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(Take the planets to have the same mass density)

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Verified by Toppr

Since mass of the object remains same

$$\therefore$$ Weight of object will be proportional to $$'g'$$ (acceleration due to gravity)

Given

$$\dfrac{W_{earth}}{W_{planar}} = \dfrac{9}{4} \dfrac{g_{earth}}{g_{planet}}$$

Also, $$g_{surface} = \dfrac{GM}{R^2}$$ ($$M$$ is mass planet, $$G$$ is universal gravitational constant, $$R$$ is radius of planet)

$$\therefore \dfrac{9}{4} = \dfrac{GM_{earth} R^2_{planet}}{GM_{planet} R^2_{earth}} = \dfrac{M_{earth}}{M_{planet}} \times \dfrac{R^2_{planet}}{R^2_{earth}} = 9\dfrac{R^2_{planet}}{R^2_{earth}}$$

$$\therefore R_{planet} = \dfrac{R_{earth}}{2} = \dfrac{R}{2}$$

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