Question

The maximum vertical distance through which a full dressed astronaut can jump on the Earth is $0.5\text{m}$. The maximum vertical distance through which astronaut can jump on the Moon, which has a mean density ${\left(\frac{2}{3}\right)}^{rd}$ that of the earth and radius one quarter that of the earth is (in $m$)

• A. 3.0
• B. 3
• C. 3.00

Answer 1

Answer: (A), (B), (C)

The acceleration due to gravity at the surface of a planet is given by
$$g=\frac{GM}{R^2}$$$\Rightarrow g=\frac{G}{R^2}\times \frac{4}{3}\pi R^3\rho [\because M=\frac{4}{3}\pi R^3\rho ]$

$\Rightarrow g=\frac{4}{3}\pi G\rho R$

Hence,$$g\propto \rho R$$Let, $g_m$ be the accelaration due to gravity at the surface of the moon and $g_e$ the accelaration due to gravity at the surface of the earth.

$\Rightarrow \frac{g_m}{g_e}=\frac{{\rho }_m}{{\rho }_e}\frac{R_m}{R_e}$

According to the given problem,

$\frac{{\rho }_m}{{\rho }_e}=\frac{2}{3}$ and $\frac{R_m}{R_e}=\frac{1}{4}$

$\Rightarrow \frac{g_m}{g_e}=\left(\frac{2}{3}\right)\left(\frac{1}{4}\right)$

$\Rightarrow g_m=\frac{g_e}{6}$

Since the astronaut has energy to jump only upto $h_e=0.5\text{m}$ on earth which will be equal to the work done by him to jump $0.5\text{m}$.

Since, he will do the same work on the both Earth and planet. So,

$m{g}_e{h}_e=m{g}_p{h}_p$

Substituting the values, in the above equation we get,

$g_e\times 0.5=\frac{g_e}{6}\times h_p$

$\therefore h_p=3\text{m}$