Calculating the time since periapse of an orbital position

Question

Given ...

• A known position along a planets orbit
• The semi major axis
• The eccentricity of the orbit
• The period of orbit

is it possible to calculate the time since periapse of this orbital position?

Answer 1

Well yes, you probably could. What you'd have to do is utilize Kepler's second law that basically says that area swept by the focus-to-body line remains constant during equal intervals of time:

                                       

            ^{Animation of the Kepler's second law: The same blue area is swept out in a given time (Source: Wikipedia)}

Your semi-major axis $a$ and eccentricity $e$ then gives you the total area $A_o$ that the one-orbit ellipse prescribes, and given your known position you could calculate the area prescribed since the periapsis, let's call it $A_p$. The fraction of $A_p$ in relation to $A_o$ is then directly proportional to the time slice since the periapsis $t$ of the total orbital period $T$.

$A_o$ equals $a\cdot b\cdot \pi$ , where $a$ is already mentioned semi-major axis, and $b$ is semi-minor axis. To calculate $b$ of an ellipse, we can use the argument of eccentricity $e$ using $b=a{\sqrt {1-e^{2}}}$, so our formula is then:

$$A_o = a^2\pi{\sqrt {1-e^{2}}}$$

How you could calculate the $A_p$, or the fraction of the $A_o$ since the periapsis comes down to what you meant with a known position along a planet's orbit, so I'll leave out this part as an exercise for the reader, but your time since periapsis $t$ would then be:

$$t = T \cdot \frac{A_p}{A_o}$$

It would of course be a lot simpler, if you calculated mean anomaly $M$ as part of your orbital elements, and then simply do $t = \frac{M}{n}$, where $n$ is mean motion that equals $\frac{2\pi}{T}$.

Answer 2

@iasksillyquestions A standard algorithm you will find typically involves a conversion from position and velocity to an orbital element set, one of which could by your true anomaly. You can then use a variety of methods to solve for your mean anomaly, which will in turn give you time since periapsis. If you are evaluating against orbits with low eccentricity, it could tend to be confusing when discussing periapsis.

As another sidenote, for the two-body problem, specifying the period is essentially the same as specifying the semi-major axis.

Answer 3

No.

First off, you also need the location of the barycenter of the system about which the body is rotating.

Assuming you have that, you are left with an ambiguity. The distance between the position and the barycenter is the current radius. Unless that radius is exactly the periapsis or exactly the apoapsis, then that radius occurs twice in the orbit. There are two possible times since periapsis for a given radius that is not exactly at the extremes.