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How to determine the orbital eccentricity of a planet, knowing the instantaneous orbital speed of it and the distance from the Sun to the planet at that moment?

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    $\begingroup$ You'll need more than just the speed; the direction is important as well. $\endgroup$ – Glorfindel Apr 16 at 21:12
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    $\begingroup$ @Glorfindel that's not the only way to work the problem. The OP is only asking for eccentricity which means we can use energy. So while there might be a bit of missing information, it would not have to be direction. Using the vis-viva equation you get the semi-major axis with $a = 1/(2/r - v^2/GM)$. All that's needed is one more bit of information, either a periapsis or an apoapsis, or a minimum or maximum velocity. $\endgroup$ – uhoh Apr 16 at 23:20
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    $\begingroup$ Here's a simple orbit sim I wrote in Python using Tkinter. It demonstrates that the period of an orbit is determined by its total energy. (The orbiting bodies don't interact with each other). So all orbits that pass through a given point with a given speed will all have the same period, it doesn't matter what direction they're heading in, unless of course they crash into the primary. :) As you can see, the eccentricity is a function of the initial heading direction. $\endgroup$ – PM 2Ring Apr 17 at 14:19
  • $\begingroup$ @uhoh, even for a point object in 2-dimensions, you most certainly need the velocity vector. Consider an object with speed |v| moving radially w.r.t. the sun vs. same object, same speed, moving tangentially. If you include the premise that you know you're at apogee or perigee, you've inherently specified the direction of motion. $\endgroup$ – Carl Witthoft Apr 17 at 17:10
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    $\begingroup$ @CarlWitthoft I was tempting to post something similar. Your argument shows that more data is needed. However, that extra data does not need to be the direction; some other parameters might suffice. $\endgroup$ – badjohn Apr 20 at 8:56
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How to determine the orbital eccentricity of a planet, knowing the instantaneous orbital speed of it and the distance from the Sun to the planet at that moment?

We are given three things:

  1. $v$ "instantaneous orbital speed" (a scalar)
  2. $r$ "distance from the Sun to the planet at that moment" (a scalar)
  3. $GM$ since the OP specifies "the Sun"

Let's see how far we can get with the vis-viva equation and look at the problem purely from an energy standpoint.

$$v^2 = GM(2/r-1/a)$$

becomes

$$a = 1/(2/r-v^2/GM).$$

Let's look at a numerical example. Say $r = 1AU \approx 1.5 \times 10^{11}$ meters, and $v = 25,000$ m/s.

We know then that the semimajor axis of the orbit must be 0.773 AU. If the orbit reaches 1AU, the velocity will be 25,000 m/s. If it reaches 1.1 AU (which it might or might not, depending on eccentricity) it's velocity will be 21,500 m/s.

We're almost there but not quite. We need one more piece of information.

The periapsis and apoapsis $r_p$ and $r_a$ are related to the semimajor axis and eccentricity by

$$r_p = (1-e)a$$ $$r_a = (1+e)a$$

So for example if you had mentioned that the apoapsis was 1.1 AU, then we could calculate eccentricity from

$$e = r_a/a -1$$

and get $e = 0.4228$ AU and a periapsis $r_p = 0.387$ AU.

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    $\begingroup$ That looks fine. But of course you may not know either the periapsis or apoapsis. And if you know both of them, the eccentricity calculation is simple. ;) $\endgroup$ – PM 2Ring Apr 18 at 0:30
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You can't do that. The object has 6 degrees of freedom and hence needs 6 variables to describe it. A common method is to use the orbital elements given here. A description is given by parameters such as eccentricity, semi-major axis and inclination.

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    $\begingroup$ A 2D planar Keplerian orbit doesn't need 6 DOF, and the OP is only asking for eccentricity which means we can use energy. So while "You can't do that." might be true, I think your reasoning is overstated. Using the vis-viva equation you get the semi-major axis with $a = 1/(2/r - v^2/GM)$. All that's needed is one more bit of information, either a periapsis or an apoapsis, or a minimum or maximum velocity. You don't need a full-blown 3D Keplerian orbit. $\endgroup$ – uhoh Apr 16 at 23:19
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    $\begingroup$ Thats right. I was lazy in my answer. Thanks for your comment. $\endgroup$ – jmh Apr 16 at 23:21

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