Answer :
Explanation:
Assume that mass of Earth is M, radius of earth orbit is R, and rotational period of Earth is T.
The angular momentum of Earth is,
[tex]L_{z} &=M R^{2} \omega \\ &=M R^{2}\left(\frac{2 \pi}{T}\right) \\ &=\frac{2 \pi M R^{2}}{T}[/tex]
The mass of mars is, mass of Earth
=0.11 M
The radius of mars orbit is, of radius of earth
=0.53 R
The rotational period of mars is, of period of Earth
=1.03 T
The angular momentum of mars is,
[tex]L_{m}=\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}[/tex]
The angular momentum of mars is,
[tex]L_{m}=\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}[/tex]
The ratio of angular momentum of mars to that of earth is,
[tex]\frac{L_{m}}{L_{E}}=\frac{\frac{2 \pi(0.11 M)(0.53 R)^{2}}{1.03 T}}{\frac{2 \pi M R^{2}}{T}} \\ \frac{L_{m}}{L_{E}}=0.03 \\ \frac{L_{m}}{L_{B}}=3.0 \times 10^{-2}[/tex]