Q. 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 $\frac{3}{4}^{t h}$ that of earth and radius is one sixth of the earth. Calculate the value of $x?$
NTA AbhyasNTA Abhyas 2022
Solution:
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 $ $\rho ^{'}=\frac{3}{4}\rho \Rightarrow \frac{\rho }{\rho ^{'}}=\frac{4}{3}$ radius: R $R^{'}=\frac{R}{6}\Rightarrow \frac{R}{R^{'}}=6$ acceleration due to gravity: $\frac{4}{3}\pi GR\rho $ $\frac{4}{3}\pi GR^{'}\rho ^{'}$
As, Gain in PE is same in both case.
$mg^{'}h^{'}=mgh$
$\Rightarrow h^{'}=\frac{gh}{g^{'}}=\frac{\frac{4}{3} \pi GR \rho }{\frac{4}{3} \pi GR^{'} \rho ^{'}}\times h=\frac{R}{R^{'}}\times \frac{\rho }{\rho ^{'}}\times h$
$=6\times \frac{4}{3}\times 1.5=12\,m$
| Earth | Planet |
| $density:\rho $ | $\rho ^{'}=\frac{3}{4}\rho \Rightarrow \frac{\rho }{\rho ^{'}}=\frac{4}{3}$ |
| radius: R | $R^{'}=\frac{R}{6}\Rightarrow \frac{R}{R^{'}}=6$ |
| acceleration due to gravity: $\frac{4}{3}\pi GR\rho $ | $\frac{4}{3}\pi GR^{'}\rho ^{'}$ |