Question

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

• A. $9$km.
• B. $4$km.
• C. $48$km.
• D. $41$km.

Answer 1

The correct option is B $4$km.

Let $p_e,p_1$ be the density of earth and the imaging planet respectively. For ball at earth to just cross river it must be thrown at angle ${45}^0$ with horizontal range of projectile becomes

$R=\frac{V_0}{\sqrt{2}}\left(2\frac{V_0}{\sqrt{2}g}\right)=\frac{V_0^2}{g}=10m$

Hence $V_0^2=10g=10\frac{GM}{R^2}=10\left(G_p\frac{4}{3}\pi R_e\right)⟶\left(1\right)$

Also if ball is thrown at $V_0$ it escape gravitational pull of imaging planet

$\frac{1}{2}{mV}_0^2=0-(-\frac{{GM}_{p}m}{R_p})$

Hence $V_0^2=2\left(G_{pp}\frac{4}{3}\pi \right)R_p^2⟶\left(2\right)$

Using eq(1) and eq(2)

$\frac{R_p^2}{R_e}=\frac{5}{2}$

$R_p=4km$.