A planet of radius R = frac{1}{10} times (rad | vedclass

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(B) Given: $R_p = \frac{R_e}{10}$,$ho_p = ho_e$,$\lambda = 10^{-3} \ kg \ m^{-1}$,$g_e = 10 \ m \ s^{-2}$,$R_e = 6 	imes 10^6 \ m$.Since density

Vedclass · Accepted Answer

$108$

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(B) Given: $R_p = \frac{R_e}{10}$,$ho_p = ho_e$,$\lambda = 10^{-3} \ kg \ m^{-1}$,$g_e = 10 \ m \ s^{-2}$,$R_e = 6 	imes 10^6 \ m$.Since density is the same,$M_p = ho \cdot \frac{4}{3} \pi R_p^3 = \frac{M_e}{10^3}$.The surface gravity of the planet is $g_p = \frac{G M_p}{R_p^2} = \frac{G (M_e / 10^3)}{(R_e / 10)^2} = \frac{G M_e}{10 R_e^2} = \frac{g_e}{10} = \frac{10}{10} = 1 \ m \ s^{-2}$.The acceleration due to gravity at a distance $r$ from the center is $g(r) = g_p \frac{r}{R_p}$.The wire extends from $r = R_p - \frac{R_p}{5} = \frac{4R_p}{5}$ to $r = R_p$.The force $F$ required to hold the wire is the weight of the wire: $F = \int_{4R_p/5}^{R_p} \lambda \cdot g(r) \, dr = \int_{4R_p/5}^{R_p} \lambda \frac{g_p}{R_p} r \, dr$.$F = \frac{\lambda g_p}{R_p} \left[ \frac{r^2}{2} ight]_{4R_p/5}^{R_p} = \frac{\lambda g_p}{2R_p} \left( R_p^2 - \frac{16R_p^2}{25} ight) = \frac{\lambda g_p}{2R_p} \left( \frac{9R_p^2}{25} ight) = \frac{9 \lambda g_p R_p}{50}$.Substituting values: $R_p = \frac{6 	imes 10^6}{10} = 6 	imes 10^5 \ m$.$F = \frac{9 	imes 10^{-3} 	imes 1 	imes 6 	imes 10^5}{50} = \frac{5400}{50} = 108 \ N$.

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