Question

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)

Answer 1

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