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A man can throw a ball at a speed on the earth which can cross the river of width 10m. The man reaches on an imaginary planet whore mean density is twice that of 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 $= 6.4 \times 10^{6} m .$

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Detailed Solution

Let the speed of the ball which can cross 10 m wide river be V. Then,

$R = \frac{V^{2} \sin \ 2 \times 45)}{g} = 10 V = \sqrt{10} g$

Let the radius of planet be R_p then mass is

$M_{p} = \frac{4}{3} {πR}_{p}^{3} \times σ \times 2 = \frac{4}{3} {πR}_{p}^{3} \times \frac{2 \times M_{e}}{\frac{4}{3} {πR}_{e}^{3}}$

$M_{p} = \frac{2 M_{e} R_{p}^{3}}{R_{e}^{3}}$ Escape velocity of the planet

$V_{e} = \sqrt{\frac{2 {GM}_{p}}{Rp}} = 10 10 = \sqrt{\frac{2 G 2 M_{e} R_{p}^{2}}{R_{e}^{3}}} {10=2R}_{p} \sqrt{\frac{g}{R_{e}}} R_{p} = 4 Km$

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