Question

A body weighs $900$N on the earth. Find its weight on a planet whose density is ${\frac{1}{3}}^{st}$ the density of earth and radius is ${\frac{1}{4}}^{th}$ that of the earth.

• A. $75$N
• B. $500$N
• C. $62$N
• D. $320$N

Answer 1

Answer: (A)

$75$N

Weight of object on earth $W=\frac{GMm}{R^2}$

Now, $M=\rho V$ =$\frac{4\pi \rho R^3}{3}$

$\therefore W=\frac{G\rho \pi R^{3}Gm}{3R^2}=\frac{4\pi \rho RGm}{3}$

$\frac{W_P}{W_E}=\left(\frac{{\rho }_P}{{\rho }_E}\right)\left(\frac{R_P}{R_E}\right)$

$=\frac{1}{3}\times \frac{1}{4}$ = $\frac{1}{12}$

$W_P=\frac{900}{12}=75N$