The radius of the Martian orbit around the Su | vedclass

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(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the semi-major axis $r$ of th

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$88$ earth days

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(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the semi-major axis $r$ of the orbit: $T^2 \propto r^3$.Given that the radius of the Martian orbit $r_M = 4 	imes r_{Me}$,where $r_{Me}$ is the radius of the Mercury orbit.The ratio of the time periods is given by: $\frac{T_{Me}}{T_M} = \left(\frac{r_{Me}}{r_M}ight)^{3/2}$.Substituting the given values: $\frac{T_{Me}}{687} = \left(\frac{1}{4}ight)^{3/2} = \frac{1}{8}$.Therefore,$T_{Me} = \frac{687}{8} = 85.875$ days.Rounding to the nearest whole number,the length of $1$ year on Mercury is approximately $88$ Earth days.

The radius of the Martian orbit around the Sun is about $4$ times the radius of the orbit of Mercury. The Martian year is $687$ Earth days. What is the length of $1$ year on Mercury?

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