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(A) The acceleration due to gravity at a distance $r$ from the center of the Earth is given by $g = \frac{GM}{r^2}$,where $r = R + h$.Given the ratio

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$3 R / 5$

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(A) The acceleration due to gravity at a distance $r$ from the center of the Earth is given by $g = \frac{GM}{r^2}$,where $r = R + h$.Given the ratio of accelerations:$\frac{g_p}{g_Q} = \frac{GM / r_p^2}{GM / r_Q^2} = \left( \frac{r_Q}{r_p} ight)^2$Substituting the given values:$\frac{36}{25} = \left( \frac{r_Q}{r_p} ight)^2$Taking the square root on both sides:$\frac{r_Q}{r_p} = \frac{6}{5} \implies r_Q = \frac{6}{5} r_p$Since $r_p = R + h_p = R + R/3 = 4R/3$:$r_Q = \frac{6}{5} 	imes \left( \frac{4R}{3} ight) = \frac{24R}{15} = \frac{8R}{5}$Now,$r_Q = R + h_Q$,so:$R + h_Q = \frac{8R}{5}$$h_Q = \frac{8R}{5} - R = \frac{3R}{5}$

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