Question

The ratio of radii of earth to another planet is $ \frac{2}{3} $ and the ratio of their mean densities is $ \frac{4}{5} $ . If an astronaut can jump to an maximum height of $1.5\,m$ on the earth, with the same effort, the maximum height he can jump on the planet is

• A. 1 m
• B. 0.8 m
• C. 0.5 m
• D. 1.25 m
• E. 2 m

Answer 1

Answer: (B)

Given parameters:
$ \frac{ R }{ R ^{\prime}}=\frac{2}{3} $
$\frac{\rho}{\rho^{\prime}}=\frac{4}{5} $
Maximum height on the earth, $h _{ e }=1.5 m$
Mass of the earth will be:
$ M _{ e }=\rho \times V$
$M _{ e }=\rho \times \frac{4}{3} \pi R ^{3} $
Gravitational acceleration at earth's surface will be -
$ g =\frac{ GM _{ e }}{ R ^{2}} $
$ g =\frac{ G \times \rho \times \frac{4}{3} \pi R ^{3}}{ R ^{2}} $
$ g=G \times \rho \times \frac{4}{3} \pi R $
Similarly, gravitational acceleration at the given planet's surface will be:
$ g ^{\prime}= G \times \rho^{\prime} \times \frac{4}{3} \pi R ^{\prime} $
Let $E$ be common energy given to astronaut of mass $m$ on both planets. At the maximum height, this energy will be converted to potential energy.
So,
$ mgh = mg ^{\prime} h ^{\prime} $
i.e. $h ^{\prime}= h \frac{ g }{ g ^{\prime}} $
Putting values of $g$ and $g$ ', we get
$h ^{\prime}= h \frac{ G \times \rho \times \frac{4}{3} \pi R }{ G \times \rho^{\prime} \times \frac{4}{3} \pi R ^{\prime}} $
$h ^{\prime}= h \times \frac{\rho}{\rho^{\prime}} \times \frac{ R }{ R ^{\prime}} $
$h ^{\prime}=1.5 \times \frac{4}{5} \times \frac{2}{3} $
$h =0.8 m $
Therefore, the astronaut can jump to max height of $o .8 m$.