Question

A body of mass \( m \) rises to a height \( h = \frac{R}{5} \) from the earth's surface, where \( R \) is the earth’s radius. If \( g \) is acceleration due to gravity at the earth’s surface, then the increase in potential energy is:

• A. \( mgh \)
• B. \( \frac{4}{5} mgh \)
• C. \( \frac{5}{6} mgh \)
• D. \( \frac{6}{7} mgh \)

Hint:

For large heights comparable to Earth's radius, always use \[ \Delta U = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right) \] instead of \( mgh \), which is only valid for small heights.

Answer 1

The Correct Option is C

To find the increase in potential energy when the body rises to a height of \( h = \frac{R}{5} \) from the Earth's surface, we start by considering the gravitational potential energy formula and the change in gravitational potential energy with respect to height from the Earth's surface.

The gravitational potential energy at height \( h \) is given by:

\(U = -\frac{GMm}{R + h}\)

where:

• \(G\) is the gravitational constant.
• \(M\) is the mass of the Earth.
• \(m\) is the mass of the body.
• \(R\) is the radius of the Earth.
• \(h\) is the height above the Earth's surface.

The increase in potential energy as the body moves from the Earth's surface to a height \( h \) is given by:

\(\Delta U = U(h) - U(0)\)

Using the potential energy formulas, this becomes:

\(\Delta U = -\frac{GMm}{R + h} + \frac{GMm}{R}\)

Substitute \( h = \frac{R}{5} \):

\(\Delta U = -\frac{GMm}{R + \frac{R}{5}} + \frac{GMm}{R} = -\frac{GMm}{\frac{6R}{5}} + \frac{GMm}{R}\)

Simplify the expression:

\(\Delta U = -\frac{5GMm}{6R} + \frac{GMm}{R} = \frac{GMm}{R} - \frac{5GMm}{6R}\)

Combine the fractions:

\(\Delta U = \frac{6GMm}{6R} - \frac{5GMm}{6R} = \frac{GMm}{6R}\)

At Earth's surface, \( g = \frac{GM}{R^2} \). Therefore, \( GM = gR^2 \).

Substituting back in the equation:

\(\Delta U = \frac{(gR^2)m}{6R} = \frac{gmR}{6}\)

Since \( h = \frac{R}{5} \), we have:

\(mgh = mg \left(\frac{R}{5}\right) = \frac{mgR}{5}\)

Finally, the increase in potential energy as a fraction of \(mgh\) is:

\(\frac{\Delta U}{mgh} = \frac{\frac{gmR}{6}}{\frac{mgR}{5}} = \frac{5}{6}\)

Therefore, the increase in potential energy is \(\frac{5}{6} mgh\).