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

Answer 1

The correct option is B ${\left(1.52\right)}^{3/2}\times 365$

According to the keplers third law:

$\frac{I_{2}m}{T_{2}E}=\frac{R_{ms}^3}{R_{rs}^3}$

Where $R_{ms}$ is the mass sun distance and $R_{es}$ is the earth sun distance,

$T_m=\left(\frac{R_{ms}}{R_{es}}\right)TE=(1.52{)}^{\frac{3}{2}}\times 365days$