I've read from a page that you can find the distance from the Sun to the Earth via the following equation:

$$r = a\frac{1-e^2}{1+e\cdot \cos(θ)}$$

where r is the distance from the Sun, a is the semi-major axis, e is the eccentricity and θ is the angle around from perihelion.

I am trying to create a model of the orbit of Earth around the Sun and I have to find the distance between the Earth and the Sun at multiple points and also its speed. Can anyone explain the maths?

  • $\begingroup$ Are you trying to figure it out on a particular day of the year, or just plot how it varies? $\endgroup$ – PearsonArtPhoto Jan 14 '18 at 21:57
  • $\begingroup$ are you asking how to calculate the orbital mechanics, or where to find values for the semi-major axis and eccentricity of Earth's orbit around the sun? $\endgroup$ – JCRM Jan 14 '18 at 22:46
  • 2
    $\begingroup$ As JCRM implies, you must have some initial information. For Terra, the semi-major axis a = 1.496E11 meters, and the eccentricity e = 0.0167. Choose your true anomaly θ and your equation lets you calculate r. Knowing r, you can find the speed v from v^2 = (2μ/r) – (μ/a), with μ = (mass of Sol)*(Newton’s constant) = 1.327E20 m^3/s^2. Finding θ, r, and v at any given time is more complicated. There are other threads that discuss Kepler’s Equation. $\endgroup$ – MBM Jan 15 '18 at 19:04
  • $\begingroup$ astronomy.stackexchange.com/questions/13488 may or may not be helpful $\endgroup$ – user7073 Jan 16 '18 at 15:30
  • $\begingroup$ As a possible simple cheat, you can just ask Mathematica to do the work for you [mostly joking because it's a bad idea to trust "black boxes"] ;) $\endgroup$ – honeste_vivere Jan 19 '18 at 15:44

It seems to me there are two genearl ways to model the earth's rotation around the sun. One is to use orbital parameters the other to use Mean orbital elements. The first is more accurate but more complex and less defined a method. Using the first, you simply look up the earth's most accurate and current velocity vector as well as position. Often these are hard to find, so you must use the second method, to precisely calculate these two parameters. Then you use your own models which aren't keplarian and include pertubations etc to work out any further time, it's position. As you can see it is more complex and really not defined. It's up to you how you model Earth's position, you can use General Relativity, model every other planet in the solar system etc to make it as accurately as you like.

The second method basically takes these very complex parameters, and averages them through various methods, into an average newtonian/keplarian orbit which minimizes error as best as possible for a known time range that the mean elements are valid. For example some J2000 elements are valid from 1950 to 2050.

All mean elments are usually specified at J2000 which is 01 January 12:00:00 GMT or januray 1st at noon Greenwhich time. This is the location of earth, earth's eccentricity, and major axis basically the specific elliptical orbit of earth which will minimize error of pertubations for the given time interval (like 1950 - 2050) at the january 1st epoch. So to work out current locations you must calculate it from January 1st noon GMT's locations.

Okay, so basically you need the true anomaly which is the angle from the perhelion of earth's orbit to it's current location at J2000. This is rarely published, instead it is either the Mean anomaly or more common the Mean longitude. You can calculate the Mean anomaly from mean longitude with additoinal J2000 parameters like argument of perhelion or longitude of ascending node etc. So basically you work out the Mean anomaly for J2000. Now you work out the degrees per second it changes, then apply the degrees it has changed in the roughly 18 years till now to the Mean anomaly at J2000 to get Mean anomaly at current time.

Having the current Mean anomaly is the start. Now you must find teh eccentric anomaly from the mean anomaly. You will need either a taylor series approximation or newtom method will work to solve $M = E - e sin(E)$ for eccentric anomaly given current Mean anomaly. You need either of the two methods to approximate it, cause there is no closed form equation which solves this.

Having eccentric anomaly its a simple matter to work out true anmoaly for the current time, and thus use it in your equation. Unlike eccentric anomaly, true anomaly is easily gotten from eccentric anomaly as there are closed form equations relating the two. This will get you a decent approximation of Earth's real time true anomaly.

| improve this answer | |
  • $\begingroup$ Again to reinforce the approximation here, this is a newtonian gravity model of earth's trajectory for typically 100 years used to approximate earth's real trajectory which as you know General Relativity was discoverd because of the short comming's of Newtonian gravity. That and n body effects are still very difficult to model calculate. So this is a newtonian 2 body model for a GR 10 body or more system, over 100 years. $\endgroup$ – marshal craft Jul 8 '18 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.