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# 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 $$\dfrac{1}{9}th$$ 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)

A
$$\dfrac{R}{4}$$
B
$$\dfrac{R}{3}$$
C
$$\dfrac{R}{2}$$
D
$$\dfrac{R}{9}$$
Solution
Verified by Toppr

#### Correct option is B. $$\dfrac{R}{2}$$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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