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Law of Floatation

Find the rela...

Question

Find the relation between the acceleration due to gravity on the surfaces of two planets $A$ and $B$ of masses $m_{A}$ and $m_{B}$ and radii $R_{A}$ and $R_{B}$ respectively. If they have same densities.

\frac{g_{A}}{g_{B}} = \frac{R_{B}}{R_{A}}

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\frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}}

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\frac{g_{A}}{g_{B}} = \frac{R_{B}^{2}}{R_{A}^{2}}

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\frac{g_{A}}{g_{B}} = \frac{R_{A}^{2}}{R_{B}^{2}}

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Solution

The correct option is B $\frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}}$

Let $ρ$ is the density of both the planets.

Mass of planet $A$ :

$m_{A} = \frac{4}{3} π R_{A}^{3} ρ$

So, acceleration due to gravity on the surface of planet $A$ :

$g_{A} = \frac{G m_{A}}{R_{A}^{2}}$

$\Rightarrow g_{A} = \frac{G \times \frac{4}{3} π R_{A}^{3} ρ}{R_{A}^{2}}$

$\Rightarrow g_{A} = \frac{4}{3} π G R_{A} ρ$

Similarly, acceleration due to gravity on the surface of planet $B$ :

$\Rightarrow g_{B} = \frac{4}{3} π G R_{B} ρ$

So, the ratio of the acceleration due to gravity is,

$\frac{g_{A}}{g_{B}} = \frac{\frac{4}{3} π G_{A} ρ}{\frac{4}{3} π G R_{B} ρ}$

$∴ \frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}}$

Hence, option (b) is the correct answer.

Why this Question?
To understand the concept of acceleration due to gravity due to continuous mass distribution.

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The ratio of the acceleration due to gravity $\frac{g_{1}}{g_{2}}$ at the surface of the two planets is:

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