# How to calculate angle in elliptic orbit?

I would like calculate angle in elliptic orbit ZCP at any time, where:

• Z (position of periapsis),
• C (center of ellipse),
• P (current position of planet).

I find topic, where is picture (in accepted answer) that exactly describes my situation. But there is calculation of true anomaly, that is angle ZFP, where:

• Z (position of periapsis),
• F (focus of ellipse),
• P (current position of planet).

So, I know this informations:

• semi-major axis,
• semi-minor axis,
• eccentricity,
• apoapsis,
• periapsis,
• time in periapsis,
• any time,
• orbital period of body,
• current distance of planet from focus,

and I would like to calculate angle ZCP. Is there any way to do that without true anomaly? Or there is any relation between true anomaly and this angle?

• I'm not familiar with "pericenter"/"apocenter". Do you mean periapsis/apoapsis, the generic terms for the lowest and highest points around an arbitrary body? Or are you referring to something to do with the center of the ellipse, as opposed to its focal points? Mar 16 '19 at 10:18
• I mean periapsis and apoapsis, I fixed it, thanks. Mar 16 '19 at 10:22
• Before I answer this question, I need to ask: Why do you want this quantity? It's not a useful quantity. True anomaly is a useful quantity because that is the key angle of concern. The positions of the two bodies are observable, and periapsis is an observable event. This makes the true anomaly an observable quantity. Mean anomaly, while not observable, is a useful quantity because it is a purely linear function of time, and time since periapsis passage is observable it is easy to calculate. Mar 16 '19 at 10:42
• Eccentric anomaly similarly is not observable. It is a useful quantity solely because it is a relatively simple intermediary that enables one to transfer from the easily calculable mean anomaly and the desired true anomaly, or from the observable true anomaly to the mean anomaly. Eccentric anomaly is a useful fiction. The angle between the line segment from the center of the ellipse to the periapsis point and the line segment from the center to the current position is not a useful quantity. Mar 16 '19 at 10:46

Draw triangle $$PFG$$ where $$G$$ is the second focus of the ellipse. You know the following:

• $$|PF|$$ = distance from planet to $$F$$.
• $$|PG|$$ = major axis minus $$|PF|$$. Sum of distances definition of an ellipse.
• $$|FG|$$ = major axis times eccentrity.

Knowing all three sides of this triangle you calculate the angles from trigonometric laws.

Next draw the perpendicular segment from $$P$$ to the major axis, hitting the major axis at $$Q$$. From right triangle $$FPQ$$ you have $$|FQ|=\mp(|PQ|)\cot\angle F$$ where a positive sign indicates displacement towards $$Z$$ and a negative sign is used for the opposite direction. Similarly right triangle $$GPQ$$ gives $$|GQ|=\pm(|PQ|)\cot\angle G$$.

Now just observe that $$C$$ is just the midpoint netween the foci. That plus the right triangle relations above give the ultimate result

$$\cot\angle ZCP = (1/2)(\cot\angle G - \cot\angle F)$$

where angles $$F$$ and $$G$$ are defined within your original triangle $$PFG$$. The above equation will have a unique solution between $$0°$$ and $$180°$$ everywhere in the orbit.