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

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${(1.52)}^{2} \times 365$

${(1.52)}^{2 / 3} \times 365$

${(1.52)}^{3} \times 365$

${(1.52)}^{3 / 2} \times 365$

answer is B.

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Detailed Solution

According to Kepler's third law,

$\frac{T_{M}^{2}}{T_{E}^{2}} = \frac{R_{M S}^{3}}{R_{E S}^{3}}$

where $R_{M S}$ is the mars-sun distance and $R_{E S}$ is the Earth-sun distance.

$∴ T_{M} = {(\frac{R_{M S}}{R_{E S}})}^{3 / 2} \times T_{E}; ∴ T_{M} = {(1.52)}^{3 / 2} \times 365 d a y s$

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