Question

The ratio of radii of the 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 a maximum height of $1.5m$ on the earth, with the same effort, the maximum height he can jump on the planet is

• A. $1m$
• B. $0.8m$
• C. $0.5m$
• D. $1.25m$

Answer 1

The correct option is B $0.8m$

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 0.8 m.

The correct option is B.