Question:
The density of a planet is three times that of earth and radius \(2.5\) times that of earth. If \(g_p\) and \(g_e\) represents acceleration due to gravity on planet and earth respectively then ratio \(g_p\) to \(g_e\) is
The density of a planet is three times that of earth and radius \(2.5\) times that of earth. If \(g_p\) and \(g_e\) represents acceleration due to gravity on planet and earth respectively then ratio \(g_p\) to \(g_e\) is
Show Hint
g is proportional to density times radius.
Updated On: Oct 1, 2026
- \(7.5\)
- \(3.5\)
- \(6.5\)
- \(9.5\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
For a spherical planet, \(g=\dfrac{GM}{R^2}\) with \(M=\dfrac43\pi R^3\rho\). So \(g=\dfrac43\pi G\rho R\).
Step 2: Compare planet and earth:
\(\dfrac{g_p}{g_e}=\dfrac{\rho_p}{\rho_e}\times\dfrac{R_p}{R_e}=3\times2.5=7.5\).
Step 3: Why the other options are wrong.
3.5, 6.5 and 9.5 are not the product of 3 and 2.5. Using \(g\propto M/R^2\) with an incorrect cube would also give a different value.
Final Answer:
The ratio g_p / g_e is 7.5. \[ \boxed{\text{(A) }7.5} \]
For a spherical planet, \(g=\dfrac{GM}{R^2}\) with \(M=\dfrac43\pi R^3\rho\). So \(g=\dfrac43\pi G\rho R\).
Step 2: Compare planet and earth:
\(\dfrac{g_p}{g_e}=\dfrac{\rho_p}{\rho_e}\times\dfrac{R_p}{R_e}=3\times2.5=7.5\).
Step 3: Why the other options are wrong.
3.5, 6.5 and 9.5 are not the product of 3 and 2.5. Using \(g\propto M/R^2\) with an incorrect cube would also give a different value.
Final Answer:
The ratio g_p / g_e is 7.5. \[ \boxed{\text{(A) }7.5} \]
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