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

Moderate

Solution

Let the speed of the ball which can cross 10 m wide river be V.
$Then R = \frac{V^{2} \sin (2 \times 45^{\circ})}{g} = 10, v = \sqrt{10 g}$
Let the radius of planet be ' R_p', then mass of the planet
$M_{p} = \frac{4}{3} {πR}_{p}^{3} \times 2 σ = \frac{4}{3} {πR}_{p}^{3} \times \frac{2 \times M_{e}}{4 / 3 {πR}_{e}^{R}} = \frac{2 M_{e} R_{p}^{3}}{R_{e}^{3}}$
Escape velocity of the Planet
$\begin{array}{l} V_{e} = \sqrt{\frac{2 {GM}_{p}}{R_{p}}} = \frac{2 G \times 2 \times M_{e} {R_{p}}^{3}}{R_{p} \times R_{e}^{3}} = \sqrt{\frac{10 {GM}_{e}}{R_{e}^{2}}} \\ 2 R_{p} = \sqrt{10 R_{e}} \Rightarrow 2 R_{p} = \sqrt{10 \times 6 .4 \times 10^{6}} \\ \Rightarrow R_{p} = \frac{8 \times 10^{3}}{2} = 4 \times 10^{3} = 4 km \end{array}$

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