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I've been wondering whether it's possible to change a satellite's eccentricity without changing its semi-major axis. (I'm assuming instantaneous changes in velocity.) At first I thought that this would be impossible, because changing eccentricity necessarily changes the size of an ellipse when trying to match the before and after ellipse at the point of the maneuver, and changing the size of the ellipse changes the semi-major axis.

But then I realized that I was picturing ellipses with the same orientation. If I change to a different ellipse of the same semi-major axis but different eccentricity than the one I'm currently on, if I rotate the orientation of that second ellipse (change argument of periapsis of the orbit) and don't change the orbital plane, I should be guaranteed to get the two ellipses to intersect at any arbitrary point along the first ellipse, right?

Phrased a different way, if I don't require that argument of periapsis remains fixed (but inclination and RAAN do), I should be able to perform an instantaneous maneuver into an orbit with different eccentricity but same semi-major axis as my current one, correct?

update: I'm looking for a general solution (being able to change eccentricity at any point in the orbit), not just at two points in the orbit.

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    $\begingroup$ Any non-instantaneous burn that remains perpendicular to the direction of motion throughout, directed within the plane of the orbit, will change eccentricity without affecting semi-major axis, inclination, or direction of the ascending node. $\endgroup$
    – notovny
    Commented Apr 11 at 13:48
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    $\begingroup$ That is, you just need to change the direction of the velocity without changing its magnitude. physics.stackexchange.com/a/676872/123208 $\endgroup$
    – PM 2Ring
    Commented Apr 11 at 13:55
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    $\begingroup$ @uhoh I'm looking for a general solution (being able to change eccentricity at any point in the orbit), not just at two points in the orbit. $\endgroup$ Commented Apr 12 at 2:52
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    $\begingroup$ @NeutronStar The comment by PM 2Ring is the general solution. You need to keep the magnitude of the before and after velocities the same. $\endgroup$ Commented Apr 12 at 11:30
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    $\begingroup$ @notovny Burning perpendicular to the direction of motion does change the semi-major axis length, at least by a bit. Whether finite or impulsive, a burn needs to have at least a slight retrograde component to keep the semi-major axis length unchanged. $\endgroup$ Commented Apr 12 at 11:37

1 Answer 1

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Is it possible to change a satellite's eccentricity without changing semimajor axis (instantaneous maneuver paradigm)?

Sure. From the vis-viva equation, $v^2 = \mu\left(\frac2r-\frac1a\right)$, you simply need to keep kinetic energy the same before and after the instantaneous $\Delta$v to keep the semi-major axis length the same. In other words, $$\|\vec v + \Delta \vec v\|^2 = v^2\tag{1}$$ It helps to break the instantaneous $\Delta$v into a component parallel (or anti-parallel) to the original velocity vector and a component orthogonal to the original velocity vector: $$\Delta \vec v = \Delta v_{\|}\hat v + \Delta v_{\perp}\hat v_{\perp}\tag2$$ where

  • $\hat v$ is the unit vector pointing along the initial velocity vector,
  • $\Delta v_{\|}$ is the component of the instantaneous $\Delta \vec v$ parallel (or anti-parallel) to the initial velocity vector,
  • $\hat v_{\perp}$ is a unit vector orthogonal to the initial velocity vector, and
  • $\Delta v_{\perp}$ is the component of the instantaneous $\Delta \vec v$ orthogonal to the initial velocity vector.

With this, equation (1) becomes $(v+\Delta v_{\|})^2 + \Delta v_\perp^2 = v^2$ or $$\Delta v_{\|}^2 + 2 \Delta v_{\|} v + \Delta v_\perp^2 = 0\tag3$$ This is a quadratic equation in $\Delta v_\|$ with solutions $$\Delta v_\| = -v\pm\sqrt{v^2-\Delta v_\perp^2}\tag4$$ Both solutions have an imaginary component if $\Delta v_\perp > v$. In other words, you can't meet the objective of keeping $a$ the same for ridiculously large burns orthogonal to the original velocity vector. If $\Delta v_\perp = 0$, the two solutions are $\Delta v_\| = 0$ (i.e., no instantaneous $\Delta$v) and $\Delta v_\| = -2v$ (i.e., completely reversing direction). I'll assume you want the smaller burn, so equation (4) becomes $$\Delta v_\| = -v+\sqrt{v^2-\Delta v_\perp^2}\tag5$$ Note that in the range $0<\Delta v_\perp\le v$, the component along the velocity vector is always negative: The burn will be at least slightly retrograde. (And, as noted above, if $v_\perp > v$, the goal is unacheivable.)

In the case that $\Delta v_\perp \ll v$, equation (5) becomes $$\Delta v_\| \approx -\frac{\Delta v_\perp^2}{2 v}\tag6$$

What else needs to "give"?

That depends on whether or not the instantaneous $\Delta$v is in-plane. If $\hat v_\perp$ is in plane, the orbital plane won't change, but the argument of periapsis and true anomaly (and hence also the mean and eccentric anomalies) will change. If $\hat v_\perp$ has an out of plane component, the burn will also change the argument of ascending node and the inclination as well.

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    $\begingroup$ Note that for any in-plane instantaneous $\Delta$v, the argument of latitude (the angle between the ascending node and the current position, also the sum of the argument of periapsis and the true anomaly) will remain unchanged because the right ascension of ascending node hasn't changed due to the in-plane nature of the maneuver. $\endgroup$ Commented Apr 12 at 12:10
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    $\begingroup$ +1 for you're godlike mastery of the SE equations feature! (and also a great answer) $\endgroup$
    – phil1008
    Commented Apr 23 at 1:02
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    $\begingroup$ @phil1008 When I worked (I'm now retired), my preferred tool for documentation was LaTeX. You can see some of that work here; look at the docs directories in each of the models subdirectories. Every model is documented. My name is on several of those documents, and my influence is on all of them. I led that project for a long time. I also spread the LaTeX religion to other projects. $\endgroup$ Commented Apr 23 at 11:21
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    $\begingroup$ Thanks for the link @DavidHammen! I didn't know Trick and JEOD was on github. $\endgroup$ Commented Apr 23 at 12:41
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    $\begingroup$ @OrganicMarble There's one chunk of JEOD that remains subject to ITAR (and should remain subject to ITAR), but I am very happy that NASA deemed the vast majority of JEOD to be NASA open source. It was my baby! $\endgroup$ Commented Apr 23 at 14:21

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