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Comparing orbits between a planet and Red Dwarf
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Hello, I have a problem, so I have an earth like planet orbiting around a sun like star and then there is a red dwarf on an elliptical orbit around the star that passes close to the planet. I want to show that sometimes when the Red Dwarf is at it's closest approach the planet may not be there so it won't gain any heat from the red dwarf. To do this I was told to compare the orbital periods of the two. So for the planet it's orbital period is just like the earth 365 days and the red dwarf's period is 1896.59 days.
I was thinking is there a way I could plot the orbits of these two and run it for say like 100 years and see the if there is points when the earth could miss the red dwarf? If this is wrong please let me know a better way of tackling this problem.
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John D'Errico
2020-4-8
I would ask if this is intended to be a classical three body problem, where gravitation is influencing each of the other planets. I'll assume not, since that is seen to result in chaotic behavior, at least over the long term. Chaotic here means the three body problem is unstable, where a tiny perturbation or uncertainty in the orbits will amplify over time, eventually making the orbits unpredictable over a very long term.
That implies you are assuming that each object lives in a purely elliptical orbit, and what you care about are the timings of those orbits. In that case, what matters are the orbital periods. As well, in this case, what will be important is group theory, and the greatest common divisor of the two orbital periods. In the end, it comes down to simple number theoretic questions.
You have two periodic objects, one with a period of 365 days, and the second with a period of 1896.59 days. Let me assume that each of them is exactly known. In fact, I'll break each day up into hundreths of a day. So now the orbiral periods are 36500 hundreths, and 189659 hundreths. The important question is now (the only MATLAB content in my comment):
gcd(189659,36500)
ans =
1
the greatest common divisor is 1. That is, that pair of orbital periods is relatively prime, since the GCD is 1. What, in turn, does that tell me? It tells me that if the two orbiting objects do not influence each other gravitationally, so their orbits do not lengthen or shorten, or change relative to each other, then EVERY possible configuration between the two orbiting objects will be observed, and over the long term, EVERY possible configuration will happen equally often.
Again, to within 0.01 days, given the system you seem to be describing, EVERY possible configuration of the two objects will be observed, and it will be observed equally often - OVER THE LONG TERM.
And all of this is a simple claim that will come from group theory, concepts from number theory, etc. One of the things I love about mathematics is how an understanding of various branches of mathematics will sometimes offer insight into a variety of processes, of physical systems. Even the most pure mathematics can sometimes offer insights too.
In fact of course, what does tend to happen is orbital periods do change. And the Solar system is an n-body problem, not even just a 3 body problem. Planets do influence each other. Looking at the solar system, for example, one can find orbital resonances, so one planet's orbital period becomes related to others in the same system. And for more number theoretic content yet, we might read about planetary orbital periods and the Fibonacci numbers, for example:
I'm sure a little digging will find lots of intriging stuff.
Mathematics can be a fascinating subject, and having a breadth of mathematical knowledge is sometimes as useful/interesting as it is to have a deep understanding of a single area of mathematics.
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James Tursa
2020-4-8
编辑:James Tursa
2020-4-8
To calculate how often stellar eclipses occur for the 2D geometry, on average, yes all you need to do is look at orbit periods.
The "inner" planet moves faster than the "outer" Red Dwarf, so you just need to calculate the average time it takes for the planet to complete one revolution and then some to catch up with the Red Dwarf. Since we are just after average results here, consider the problem as two circular orbits with the same periods:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/282934/image.jpeg)
perp = period planet (days)
perd = period red dwarf (days)
np = 2*pi / perp; % mean motion of planet (rad/day)
nd = 2*pi / perd; % mean motion of red dwarf (rad/day)
Then you simply need to solve for the time T that it takes for the next alignment. So you have two equations:
np * T = 2*pi + a
nd * T = a
Solving for T yields
T = 2*pi / (np - nd); % (days)
This is the average time between stellar eclipses.
2 个评论
James Tursa
2020-4-8
I put it together myself with powerpoint. Use it all you want. No citation needed.
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