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 $1.5m$ 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, $h_e=1.5m$
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 -
$g=\frac{G{M}_e}{R^2}$
$g=\frac{G\times \rho \times \frac{4}{3}\pi R^3}{R^2}$
$g=G\times \rho \times \frac{4}{3}\pi R$
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,
$mgh=m{g}^{\prime }{h}^{\prime }$
i.e. $h^{\prime }=h\frac{g}{g^{\prime }}$
Putting values of $g$ and $g$ ', we get
$h^{\prime }=h\frac{G\times \rho \times \frac{4}{3}\pi R}{G\times {\rho }^{\prime }\times \frac{4}{3}\pi R^{\prime }}$
$h^{\prime }=h\times \frac{\rho }{{\rho }^{\prime }}\times \frac{R}{R^{\prime }}$
$h^{\prime }=1.5\times \frac{4}{5}\times \frac{2}{3}$
$h=0.8m$
Therefore, the astronaut can jump to max height of $o.8m$.