MHT CET · Physics · Gravitation
If two planets have their radii in the ratio \(x: y\) and densities in the ratio \(m: n\), then the acceleration due to gravity on them are in the ratio
- A \(\mathrm{ny} / \mathrm{mx}\)
- B \(\mathrm{my} / \mathrm{nx}\)
- C \(\mathrm{nx} / \mathrm{my}\)
- D \(\mathrm{mx} / \mathrm{ny}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{mx} / \mathrm{ny}\)
Step-by-step Solution
Detailed explanation
We know,
\(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\)
If the density is d, then
\(\Rightarrow M=\frac{4}{3} \pi R^3 d \quad \quad \quad \ldots(\because M=V \times d)\)
\(\mathrm{g} \leqslant \frac{4}{3} \pi \mathrm{GRd}\)
Given,
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{x}}{\mathrm{y}}\) and \(\frac{\mathrm{d}_1}{\mathrm{~d}_2}=\frac{\mathrm{m}}{\mathrm{n}}\)
\(\mathrm{g}_1=\frac{4}{3} \pi \mathrm{GR}_1 \mathrm{~d}_1\)
\(\mathrm{g}_2=\frac{4}{3} \pi \mathrm{GR}_2 \mathrm{~d}_2\)
Therefore,
\(\frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\mathrm{R}_{\mathrm{l}}^{\prime}}{\mathrm{R}_2^{\prime}} \times \frac{\mathrm{d}_1^{\prime}}{\mathrm{d}_2}\)
\(\therefore \quad \frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\mathrm{xm}}{\mathrm{yn}}\)
\(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}\)
If the density is d, then
\(\Rightarrow M=\frac{4}{3} \pi R^3 d \quad \quad \quad \ldots(\because M=V \times d)\)
\(\mathrm{g} \leqslant \frac{4}{3} \pi \mathrm{GRd}\)
Given,
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{x}}{\mathrm{y}}\) and \(\frac{\mathrm{d}_1}{\mathrm{~d}_2}=\frac{\mathrm{m}}{\mathrm{n}}\)
\(\mathrm{g}_1=\frac{4}{3} \pi \mathrm{GR}_1 \mathrm{~d}_1\)
\(\mathrm{g}_2=\frac{4}{3} \pi \mathrm{GR}_2 \mathrm{~d}_2\)
Therefore,
\(\frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\mathrm{R}_{\mathrm{l}}^{\prime}}{\mathrm{R}_2^{\prime}} \times \frac{\mathrm{d}_1^{\prime}}{\mathrm{d}_2}\)
\(\therefore \quad \frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\mathrm{xm}}{\mathrm{yn}}\)
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