# Difference between true anomaly and mean anomaly

What is the difference between true anomaly and mean anomaly in orbital mechanics? I want to understand the physical meaning of mean anomaly, for example true anomaly is the angular position of a satellite/object when viewed from earth/focus but what is the corresponding mean anomaly for that true anomaly. I don't want formulas, but the physical explanation.

• Welcome to the site. I'm afraid this question lacks context, you need more detail for it to be answered.
– GdD
Jun 30, 2017 at 15:39
• Like GdD said, this question could benefit from being fleshed out a little. Particularly in this case, if you are trying to understand the difference between two related terms, then at the very least look over the relevant Wikipedia articles, and point out the part of the difference that you don't understand. Not only might you actually in doing so learn the answer on your own, but it also helps us to give you an answer that you are likely to understand and find useful. Show us that you have tried!
– user
Jun 30, 2017 at 18:10
• added more detail. Jul 2, 2017 at 2:41
• The question has been improved and I hope it is now re-opened. I have to admit that I don't know and can never remember what those "anomaly" terms mean either. I think it's the word "anomaly" which taken by it self sounds like something is unusual or wrong (or lost contact or blew up). I always have to go spend five minutes and look at complicated diagrams that send you searching for "the Greek letters $\nu$ or $\theta$, or the latin letter $f$" to find out just what angle it is that "true anomaly" denotes for example.
– uhoh
Jul 2, 2017 at 3:20
• @uhoh my code interchanged true and mean anomaly b/c of that very reason, mistook which symbol meant which and it took me a long time to figure out what was up. Nov 9, 2018 at 18:37

The two are related through the eccentric anomaly: let $$M$$ be the mean anomaly, $$E$$ be the eccentric anomaly, $$e$$ be the eccentricity, and $$\nu$$ be the true anomaly. Then,
$$M = E - e \sin E$$ and $$\cos \nu = \frac{\cos E - e}{1 - e \cos E}$$ or perhaps more clearly (thanks to David Hammen) $$\tan\frac{\nu}{2}=\sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}$$