Question

The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is:

• A. \(\mgR\)
• B. \(\2mgR\)
• C. \(\frac{1}{2}mgR\)
• D. \(\frac{3}{2}mgR\)

Answer 1

The Correct Option is C

To find the work done in raising a mass \( m \) from the surface of the Earth to a height \( h \) equal to the radius of the Earth \( R \), we can use the concept of gravitational potential energy. The formula for the gravitational potential energy difference when raising a mass in a gravitational field from one height to another is:

\(\Delta U = U_{\text{final}} - U_{\text{initial}}\)

The gravitational potential energy \( U \) at a height \( r \) from the center of the Earth is given by:

\( U = -\frac{G M m}{r} \)

Where:

• G is the gravitational constant.
• M is the mass of the Earth.
• m is the mass being raised.
• r is the distance from the center of the Earth.

Initially, the mass is at the Earth's surface, so the initial potential energy is:

\( U_{\text{initial}} = -\frac{G M m}{R} \)

Finally, the mass is at a height \( R \), i.e., at \( r = 2R \), so the final potential energy is:

\( U_{\text{final}} = -\frac{G M m}{2R} \)

The work done \( W \) to raise the mass is equal to the change in gravitational potential energy:

\( W = U_{\text{final}} - U_{\text{initial}} \)

Substitute the expressions for \( U_{\text{initial}} \) and \( U_{\text{final}} \):

\[ W = \left(-\frac{G M m}{2R}\right) - \left(-\frac{G M m}{R}\right) = \frac{G M m}{R} - \frac{G M m}{2R} = \frac{G M m}{2R} \]

Using the relation \( g = \frac{G M}{R^2} \) where \( g \) is the acceleration due to gravity at Earth's surface, we can rewrite \( \frac{G M m}{R} = mgR \). Thus, the work done is:

\( W = \frac{1}{2} mg R \)

Therefore, the correct answer is \(\frac{1}{2}mgR\).