Question

Assuming that earth and mars move in circular orbits around the sun, with the martian orbit being $1.52$ times the orbital radius of the earth. The length of the martian year in days is

• A. $(1.52)^{2/3} \times 365$
• B. $(1.52)^{3/2} \times 365$
• C. $(1.52)^{2} \times 365$
• D. $(1.52)^{3} \times 365$

Answer 1

Answer: (B)

According to Kepler s third law
$\frac{T^{2}_{M}}{T_{E}^{2}} = \frac{R^{3}_{MS}}{R^{3}_{ES}}$
where $R_{MS}$ is the mars-sun distance and $R_{ES}$ is the earth sun distance.
$\therefore T_{M} = \left(\frac{R_{MS}}{R_{ES}}\right)^{3/2} \times T_{E}$;
$\therefore T_{M} = \left(1.52\right)^{3/2} \times 365$ days