Two planets of radii $r_{1}$ and $r_{2}$ are made from the same material. The ratio of the acceleration due to gravity $\frac{g_{1}}{g_{2}}$ at the surface of the two planets is:

$\frac{r_{1}}{r_{2}}$

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${(\frac{r_{1}}{r_{2}})}^{2}$

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$\frac{r_{2}}{r_{1}}$

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${(\frac{r_{2}}{r_{1}})}^{2}$

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Solution

The correct option is A
$\frac{r_{1}}{r_{2}}$

Since both the planets are made of the same material, the density of both the planets would be the same.
We know, mass of the planet = volume $\times$ density
Let, $ρ$ be the density of the material.
Volume = $\frac{4}{3}$ $π$ $r^{3}$ , where 'r' is the radius of the planet.
We know, $F = m g = \frac{G M m}{r^{2}}$ $\Rightarrow g = \frac{G M}{r^{2}}$ $= \frac{G \frac{4}{3} π r^{3} ρ}{r^{2}}$
Let radius of first planet be $r_{1}$ and acceleration due to gravity be $g_{1}$
$\Rightarrow$ $g_{1} = G \times \frac{\frac{4}{3} π {r_{1}}^{3} ρ}{{r_{1}}^{2}} = \frac{4}{3} π G r_{1} ρ$ and
for the second planet, let the radius be $r_{2}$ and acceleration due to gravity be $g_{2}$
$\Rightarrow$ $g_{2} = G \times \frac{\frac{4}{3} π {r_{2}}^{3} ρ}{{r_{2}}^{2}} = \frac{4}{3} π G r_{2} ρ$
On dividing $g_{1}$ by $g_{2}$ , we get,
$\frac{g_{1}}{g_{2}} = \frac{r_{1}}{r_{2}}$

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