Question:
A body of mass m is taken from earth surface to a height h equal to twice the radius of earth, the increase in potential energy will be (\(g\) = acceleration due to gravity on earth's surface) (R - radius of earth)
A body of mass m is taken from earth surface to a height h equal to twice the radius of earth, the increase in potential energy will be (\(g\) = acceleration due to gravity on earth's surface) (R - radius of earth)
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Use U = -GMm/r and compare the values at R and 3R.
Updated On: Oct 1, 2026
- \(\frac{2}{3}\text{mgR}\)
- \(\frac{1}{3}\text{mgR}\)
- \(\frac{1}{2}\text{mgR}\)
- \(3\text{mgR}\)
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The Correct Option is A
Solution and Explanation
Step 1: Potential energy
At distance \(r\) from the centre, \(U = -\frac{GMm}{r}\). Also \(GM = gR^2\).
Step 2: Initial and final
At the surface \(r=R\): \(U_i = -mgR\). At height \(h=2R\), \(r=3R\): \(U_f = -\frac{mgR}{3}\).
Step 3: Increase
\[ \Delta U = -\frac{mgR}{3} + mgR = \frac23mgR \]
Option (A). Using \(mgh = 2mgR\) is only valid for small heights.
Final Answer:
The increase in potential energy is 2/3 mgR. \[ \boxed{\text{(A)}\ \frac23mgR} \]
At distance \(r\) from the centre, \(U = -\frac{GMm}{r}\). Also \(GM = gR^2\).
Step 2: Initial and final
At the surface \(r=R\): \(U_i = -mgR\). At height \(h=2R\), \(r=3R\): \(U_f = -\frac{mgR}{3}\).
Step 3: Increase
\[ \Delta U = -\frac{mgR}{3} + mgR = \frac23mgR \]
Option (A). Using \(mgh = 2mgR\) is only valid for small heights.
Final Answer:
The increase in potential energy is 2/3 mgR. \[ \boxed{\text{(A)}\ \frac23mgR} \]
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