Question

Ravi can throw a ball at a speed on the earth which can cross a river of width 10 m. Ravi reaches on a imaginary planet whose mean density is twice that of the earth. The maximum possible radius of the planet (in km)so that if Ravi throws the ball at the same speed it may escape from the planet is
Given radius of the earth = $6.4\times {10}^{6}m$

Answer 1

Let ${\rho }_e,{\rho }_p$ be the density of Earth and the imaginary planet respectively.

For the ball at Earth to just cross the river, it must be thrown at angle of 45${}^{\circ }$ to horizontal.

The range of the projectile becomes
$R=\frac{v_0}{\sqrt{2}}\left(2\frac{v_0}{\sqrt{2}g}\right)=\frac{v_0^2}{g}=10m$

Hence $v_0^2=10g=10\frac{GM}{R^2}=10\left(G{\rho }_e\frac{4}{3}\pi R_e\right)$........................Equation 1

Also if the the ball thrown at $v_0$ just escapes the gravitational pull of the imaginary planet,

$\frac{1}{2}m{v}_{0^2}=0-(-\frac{G{M}_{p}m}{R_p})$

Hence, $v_0^2=2\left(G{\rho }_p\frac{4}{3}\pi \right)R_p^2$................................... Equation 2

Using Equation 1 and Equation 2,

$\frac{R_p^2}{R_e}=\frac{5}{2}$
Hence, $R_p=4km$