The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are $\rho$ and $\rho / 3$ respectively. The ratio of acceleration due to gravity at their surfaces $\left(g_{A}: g_{B}\right)$ will be:
Solution
The acceleration due to gravity at the surface of a planet can be expressed as:
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$g \propto \rho R$
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Now let's find the ratio of acceleration due to gravity at the surfaces of planets A and B:
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$\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B}$
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Given that the densities are $\rho$ and $\frac{\rho}{3}$ and the radii are $R$ and $4R$ for planets A and B, respectively, we have:
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$$\frac{g_A}{g_B} = \frac{\rho \cdot R}{\left(\frac{\rho}{3}\right) \cdot (4R)} = \frac{3}{4}$$
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So, the ratio of acceleration due to gravity at their surfaces is 3 : 4
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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