The ratio of the weights of a body on Earth’s surface to that on the surface of a planet is 9 : 4 The mass of the planet is 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)

Weight = m G M R 2 ∴ 9 4 = G M e R e 2 × R p 2 G M p Given that, M p = M e 9 ∴ 9 4 = M e M e 9 × R p 2 R e 2 R p 2 = R e 2 × 1 4 R p = R e 2 As, R e = R ⇒ R p = R 2

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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceMedium2 minPYQ_2019

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The ratio of the weights of a body on 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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Answer:

Solution:

Weight $= m (\frac{G M}{R^{2}})$
$∴ \frac{9}{4} = \frac{G M_{e}}{R_{e}^{2}} \times \frac{R_{p}^{2}}{G M_{p}}$
Given that, $M_{p} = \frac{M_{e}}{9}$
$∴ \frac{9}{4} = \frac{M_{e}}{(\frac{M_{e}}{9})} \times \frac{R_{p}^{2}}{R_{e}^{2}}$
$R_{p}^{2} = R_{e}^{2} \times \frac{1}{4}$
$R_{p} = \frac{R_{e}}{2}$
As, $R_{e} = R$
$\Rightarrow R_{p} = \frac{R}{2}$

Stream:JEESubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2019

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