### Can one approximate the semi-major axis of an orbit as the average orbital distance for eccentric orbits?

Kepler's 3rd Law (regarding the relation between orbital period and the semi-major axis of an orbit) applies to all elliptical orbits. But as I understand it, the only reason it is safe to use the approximation that the semi-major axis of an orbit is approximately equal to the average orbital distance is because the planets of our solar system are generally not very eccentric.

As an example, the Earth is about 1 AU away from the Sun and has a very low orbital eccentricity, so one can say that the semi-major axis of the Earth's orbit is also approximately 1 AU in length. Since

`P^2 (yr) = a^3 (AU)`

, one can deduce that the orbit of Earth should take about a year. But, what if the Earth had a higher eccentricity - If the orbital eccentricity of the Earth were higher, would it still be wise to approximate the orbital distance as the semi-major axis of an orbit?userLTK Correct answer

4 years agoThe gist of this, is that your assumption is incorrect. It's the semi-major axis that defines the period, not the average distance. Newton worked this out when he invented calculus and derived Kepler's laws. (Look up Derivation of Kepler's laws for some explanations). Here's the Wikipedia version of the Math.

I should add that Kepler's three laws, and Newton's derivation of them still don't perfectly define orbits. They assume that only the larger mass matters, when in reality, the two objects tug on Each other. The Sun doesn't just tug on the Earth but the Earth tugs on the Sun, which speeds up the orbit a tiny bit (the mass ratio of 330,000 to 1 makes this close to negligible, but it's there). More accurate Newtonian orbital calculations between two massive objects require more advanced mathematics. Kepler's laws only work if the central object is much more massive than the 2nd object.

Ignoring the finer details and just looking at the three laws, Newton's derivation works out exactly to the semi-major axis. The average distance doesn't enter into the equation.

Average distance of an orbiting object to it's Sun or central object is problematic anyway because the planet moves more slowly when it's more distant, so there's a few ways to calculate average distance.

**By time**the planet spends in each part of it's orbit (same as area as defined by Kepler's 3rd law)Or, you can measure average distance

**by arc**or angle, taking smaller and smaller angles and running an average.Or you can take average distance

**by length**of the ellipse to the specific foci.That's three separate methods for measuring average distance. Needless to say, this gets a little complicated. The good news is, you don't need to do any calculation for average distance because the Semi-major axis is actually correct.

Does the semi-major axis equal the average distance, perhaps by arc or by length? Not sure. I'll try to work that out. I know that it doesn't equal average by time.

Does that make sense. I'm not sure I explained that as well as I should.

Edit

On average distance. The standard average distance with orbits is by time or area. Kepler's 3rd law says equal areas over equal time, so it's really two ways of saying the same thing, equal time and equal surface area.

I think your question is interesting and I'll try to work out a relation between semi major axis and average distance. Off hand, it's a little complicated, but it seems to me that ratio between average distance and semi major axis changes with eccentricity but, I'd rather work that out before I say for certain.

If I understand correctly, I incorrectly assumed the law related average orbital distance with orbital period whereas the “simple math” - derivation yields the result that the period is related to the semi-major axis; The actual case is more complicated as smaller masses are no longer negligible. Is that correct? Slightly off-topic, but of the three methods used to calculate average orbital distance, would the most convenient be the first method since the swept area (or elapsed time in orbit) would be constant?

@mikey Smaller masses are closer to negligible. It's when the two masses approach being more equal then Kepler's laws become less accurate. As to your first point, yes, the derivation points to the semi-major axis. As to your 3rd, I'll edit the answer above.

I’m guessing that you integrate along the orbital curve varying the angle for the second method. But I don’t yet have an idea for the third method.

Kepler's Third law applies just fine to binary stars. No need for one to be much more massive than the other.

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user15317 4 years ago

If I understand correctly, I incorrectly assumed the law related average orbital distance with orbital period whereas the “simple math” - derivation yields the result that the period is related to the semi-major axis; The actual case is more complicated as smaller masses are no longer negligible. Is that correct? Slightly off-topic, but of the three methods used to calculate average orbital distance, would the most convenient be the first method since the swept area (or elapsed time in orbit) would be constant?