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I've been trying to make sense of the formula for an orbital inclination change for an elliptic orbit. This is for an orbit that is not altered in any way apart from its inclination.

The equation from Wikipedia is as follows:

$$\Delta v_i = {2\sin(\frac{\Delta{i}}{2})\sqrt{1-e^2}\cos(\omega+f)na \over {(1+e\cos(f))}}$$

$\omega$ is the argument of periapsis, in other words the angle between the direction of the shortest side of the ellipse and the line where the two orbital planes intersect.

$f$ is the true anomaly. This is the angle of how far around the orbiting object is relative to the shortest side.

I don't get why the value $f$ appears in this equation

There are only two possible places where you can perform a burn to change an orbital inclination, and those are the ascending and descending nodes, the opposite sides where the orbit intersects with the destination orbital plane. The angles $\omega$ and $\omega + 180$ ought to be sufficient to denote where these two nodes occur on the orbit.

So why then is $f$ in the equation? Have I missed something? $f$ isn't used to describe an orbit, it is used to describe how far along an object is in an orbit. Since there are only two possible locations where an inclination change burn can be made, it seems irrelevant to me to have this value in the equation at all.

EDIT: Notovny's comment and deleted answer suggests that the wikipedia eqn might just be referring to a situation where any inclination change is the goal. In particular, I was after an equation that matches an inclination of a known object. Naturally, this can only be done at the ascending and descending nodes. That suggests to me that the equation I want is simply one where $f$ is zero. Could anyone clear this up for me?

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  • $\begingroup$ I suspect that the provided equation only cares about keeping the shape of the final orbit (semimajor axis, eccentricity), and the final inclination, but presumes you chose a True Anomaly that puts you in the orbital plane you selected, allowing longitude of the ascending node (and possibly argument of periapsis) to change to what's needed for that. $\endgroup$
    – notovny
    Sep 27 at 12:43
  • $\begingroup$ @PM2Ring Oh, is Latex and MathJax the same thing? $\endgroup$
    – Ingolifs
    Sep 27 at 21:01
  • $\begingroup$ "MathJax can display mathematical notation written in LaTeX or MathML markup. Because MathJax is meant only for math display, whereas LaTeX is a document layout language, MathJax only supports the subset of LaTeX used to describe mathematical notation". en.wikipedia.org/wiki/MathJax $\endgroup$
    – PM 2Ring
    Sep 27 at 23:52
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    $\begingroup$ @PM2Ring Cool, thanks. Now I know I can copy paste the maths directly from Wikipedia rather than rewrite it each time. $\endgroup$
    – Ingolifs
    Sep 28 at 0:40

2 Answers 2

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Let's see.

What if we have a circular orbit ($e = 0$) with no argument of periapsis ($\omega = 0$), normalised to a unit mass, unit radius, so $a = 1$ and $n = 1$?

The formula then simplifies to

$$\Delta v_i = 2\sin\left(\frac{\Delta{i}}{2}\right)\cos(f)$$

Which says any inclination change is free if our true anomaly is 90º. That can't be true.

Indeed, the "general" formula too suffers from this problem, whenever $\cos(\omega + f) = 0$, implying any orbit has two locations were inclination changes can be performed for free.

Looks to me like bogus unsourced material from 2006.

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This answer addresses the possibilities for altering inclination outside of the context of the particular equation in the OP.

In isolation it is incorrect to say that there are "only two possible places where you can perform a burn to change an orbital inclination".

It would be better to say that it is more efficient to do so at the nodes but still possible anywhere else if one accepts other side effects. The restriction of only two such places arises if one wants to change inclination without changing the right ascension of the ascending node.

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