Gravitational acceleration on the surface of | vedclass

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(B) The gravitational acceleration on the surface of a planet is given by $g = \frac{GM}{R^2} = \frac{G}{R^2} \cdot \frac{4}{3} \pi R^3 \rho = \frac{4

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The gravitational acceleration on the surface of a planet is given by $g = \frac{GM}{R^2} = \frac{G}{R^2} \cdot \frac{4}{3} \pi R^3 ho = \frac{4}{3} \pi G R ho$.Thus,the ratio of gravitational acceleration is $\frac{g_p}{g_e} = \frac{R_p ho_p}{R_e ho_e}$.Given $\frac{g_p}{g_e} = \frac{\sqrt{6}}{11}$ and $\frac{ho_p}{ho_e} = \frac{2}{3}$,we have $\frac{R_p}{R_e} = \frac{g_p}{g_e} \cdot \frac{ho_e}{ho_p} = \frac{\sqrt{6}}{11} \cdot \frac{3}{2} = \frac{3\sqrt{6}}{22}$.The escape velocity is $v_e = \sqrt{2gR}$.Therefore,$\frac{v_p}{v_e} = \sqrt{\frac{g_p R_p}{g_e R_e}} = \sqrt{\frac{g_p}{g_e} \cdot \frac{R_p}{R_e}} = \sqrt{\frac{\sqrt{6}}{11} \cdot \frac{3\sqrt{6}}{22}} = \sqrt{\frac{3 \cdot 6}{11 \cdot 22}} = \sqrt{\frac{18}{242}} = \sqrt{\frac{9}{121}} = \frac{3}{11}$.Given $v_e = 11 \ km/s$,we get $v_p = \frac{3}{11} 	imes 11 = 3 \ km/s$.

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