If two planets have their radii in the ratio $x: y$ and densities in the ratio $m: n$, then the acceleration…
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
- $\mathrm{ny} / \mathrm{mx}$
- $\mathrm{my} / \mathrm{nx}$
- $\mathrm{nx} / \mathrm{my}$
- $\mathrm{mx} / \mathrm{ny}$
Solution
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,
$\frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\mathrm{R}_{\mathrm{l}}^{\prime}}{\mathrm{R}_2^{\prime}} \times \frac{\mathrm{d}_1^{\prime}}{\mathrm{d}_2}$
$\therefore \quad \frac{\mathrm{g}_1}{\mathrm{~g}_2}=\frac{\mathrm{xm}}{\mathrm{yn}}$
Asked in: MHT CET 2023 (13 May Shift 2)
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