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
$\frac{T_M^2}{T_E^2}=\frac{R_{MS}^3}{R_{ES}^3}$
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