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

To solve this problem, we need to determine the velocity of a body when it strikes the surface of the Earth after being released from a height equal to the radius of the Earth, denoted as \( R \).

Start by considering the potential energy and kinetic energy concepts. At the height of release, the potential energy of the body is at a maximum, while the kinetic energy is zero since the body is at rest.

Initially, the gravitational potential energy \( U \) at height \( R \) above the Earth's surface is:

\(U_i = -\frac{G M m}{2R}\)

where:

• \(G\) is the gravitational constant,
• \(M\) is the mass of the Earth,
• \(m\) is the mass of the body.

At the surface of the Earth, the potential energy is:

\(U_f = -\frac{G M m}{R}\)

The change in potential energy as the body falls is:

\(\Delta U = U_f - U_i = -\frac{G M m}{R} + \frac{G M m}{2R} = -\frac{G M m}{2R}\)

When the body strikes the Earth, all potential energy change is converted into kinetic energy:

\(elta K = \frac{1}{2} m v^2\), where \( v \) is the velocity of the body at the surface of the Eart\)

Equating the change in potential energy to the change in kinetic energy, we have:

\(\frac{1}{2} m v^2 = -\Delta U = \frac{G M m}{2R}\)

Solving for \( v \), we find:

\(v^2 = \frac{G M}{R}\)

We know that the acceleration due to gravity \( g \) is given by:

\(g = \frac{G M}{R^2}\)

Thus, substituting \( G M = g R^2 \) into the equation for \( v^2 \), we get:

\(v^2 = g R\)

Therefore, the velocity \( v \) is:

\(v = \sqrt{gR}\)

This matches the option: \(\sqrt{gR}\), confirming it as the correct answer.