfrac{GMm}{R_e + h} = -\dfrac{GMm}{R_e + 2R_e} = -\dfrac{GMm}{3R_e}\)

The increase in potential energy is:

\(\Delta U = U_f - U_i = -\dfrac{GMm}{3R_e} - \left(-\dfrac{GMm}{R_e}\right)\)

\(\Delta U = \dfrac{GMm}{R_e} \left(1 - \dfrac{1}{3}\right) = \dfrac{2GMm}{3R_e}\)

Acceleration due to gravity at the surface of the earth is \(g = \dfrac{GM}{R_e^2}\), which gives \(GM = gR_e^2\).

Substituting this value, we get:

\(\Delta U = \dfrac{2(gR_e^2)m}{3R_e} = \dfrac{2}{3}mgR_e\)

Answer: \(\dfrac{2}{3}mgR_e\)