Question

A man can jump vertically to a height of $1.5m$ on the earth. Calculate the radius of a planet of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is $6.41\times {10}^{6}m$.

Answer 1

$h=\frac{u^2}{2g_e}$
$\therefore$ $u=\sqrt{2g_{e}h}$ (i)
For the asked planet this u sbould be equal to the escape velocity from its surface.
$\sqrt{2g_{e}h}=\sqrt{2g_p{R}_p}$
or $g_{e}h=g_p{R}_p$
$\frac{G{M}_e}{R_e^2}.h=\frac{G{M}_p}{R_p^2}{R}_p$
or $\frac{\left(\frac{4}{3}\pi R_e^3\right)\rho h}{R_e^2}=\frac{\left(\frac{4}{3}\pi R_p^3\right)\rho R_p}{R_p^2}$
or $R_p=\sqrt{R_{e}h}$
$=\sqrt{(6.41\times {10}^6)\left(1.5\right)}$
$=3.1\times {10}^3$ m