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

Answer 1

Answer: (B)

Using Kepler's $ III^{rd} $ law, $\frac{T_{M}^{2}}{T_{E}^{2}}=\frac{R_{M s}^{3}}{R_{E}^{3}}$
$\Rightarrow T_{M}=\left(\frac{R_{M s}}{R_{E s}}\right)^{\frac{3}{2}} \times T_{E}=(1.52)^{3 / 2} \times 365$ days.