Question

A man can jump upto $1.5,m$ height on Earth. He can jump on a planet to a height $x$ metre. If the density of planet is ${\frac{3}{4}}^{th}$ that of earth and radius is one sixth of the earth. Calculate the value of $x?$

Answer 1

Answer: 12

As $g=\frac{GM}{R^2}=\frac{G}{R^2}\times \left[\frac{4}{3}\pi R^3\rho \right]=\frac{4}{3}\pi GR\rho$

Earth	Planet
$density:\rho$	${\rho }^{{}^{\prime }}=\frac{3}{4}\rho \Rightarrow \frac{\rho }{{\rho }^{{}^{\prime }}}=\frac{4}{3}$
radius: R	$R^{{}^{\prime }}=\frac{R}{6}\Rightarrow \frac{R}{R^{{}^{\prime }}}=6$
acceleration due to gravity: $\frac{4}{3}\pi GR\rho$	$\frac{4}{3}\pi G{R}^{{}^{\prime }}{\rho }^{{}^{\prime }}$

As, Gain in PE is same in both case.
$m{g}^{{}^{\prime }}{h}^{{}^{\prime }}=mgh$
$\Rightarrow h^{{}^{\prime }}=\frac{gh}{g^{{}^{\prime }}}=\frac{\frac{4}{3}\pi GR\rho }{\frac{4}{3}\pi G{R}^{{}^{\prime }}{\rho }^{{}^{\prime }}}\times h=\frac{R}{R^{{}^{\prime }}}\times \frac{\rho }{{\rho }^{{}^{\prime }}}\times h$
$=6\times \frac{4}{3}\times 1.5=12m$