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(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.For the Earth,$v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \, km/s$.For the Moo

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$2.5$

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(C) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.For the Earth,$v_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \, km/s$.For the Moon,the mass $M_m = \frac{M_e}{81}$ and the radius $R_m = \frac{R_e}{4}$.The escape velocity on the Moon is $v_m = \sqrt{\frac{2GM_m}{R_m}} = \sqrt{\frac{2G(M_e/81)}{(R_e/4)}} = \sqrt{\frac{2GM_e}{R_e} 	imes \frac{4}{81}}$.Substituting the values,$v_m = \sqrt{\frac{4}{81}} 	imes \sqrt{\frac{2GM_e}{R_e}} = \frac{2}{9} 	imes 11.2 \, km/s$.$v_m = 0.222 	imes 11.2 \approx 2.488 \, km/s$,which is approximately $2.5 \, km/s$.

Given the mass of the moon is $1/81$ of the mass of the earth and its radius is $1/4$ of the earth's radius. If the escape velocity on the earth's surface is $11.2 \, km/s$,the value of the escape velocity on the surface of the moon is ......... $km/s$.

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