The ratio of the weights of a body on the earth’s surface to that on the surface of a planet is 9:4. The mass of the planet is $\frac{1}{9} t h$ of that of the earth, If R is the radius of the earth. What is the radius of the planet? (Take the planets to have the same mass density)

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$\frac{R}{4}$

$\frac{R}{2}$

$\frac{R}{3}$

$\frac{R}{9}$

answer is D.

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Detailed Solution

Mass of object remains same

Weight of object $α$ acceleration due to gravity

$\frac{W_{(e a r t h)}}{W_{(p l a n e t)}} = \frac{9}{4} = \frac{g_{(e a r t h)}}{g_{(p l a n e t)}}$

$∴ \frac{9}{4} = \frac{G M_{(e a r t h)}}{G M_{(p l a n e t)}} \times \frac{{R^{2}}_{(p l a n e t)}}{{R^{2}}_{(e a r t h)}} = \frac{M_{(e a r t h)}}{M_{(p l a n e t)}} \times \frac{{R^{2}}_{(p l a n e t)}}{R^{2} (e a r t h)}$

$= 9 \times \frac{{R^{2}}_{(p l a n e t)}}{{R^{2}}_{(e a r t h)}} ∴ {R^{2}}_{(p l a n e t)} = \frac{{R^{2}}_{(e a r t h)}}{2} = \frac{R}{2}$

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