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Sunday, February 8, 2015

Worksheet 3, Problem 1: A Day in the Life of a Martian

The problem: What is the difference between sidereal and solar day on Mars if Mars has the same rotation period and orbits at 1.5 AU.

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: P2αa3
and from the problem: Pe=1year ae=1AU am=32AU(where Pe and Pm is the rotation period of the Earth and Mars respectively and ae and am are the distance between the Sun and Earth and Mars respectively).

Solve:
From Kepler's Law we can set up the ratio: P2eP2m=a3ea3m and isolate the rotation period of Mars: Pm=P2ea3ma3e plugging in the provided values give:Pm=1232313 which simplifies to: Pm=(32)3295Earthyears when converting to days we get 95Earthyears365days660daysinaMartianyear 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 360660days0.55day to show that a sidereal day is about half a degree less than a solar day. Then we can convert to days per minute: 360660days1hr1560min1hr2minday

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.

3 comments:

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  2. Very 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 32=5.22.8 .

    By 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 (eiπeiθcosθsinθ)=1

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  3. Fixed! Thank you, that trick was very useful in future blog posts.

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