Question

An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth Its velocity when it strikes the earth's surface, ignoring air resistance, will be

• A. $\sqrt{g R}$
• B. $2 \sqrt{g R}$
• C. $\sqrt{2 g R}$
• D. $\sqrt{\frac{g R}{2}}$

Answer 1

The Correct Option is A

To determine the velocity of an object when it strikes the Earth's surface after falling from a height equal to the Earth's radius, we need to apply the principles of gravitational potential energy and kinetic energy in a gravitational field.

1. Initially, the object is at rest at a distance \(d = 2R\) from the center of the Earth, where \(R\) is the radius of the Earth. Its initial velocity is \(0\).
2. As the object falls due to gravity, its potential energy is converted into kinetic energy. The gravitational potential energy at height \(R\) from the Earth's surface (which is \(2R\) from the center of the Earth) is given by:

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

1. When the object reaches the Earth's surface, its potential energy is now:

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

1. The change in potential energy is converted into kinetic energy (\(K\)), thus:

\(K = \left( \frac{GMm}{2R} - \frac{GMm}{R} \right) = \frac{GMm}{2R}\)

1. The kinetic energy equation is given by:

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

1. Solving for the velocity \(v\), we get:

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

1. Using the relation where \(g = \frac{GM}{R^2}\), we substitute and simplify:

\(v^2 = gR\)

\(v = \sqrt{gR}\)

1. Thus, the velocity of the object when it strikes the Earth's surface is \(\sqrt{gR}\), which matches the correct answer provided.

The correct answer is: \(\sqrt{g R}\)