Given the Keplerian elements for an object for current time. The object is on ellyptic orbit.

How to calculate, at what time the object will be on one of the orbital nodes? That's required for inclination change maneuver.


1 Answer 1


Using the definition of Keplerian elements: enter image description here

Then it can be seen that the angle between the ascending- and descending node and the periapsis will be equal to $-\omega + 2\,k\,\pi$ and $\pi-\omega + 2\,k\,\pi$ for all integers $k$ respectively. The first node that you will come across will be the first of these two values which becomes larger than $\nu$ as you increase $k$. The time can be calculated with Kepler's equation,

$$ M = E + e \sin E, \tag{1} $$

but for this you do first need to the convert these angles (in radians) into the eccentric anomaly. Another way is rewriting $(1)$ into the original angles, which yields,

$$ t=\sqrt{\frac{a^3}{\mu}}\left[2\tan^{-1}\left(\frac{\sqrt{1-e^2}\tan{\frac{\nu}{2}}}{1+e}\right)-\frac{e\, \sqrt{1 - e^2}\sin{\nu}}{1 + e\cos{\nu}}\right] \tag{2}. $$

Both equations can give you the time since periapsis passage, which is positive when $E$ or $\nu$ is between $0$ and $\pi$ and negative when between $-\pi$ and $0$. In both cases you need to take the difference between the two times (at the current position and at the node). However you do have to be a bit careful when crossing $\pm\pi$ radians, because you might have to add or subtract the total period of the orbit once to get the actual time till node.

  • $\begingroup$ The first node that you will come across.. Should I compare my true anomaly with the angles you showed? $\endgroup$ Commented May 1, 2017 at 6:19
  • 2
    $\begingroup$ @TarlanMammadzada Yes, because your true anomaly is your current position. However if you want to be perform an inclination change you might want to do that at the node with the highest altitude and not necessary the first node that you come across. And a last side note if your inclination change is sufficiently large it might be more efficient fuel wise to perform a bi-elliptic transfer. $\endgroup$
    – fibonatic
    Commented May 1, 2017 at 10:08

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