What is the increase in the potential energy of the body with mass m if the body is taken at the height h which is equal to the radius of the earth?

The correct option is (1/2) mgR.

To find the increase in the gravitational potential energy of a body of mass m when it is raised from the surface of the Earth to a height h equal to the radius of the Earth (R), we must use the general formula for gravitational potential energy.

The gravitational potential energy U of a system consisting of the Earth of mass M and a body of mass m at a distance r from the center of the Earth is given by the formula:

$$U=-\frac{GMm}{r}$$

where G is the universal gravitational constant.

Let's define the initial and final states of the body:
1. Initially, the body is on the surface of the Earth, so its distance from the center is r₁ = R.
Therefore, the initial potential energy U_i is:

$$U_i=-\frac{GMm}{R}$$

2. Finally, the body is taken to a height h = R above the surface. Its distance from the center of the Earth becomes r₂ = R + h = R + R = 2R.
Therefore, the final potential energy U_f is:

$$U_f=-\frac{GMm}{2R}$$

The increase in potential energy, ΔU, is the difference between the final potential energy and the initial potential energy:

$$\Delta U=U_f-U_i$$

Substituting the expressions for U_i and U_f:

$$\Delta U=-\frac{GMm}{2R}-\left(-\frac{GMm}{R}\right)$$

Simplifying the signs:

$$\Delta U=\frac{GMm}{R}-\frac{GMm}{2R}$$

Factoring out the common terms:

$$\Delta U=\frac{GMm}{R}\left(1-\frac{1}{2}\right)=\frac{1}{2}\frac{GMm}{R}$$

We know that the acceleration due to gravity on the surface of the Earth, g, is related to the gravitational constant G by the relation:
g = GM / R²
This can be rewritten as:
GM = gR²

Substituting GM = gR² into our expression for ΔU:

$$\Delta U=\frac{1}{2}\frac{\left(g{R}^2\right)m}{R}$$

Canceling one R from the numerator and denominator, we get:

$$\Delta U=\frac{1}{2}mgR$$

Thus, the increase in the potential energy of the body is indeed (1/2) mgR.