Question

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

• A. mg2R
• B. ⅔ mgR
• C. 3mgR
• D. ⅓ mgR

Answer 1

The Correct Option is B

To determine the change in potential energy when a body is taken from the Earth's surface to a height equal to twice the Earth's radius, we can use the gravitational potential energy formula:

The gravitational potential energy at a distance r from the center of the Earth is given by:

U = -\frac{GMm}{r}

where:

• G is the gravitational constant,
• M is the mass of the Earth,
• m is the mass of the body,
• r is the distance from the center of the Earth to the body.

Initially, the body is on the surface of the Earth, so:

U_1 = -\frac{GMm}{R}

When the body is at a height equal to twice the radius of the Earth, the distance from the center of the Earth becomes 3R. Therefore, the potential energy at this height is:

U_2 = -\frac{GMm}{3R}

The change in potential energy (\Delta U) is the difference between the final potential energy U_2 and the initial potential energy U_1:

\Delta U = U_2 - U_1 = \left(-\frac{GMm}{3R}\right) - \left(-\frac{GMm}{R}\right)

Simplifying the expression yields:

\Delta U = -\frac{GMm}{3R} + \frac{GMm}{R}

\Delta U = \frac{GMm}{R} \left(1 - \frac{1}{3}\right)

\Delta U = \frac{GMm}{R} \times \frac{2}{3}

\Delta U = \frac{2}{3} \times \frac{GMm}{R}

Now, using the fact that the gravitational force mg = \frac{GMm}{R^2}, we can rewrite \frac{GMm}{R} as mgR. Thus, the change in potential energy is:

\Delta U = \frac{2}{3} mgR

Therefore, the correct answer is \frac{2}{3} mgR, which corresponds to the option ⅔ mgR.