We are a few hours away from spring equinox. I was thinking if there is a mathematical expression that would define the path of a given point on the Earth, given by longitude and latitude say 34.05, 118.25 Los Angeles, around the Sun. Considering the angle of axis of the Earth, this should look like a stretched slinky or an uneven sinusoidal (because of the superposition of additive and subtractive Earth daily rotation). I guess this path would be on a region of a corrugated cylinder slanted 23.5 to ecliptic plane. I am more interested at an answer at the freshman college level ignoring for example Earth's axial precession. I would imagine there is already expressions or charts available to astronomers, but I am interested more of an elegant, simple solution.
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I can see the path in my mind. because the latitude of Los Angeles is not zero by assuming a circle orbit of earth we could immediately see the path as intersection of a cylinder slanted at 66.5 degrees with another cylinder vertical with respect to ecliptic plane but corrugated into 365.25 waves that are stretched and contracted a bit ( speed of rotation of Los Angeles compared to speed of earth in its orbit around sun is in the order of 1/80). two things make writing this into a mathematical expression a bit hard. – kamran Mar 19 '16 at 21:02
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My guess is that you would start with the path of the Earth being an elliptical orbit. Perhaps you could put that in polar coordinates and attempt to add on a sinusoidal nature to the movement as it goes around or perhaps a parametric equation? – Simply Beautiful Art Mar 19 '16 at 23:27
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the path of Los Angeles is not symmetrical with center earth. it is offset by a 800- 900 miles because of the tilt angle of axis of earth. it is rotating on a circle with a radius of R.sin(34.05) . the plane of this circle is sloped at 23.4 and the center of it is offset from the Earth center by R.cos(34.05). this offset is not rotating while Earth rotates as well as orbits the sun. At the same time discounting the wobbling of earth about its barycenter with moon. – kamran Mar 20 '16 at 00:01
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Since it is not rotating, I would add it in as a Cartesian coordinate type of thing. On the other hand, it is time based, so parametric equations sound more right. I'll think about this. – Simply Beautiful Art Mar 20 '16 at 00:54
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so far I have come up with this very rough sketch and will greatly appreciate help/ correction/comment. – kamran Mar 20 '16 at 20:15
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so far I have come up with this very rough sketch and will greatly appreciate help/ correction/comment. assuming the origin of coordinate system at a point on ecliptic plane at the center of Los Angeles rotation circle, we can write the parametric equations of the ellipse which is the projection of this circle to ecliptic plane as: X= a.cos(n.2pi/365.25) and Y= b.sin(n.2pi/365.25), for the ellipse. for the rotation of earth around sun we have:X=R.cos(phi)-r.cos(365.25phi) and likewise Y=R.sin(phi)-r.sin(365.25phi). assuming the path of rotation of Los Angeles a circle! needs more work. thanks. – kamran Mar 20 '16 at 20:43
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continuing from my last comment. R is distance from earth to sun, r is the radius of circle of Los Angeles, LA from here on, about earth axis. the above parametric equation will define a rough sketch of the path assuming both orbit of earth and LA's path circle. all we need to add is another parameter which is Z= vertical coordinate of path of LA. I am a retired engineer with rusty math skills, would appreciate any help. – kamran Mar 20 '16 at 20:55
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This just seems like a very messy 'guess and see how it works' problem, so I don't think I will be answering it. – Simply Beautiful Art Mar 21 '16 at 10:55
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i will clean it up. i need to find an app to draw a sketch. but i think the motivation is in the right path. both earth orbit its daily rotation are counter clockwise. so it makes sense to assume X of the path is algebraic addition of the X of earth orbit and x of LA. i may even draw a sketch on paper and take photo of it. although rotation of LA around earth axis is an ellipse, its angular rotation is constant as opposed to a planetary orbit so it is easy to parametrize. – kamran Mar 21 '16 at 22:27
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I do not think there is an easy way to express it. You can approximate it as an average between curtate elliptical hypocycloid and a curtate elliptical epicycloid. (It is kinda contradictory because a cycloid is usually defined around a circle, but I assume you can generalize it.) It would take a lot of work. – KR136 Mar 22 '16 at 08:10
1 Answers
This sketch is an integral part of the answer. Please make sure you have it open.
For the sake of simplicity I have made these assumptions: 1- The orbit of earth around sun is a circle. 2- The projection of Los Angeles (LA) parallel circle onto the ecliptic plane is still circle not ellipse, which is not to far fetched because the minor axis is 92% of major axis and as opposed to planetary orbit its angular velocity is constant with a period of 1/day. 3- The coordinate system has its origin at center of sun and x axis crosses LA on calendar time of spring equinox. 4- rotation of earth is synchronized with annual orbit of earth around sun at a ratio of 365.25 revolutions/year, exactly like the arms of a clock! All of the above assumptions are reasonable within our level of approximation. by inspection of the sketch I have attached it is easily observed the path of city of LA around sun is a meandering course which wobbles around the orbit of earth with adding and subtracting maximum of 4000.cos(34.05) = 3289 miles. This is but small fraction of the ~93000000 miles distance of earth to sun, something in the order of 1/30000. the coordinates of graph are obtained by adding the coordinate of earth in its orbit to that of city of LA on its parallel rotating around the earth axis. $$ X=AU.cos(\theta)-1660.cos(365.25.\theta)$$ and $$ Y=AU.sin(\theta)-1660.sin(365.25.\theta)$$ with 1660 miles being the radius of city of LA's latitude. by checking the sketch you will notice that I have measured the distance from the sun immediately to LA even though its latitude is offset from the center of the earth by about 600 miles, but that is ok because this offset is constant and does not rotate with earth. it's a big gyro. This graph could be refined farther and be given to a computer for plotting. Any help and correction is appreciated.
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