Question

A body is released from a height equal to the radius (r) of the earth. The velocity of the body when it strikes the surface of the earth will be:

• A. \(\sqrt{gR}\)
• B. \(\frac{\sqrt{gR}}{2}\)
• C. \(\sqrt{2gR}\)
• D. \(\sqrt{4gR}\)

Answer 1

The Correct Option is A

Solution:

Using the principle of energy conservation:

\( K_1 + U_1 = K_2 + U_2 \)

Where:

• \( K_1 \) is the initial kinetic energy,
• \( U_1 \) is the initial potential energy,
• \( K_2 \) is the final kinetic energy,
• \( U_2 \) is the final potential energy.

At the initial height, potential energy is:

\( U_1 = -\frac{GMm}{2R} \)

At the final height, the body strikes the surface, and the potential energy is:

\( U_2 = -\frac{GMm}{R} \)

The body is released, meaning its initial velocity is zero. Therefore, \( K_1 = 0 \).

Using conservation of energy:

\( 0 + \left(-\frac{GMm}{2R}\right) = \frac{1}{2}mv^2 + \left(-\frac{GMm}{R}\right) \)

Simplifying for velocity \( v \):

\( -\frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \)

\( \frac{GMm}{R} - \frac{GMm}{2R} = \frac{1}{2}mv^2 \)

\( \frac{GMm}{2R} = \frac{1}{2}mv^2 \)

\( \frac{GM}{R} = v^2 \)

\( v = \sqrt{\frac{GM}{R}} \)

Substitute \( g = \frac{GM}{R^2} \):

\( GM = gR^2 \)

\( v = \sqrt{\frac{gR^2}{R}} \)

\( v = \sqrt{gR} \)

Final Answer:

The velocity with which the body strikes the surface is \( v = \sqrt{gR} \).