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 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:
Show 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.
Updated On: Apr 16, 2026
- \( mgh \)
- \( \frac{4}{5} mgh \)
- \( \frac{5}{6} mgh \)
- \( \frac{6}{7} mgh \)
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The Correct Option is C
Solution and Explanation
Concept:
Gravitational potential energy at a distance \( r \) from the center of Earth is:
\[
U = -\frac{GMm}{r}
\]
Thus, increase in potential energy when moving from \( r_1 \) to \( r_2 \) is:
\[
\Delta U = GMm \left( \frac{1}{r_1} - \frac{1}{r_2} \right)
\]
Also, \( g = \frac{GM}{R^2} \Rightarrow GM = gR^2 \).
Step 1: Identify initial and final positions.
Initial distance from Earth's center: \[ r_1 = R \] Final distance: \[ r_2 = R + \frac{R}{5} = \frac{6R}{5} \]
Step 2: Apply the potential energy formula.
\[ \Delta U = GMm \left( \frac{1}{R} - \frac{1}{\frac{6R}{5}} \right) \] \[ = GMm \left( \frac{1}{R} - \frac{5}{6R} \right) = GMm \cdot \frac{1}{6R} \]
Step 3: Substitute \( GM = gR^2 \).
\[ \Delta U = \frac{gR^2 m}{6R} = \frac{mgR}{6} \]
Step 4: Express in terms of \( mgh \).
Since \( h = \frac{R}{5} \Rightarrow R = 5h \), \[ \Delta U = \frac{mg(5h)}{6} = \frac{5}{6} mgh \]
Step 1: Identify initial and final positions.
Initial distance from Earth's center: \[ r_1 = R \] Final distance: \[ r_2 = R + \frac{R}{5} = \frac{6R}{5} \]
Step 2: Apply the potential energy formula.
\[ \Delta U = GMm \left( \frac{1}{R} - \frac{1}{\frac{6R}{5}} \right) \] \[ = GMm \left( \frac{1}{R} - \frac{5}{6R} \right) = GMm \cdot \frac{1}{6R} \]
Step 3: Substitute \( GM = gR^2 \).
\[ \Delta U = \frac{gR^2 m}{6R} = \frac{mgR}{6} \]
Step 4: Express in terms of \( mgh \).
Since \( h = \frac{R}{5} \Rightarrow R = 5h \), \[ \Delta U = \frac{mg(5h)}{6} = \frac{5}{6} mgh \]
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