Q. If the earth is a point mass of $6 \times 10^{24}\ kg$ revolving around the sun at a distance of $1.5\ \times 10^{8}\ km$ and in time $T = 3.14 \times 10^{7}s$, then the angular momentum of the earth around the sun is
A) $1.2 \times 10^{18}\ kg\ m^{2}/s$
B) $1.8 \times 10^{29}\ kg\ m^{2}/s$
C) $1.5 \times 10^{37}\ kg\ m^{2}/s$
D) $2.7 \times 10^{40}\ kg\ m^{2}/s$
Explanation
(d) Angular momentum, $L = mvr = m\omega r^{2} = m \times \frac{2\pi}{T} \times r^{2}$
$= \frac{2 \times 3.14 \times 6 \times 10^{24} \times \left( 1.5 \times 10^{11} \right)^{2}}{3.14 \times 10^{7}} = 2.7 \times 10^{40}kg - m^{2}/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.