What we know: We know a solar day on Mars is how long it takes for the sun to make a complete "rotation" in the sky and return to the same point. Because the Sun is not a fixed point in the sky (due to the fact that Earth and Mars are rotating it) a solar day will differ from a sidereal day, which is how long it takes for a fixed point (typically a star) to reach its same location in the sky.
Image from: http://ottawa-rasc.ca/articles/earl_mike/Satellite_Tracking/Dishes/Satellite_Dishes.html
We also know from Kepler's Law: \[P^2 \: \alpha \: a^3 \]
and from the problem: \[P_e= 1 \: year\] \[a_e= 1\: AU\] \[a_m= \frac{3}{2} \: AU\](where \(P_e\) and \(P_m\) is the rotation period of the Earth and Mars respectively and \(a_e\) and \(a_m\) are the distance between the Sun and Earth and Mars respectively).
Solve:
From Kepler's Law we can set up the ratio: \[\frac{P_e^2}{P_m^2} = \frac{a_e^3}{a_m^3}\] and isolate the rotation period of Mars: \[P_m =\sqrt{\frac{P_e^2a_m^3}{a_e^3}}\] plugging in the provided values give:\[P_m =\sqrt{\frac{1^2\cdot \frac{3}{2}^3}{1^3}}\] which simplifies to: \[P_m=\left( \frac{3}{2} \right)^{\frac{3}{2}} \approx \frac{9}{5}\: Earth \:years\] when converting to days we get \[\frac{9}{5}\: Earth \:years \cdot 365 \: days \approx 660 \: days\: in \:a \:Martian\: year\] But this is not enough, we want the difference between the two days in minutes per day, not the length of a Martian year. To find the difference we can use the fact that Mars traverses an entire orbit in a 660 days \[\frac{360\,^{\circ}}{660 \:days} \approx \frac{0.55\,^{\circ}}{day}\] to show that a sidereal day is about half a degree less than a solar day. Then we can convert to days per minute: \[\frac{360\,^{\circ}}{660 \:days}\cdot \frac{1 \:hr}{15\,^{\circ}} \cdot \frac{60 \: min}{1 \: hr} \approx 2 \frac{min}{day}\]
Conclusion: with the given values, the difference between sidereal and solar day is about 2 minutes. This means that Mars's sidereal day is 2 minutes shorter than its solar day.
Acknowledgements: Many thanks to Barra for working on this worksheet with me.
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ReplyDeleteVery nice job working this through. I particularly like how you converted 1.5 to a fraction. It's pretty cool that using an approximation as rough as you did actually works. The actual fraction is \(\frac{3}{2}=\frac{5.2}{2.8}\) .
ReplyDeleteBy the by, to get well proportioned parenthesis the \left( \right is how to get them to scale with what is in the problem. e.g \left(\frac{\sqrt{e^{i \pi}}}{\frac{e^{i \theta} - \cos{\theta}}{\sin{\theta}}}\right) = 1 is \[ \left(\frac{\sqrt{e^{i \pi}}}{\frac{e^{i \theta} - \cos{\theta}}{\sin{\theta}}}\right) = 1 \]
Fixed! Thank you, that trick was very useful in future blog posts.
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