The distance between the Sun and the Earth is | vedclass

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(C) Given: Distance between Sun and Earth, $r = 1.6 	imes 10^{11} \,m$. Radius of Earth, $R_e = 6.4 	imes 10^6 \,m$. Mass of Earth, $M_e = 6.0 	ime

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$4.3 	imes 10^6$

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(C) Given: Distance between Sun and Earth, $r = 1.6 	imes 10^{11} \,m$. Radius of Earth, $R_e = 6.4 	imes 10^6 \,m$. Mass of Earth, $M_e = 6.0 	imes 10^{24} \,kg$.Angular momentum of Earth around the Sun $(L_1)$: $L_1 = M_e v r = M_e (\frac{2 \pi r}{T_1}) r = \frac{2 \pi M_e r^2}{T_1}$.Here, $T_1 = 365 	imes 24 	imes 3600 \,s \approx 3.15 	imes 10^7 \,s$.$L_1 = \frac{2 	imes 3.14 	imes 6.0 	imes 10^{24} 	imes (1.6 	imes 10^{11})^2}{3.15 	imes 10^7} \approx 3.06 	imes 10^{40} \,kg \cdot m^2/s$.Angular momentum of Earth about its own axis $(L_2)$: $L_2 = I \omega = (\frac{2}{5} M_e R_e^2) (\frac{2 \pi}{T_2})$.Here, $T_2 = 24 	imes 3600 \,s = 8.64 	imes 10^4 \,s$.$L_2 = \frac{2}{5} 	imes 6.0 	imes 10^{24} 	imes (6.4 	imes 10^6)^2 	imes \frac{2 	imes 3.14}{8.64 	imes 10^4} \approx 7.15 	imes 10^{33} \,kg \cdot m^2/s$.Ratio $\frac{L_1}{L_2} = \frac{3.06 	imes 10^{40}}{7.15 	imes 10^{33}} \approx 4.28 	imes 10^6 \approx 4.3 	imes 10^6$.

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