The ratio of the weights of a body on the Ear | vedclass

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(D) Since the mass of the object remains constant,the weight of the object is proportional to the acceleration due to gravity $g$.Given: $\frac{W_{ear

Vedclass · Accepted Answer

$\frac{R}{2}$

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(D) Since the mass of the object remains constant,the weight of the object is proportional to the acceleration due to gravity $g$.Given: $\frac{W_{earth}}{W_{planet}} = \frac{9}{4} = \frac{g_{earth}}{g_{planet}}$.We know that $g = \frac{GM}{R^2}$. Since density $\rho$ is constant,$M = \rho \cdot \frac{4}{3}\pi R^3$,so $g = \frac{G \cdot \rho \cdot \frac{4}{3}\pi R^3}{R^2} = G \cdot \rho \cdot \frac{4}{3}\pi R \propto R$.However,the problem states the mass of the planet is $\frac{1}{9}$ of the Earth's mass. Let $M_p = \frac{M_e}{9}$.Using $g = \frac{GM}{R^2}$,we have $\frac{g_e}{g_p} = \frac{M_e}{M_p} \cdot \frac{R_p^2}{R_e^2} = 9 \cdot \frac{R_p^2}{R^2}$.Given $\frac{g_e}{g_p} = \frac{9}{4}$,so $\frac{9}{4} = 9 \cdot \frac{R_p^2}{R^2}$.$\frac{1}{4} = \frac{R_p^2}{R^2} \implies \frac{R_p}{R} = \frac{1}{2}$.Therefore,$R_p = \frac{R}{2}$.

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 $\frac{1}{9}^{th}$ of that of the Earth. If $R$ is the radius of the Earth,what is the radius of the planet? (Assume the planets have the same mass density)

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