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(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Given that the mass of the planet $M_p = M_e$ (mass of Earth) and the rad

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$22.4$

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(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Given that the mass of the planet $M_p = M_e$ (mass of Earth) and the radius $R_p = \frac{R_e}{4}$ (where $R_e$ is the radius of Earth).The escape velocity for the planet is $v_p = \sqrt{\frac{2GM_p}{R_p}} = \sqrt{\frac{2GM_e}{R_e/4}}$.This simplifies to $v_p = \sqrt{4 	imes \frac{2GM_e}{R_e}} = 2 	imes \sqrt{\frac{2GM_e}{R_e}}$.Since the escape velocity of Earth is $v_e = 11.2 \ km/s$,the escape velocity for the planet is $v_p = 2 	imes 11.2 \ km/s = 22.4 \ km/s$.

For a planet having mass equal to the mass of the Earth but a radius that is one-fourth of the radius of the Earth,the escape velocity for this planet will be (in $km/s$): (in $km/s$)

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