Question:
The ratio between the gravitational potential energies of 1 kg of mass on the surface of two planets having masses and radii in the ratio \(1 : 2\) and \(1 : 3\) respectively, is
The ratio between the gravitational potential energies of 1 kg of mass on the surface of two planets having masses and radii in the ratio \(1 : 2\) and \(1 : 3\) respectively, is
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Gravitational potential energy on surface \(U \propto \frac{M}{R}\).
Updated On: Apr 24, 2026
- \(1:1\)
- \(2:3\)
- \(3:2\)
- \(9:4\)
- \(4:9\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Gravitational potential energy \(U = -\frac{GMm}{R}\). For 1 kg mass, \(U = -\frac{GM}{R}\).
Step 2: Detailed Explanation:
\(M_1 : M_2 = 1 : 2\) and \(R_1 : R_2 = 1 : 3\)
\(\frac{U_1}{U_2} = \frac{M_1/R_1}{M_2/R_2} = \frac{M_1}{M_2} \times \frac{R_2}{R_1} = \frac{1}{2} \times \frac{3}{1} = \frac{3}{2}\)
Ratio = \(3 : 2\)
Step 3: Final Answer:
The ratio is \(3:2\).
Gravitational potential energy \(U = -\frac{GMm}{R}\). For 1 kg mass, \(U = -\frac{GM}{R}\).
Step 2: Detailed Explanation:
\(M_1 : M_2 = 1 : 2\) and \(R_1 : R_2 = 1 : 3\)
\(\frac{U_1}{U_2} = \frac{M_1/R_1}{M_2/R_2} = \frac{M_1}{M_2} \times \frac{R_2}{R_1} = \frac{1}{2} \times \frac{3}{1} = \frac{3}{2}\)
Ratio = \(3 : 2\)
Step 3: Final Answer:
The ratio is \(3:2\).
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