3
$\begingroup$

I've searched online for an answer, but they all seem to require other parameters (e.g, eccentricity and semi major axis) to be able to mathematically solve for the perihelion/aphelion. If we don't have those parameters, how can we calculate them in order to obtain the perihelion/aphelion?

My broader question really is: if something (or a number of things) needs to be measured first, perhaps through observation or angle measurement, in order for us to be able to mathematically derive the other orbital elements (such as semi major axis and aphelion), then what is that something and how is it obtained?

$\endgroup$
4

2 Answers 2

5
$\begingroup$

You begin with the sky. From this you measure the location of the planet. You want the location relative to the stars, and there are a number of tricks, such as timing when the planet passes the meridian (due south) that can enable a precise location relative to the stars to be found.

Then you do that again at a different time

Then you do it again.

Then you try to fit a model of the motion of an object that follows a Keplarian ellipse to your observations. In doing so, you will determine all six of the "orbital elements", and from these you can find the perihelion. This involves solving (numerically) an order 8 polynomial equation, but you can usually find an elliptical model that fits the motion in the sky.

If you have more observations, you can use those to improve the accuracy of your model.

In practice, one would use software to do the maths for you.

The things you measure first is the sky position of the planet, but you need at least three observations to determine the orbit.

$\endgroup$
2
  • $\begingroup$ Thank you, James. Your answers are always a great help. I was hoping for a more in-depth analysis suited for a beginner like myself, but with your illuminating answer I will try to dig deeper for a full understanding of these measurements. Is there any reference you recommend? $\endgroup$
    – William
    Oct 10, 2022 at 3:42
  • 1
    $\begingroup$ Here is a resource. stardust2013.eu/Portals/63/Images/Training/OTS%20Repository/… It describes the classical method of orbital determination. $\endgroup$
    – James K
    Oct 10, 2022 at 18:33
3
$\begingroup$

The simple answer is you measure the orbit, then the orbit gives you the perihelion/aphelion. James K said this in a longer way; you had already postulated this, and found it unsatisfying.

If you want to get pedantic, you blast the object with radar (possibly lidar in some future time), directly measure range (from time delay and speed of light), and then try to get range rate from Doppler shift (change in frequency of return signal). Given a crude orbit (a ballpark r value, or distance from Sun- not hard if you don’t need decimal places), and the known state of the Earth (in its orbit relative to the target, at the time of the observation), the value for range rate will likely give ‘increasing r’ (not yet at aphelion) or ‘decreasing r’ (pre-perihelion). Of course, flatlining r per se tells you nothing (at that time).

In practice, planetary radar only worked out to Saturn due to design limits, and that was Arecibo (RIP). Inside of Saturn, planets are known fairly well, I’d say. Small bodies are points of light in all but the most humongous (optical) telescopes, so James K’s process works fine; we simply spend two or more lunations to establish an overall orbit (depending on how many decimal places are necessary). Few things compel us to hurry any more than this; the exception is a small body that, after a first, crude orbit solution, may be Earth-crossing, and somewhat close at that.

In that case, we would resort to precovery- searching archived images for the object, to extend the observing arc, and put more decimal places on the orbit solution. If even that still gives a close pass by Earth, then the schedule of a planetary radar might be overruled to observe the target. Combining both the optical solution (precise in 2D, plane-of-sky) and the radar values (lousy in 2D, but precise in that third D) will give plenty of decimal places on the next orbit solution, and thus the apsides. Of course, these small bodies are not planets (the term “minor planet” is now deprecated) but should some hollywood planet magically appear, that’s what we’d do.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .