The Earth turns a little more than once with respect to the stars in order to complete one rotation with respect to the Sun. The "little extra" is just the angle through which the Earth has moved around the Sun in a day's time. Figure 1. The Earth must rotate degrees plus a , a very small angle, for observer at A to return to the same position relative to the Sun at B. The time for the Earth to turn this small angle is about four minutes.
This little difference would cause no concern if it were always the same, but it is not! Recalling that the Earth moves in an elliptical path much exaggerated in Fig.
The difference is about three million miles out of an average distance of ninety-three million miles. The speed of the Earth in its orbit increases as it gets nearer to the Sun. Since the Earth is closest to the Sun in January and furthest in July, it follows that the Earth is moving more rapidly in its orbit in January than in July!
Thus, the Earth must rotate a little more each day from October to April to return to a chosen spot to face the Sun again. However, there appears to be a problem with the expression for time-averaged distance in that answer. I think it's great to seek a mathematical expression, but it should be confirmed numerically.
I did a quick numerical double check and verified those general trends, but there may still be a problem with one of the expressions there. I think both of us should now add the path-average for completeness by averaging over ds.
Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Can one approximate the semi-major axis of an orbit as the average orbital distance for eccentric orbits? Ask Question. Asked 3 years, 11 months ago.
Active 4 months ago. Viewed 6k times. Improve this question. Add a comment. Active Oldest Votes. Average distance on geometric basis. In a way this is the most obvious one. This is a tricker calculation, but in fact it's the most "human" average. The geometric expression gets smaller as eccentricity gets larger. Improve this answer. StephenG StephenG 5, 1 1 gold badge 10 10 silver badges 27 27 bronze badges.
Are you planning on addressing this? Right now I think there is a mistake. To be honest I've lost track of this question mentally and if your maths is kosher, go for it. Show 1 more comment. Markus Schmassmann Markus Schmassmann 2 2 bronze badges. 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. Does that make sense. I'm not sure I explained that as well as I should. Edit On average distance. 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? 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. No need for one to be much more massive than the other. T[:2] orbits. Featured on Meta. Now live: A fully responsive profile. Version labels for answers. Related 2. Hot Network Questions. Question feed.
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