Question

If the earth is a point mass of 6×1024 kg revolving around the sun at a distance of 1.5×108 km and in time T = 3.14×107 s, then the angular momentum of the earth around the sun is:

• A. 1.2×1018 kgm2/s
• B. 1.8×1029 kgm2/s
• C. 1.5×1037 kgm2/s
• D. 2.7×1024 kgm2/s.

Answer 1

Answer: (D)

2.7×1024 kgm2/s.

Answer 2

Angular momentum $L = mvr = m\omega r^{2}$

= $m \times \frac{2\pi}{T} \times r^{2}$

= $\frac{2 \times 3.14 \times 6 \times 10^{24} \times (1.5 \times 10^{11})^{2}}{3.14 \times 10^{7}} = 2.7 \times 10^{40}kg ⥂ - ⥂ m^{2} ⥂ / ⥂ s$