Question

Planet X has twice the mass as the earth and orbits its star as the same orbital distance that the earth orbits the sun. If Planet X takes 365 and $1/4$ days to complete one orbit around its star, how does the angular momentum of Planet X compare to the angular momentum of the earth?
It can be assumed that both orbits are circular.

• A. $L_X=2L_{Earth}$
• B. $L_X=\frac{L_{Earth}}{2}$
• C. $L_X=L_{Earth}$
• D. $L_X=4L_{Earth}$
• E. $L_X=\frac{L_{Earth}}{4}$

Answer 1

Answer: (A)

$L_X=2L_{Earth}$

$m_X=2m_E$

$r_X=r_E$

The time period of revolution of both the planets is same.

Hence $\frac{2\pi r_X}{v_X}=\frac{2\pi r_E}{v_E}$

$⟹v_X=v_E$

The ratio of angular moementa=$\frac{L_X}{L_E}=\frac{m_X{v}_X{r}_X}{m_E{v}_E{r}_E}$

$=\frac{m_X}{m_E}$

$=2$