The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is $687\text{ Earth days}$. Then which of the following is the length of 1 year on Mercury?
A
88 earth days
B
225 earth days
C
172 earth days
D
124 earth days
The answer is A.
Solution
From Kepler's third law $T^2 \propto r^3$, $T_{\text{Mercury}} = \frac{T_{\text{Mars}}}{4^{3/2}} = \frac{687}{8} \approx 86\text{ days}$, closest to $88\text{ earth days}$.
According to Kepler's Third Law of planetary motion: $T^2 \propto r^3 \implies \left(\frac{T_{\text{Mars}}}{T_{\text{Mercury}}}\right)^2 = \left(\frac{r_{\text{Mars}}}{r_{\text{Mercury}}}\right)^3$. Given $r_{\text{Mars}} = 4\, r_{\text{Mercury}}$: $\left(\frac{T_{\text{Mars}}}{T_{\text{Mercury}}}\right)^2 = 4^3 = 64 \implies \frac{T_{\text{Mars}}}{T_{\text{Mercury}}} = \sqrt{64} = 8$. Thus, $T_{\text{Mercury}} = \frac{T_{\text{Mars}}}{8} = \frac{687\text{ days}}{8} \approx 85.875\text{ days}$. Among the choices, the correct option is $88\text{ earth days}$.
ConceptKepler's Law of Periods
Gravitation in past papers
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