Basically I want to calculate the elapsed time since passing the last ascending node crossing for a satellite in Earth orbit (assuming two-body motion).

If the orbit is circular, then we can simply use the argument of latitude in terms of true anomaly and find time since perigee. However when eccentricity is > 0 then it doesn't work due to the rotation of the Earth (it has rotated so the intersection of the inertial equator and the orbital plane has moved as well).

Does anyone know the exact equation that would give me the value that I want?

  • 2
    $\begingroup$ There is no exact equation, this is essentially solving Kepler's equation in reverse, which itself is an iterative process. You can either solve for an inverse of Kepler's equation, or iteratively search for the crossing point. $\endgroup$ Apr 11, 2022 at 3:15
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    $\begingroup$ Also, neither longitude of periapsis, nor the direction of the ascending node change with the planet's rotation, unless you're dealing with atmospheric friction, perturbation due to the oblateness of the planet, or other perturbations. In the two-body Keplerian simplification, you work out the true anomaly of the ascending node from the argument of periapsis, find solve Kepler's equations to find time since periapsis for that and for your chosen true anomaly, and do the appropriate subtraction. $\endgroup$
    – notovny
    Apr 11, 2022 at 11:31

1 Answer 1


If you aren't dealing with some sort of perturbation of the orbit (from atmosphere, oblateness of the Earth, the satellite thrusting), and are not dealing with extreme timescales, and are only doing a two-body Keplerian/Newtonian simulation, the equator of the earth keeps the same orientation, as does the orbital plane of the satellite.

In that situation, you can completely ignore the rotation of the Earth as it has no effect on any of the standard orbital parameters.

Parameter Symbol
Semi-major axis of the orbit $a$
Orbital Eccentricity $e$
Argument of Perigee of orbit $\omega$
Earth's Gravitational Parameter $\mu$
True anomaly of the satellite's current position $f_t$

Calculations below will use radians for all angular measurements.

1. Find the True Anomaly of the Ascending Node, $f_{\Omega}$.

The Argument of Perigee $\omega$ is simply the angle, measured in the direction of travel around the orbit, from the ascending node, through the center of the body being orbited, to the periapsis. As a result: $$f_{\Omega} = 2\pi - \omega$$

2. Calculate the Eccentric Anomaly $E$ for both true anomalies

The Eccentric anomaly is a quasi-angular parameter used in Kepler's equations to convert true anomaly angle into the Mean anomaly. In this direction, the calculation uses closed-form equations.

$$0 <= f < \pi$$ $$\pi <= f < 2\pi$$
$$E = \arccos\left(\frac{e + \cos(f)}{e \cos(f) + 1}\right)$$ $$E = 2\pi - \arccos\left(\frac{e + \cos(f)}{e \cos(f) + 1}\right)$$

3. Calculate the Mean Anomaly $M$ for both eccentric anomalies.

Mean anomaly is another quasi-angular parameter which represents the angle swept out by a hypothetical object in a circular orbit with the same semi-major axis since periapsis passage.

$$M = E - e \sin E $$

4. Calculate the Mean Motion $n$ for the orbit.

Mean motion is the average angular velocity of an orbit with the semi-major axis $a$. $$n = \sqrt{\frac{\mu}{a^3}}$$

5 Using the difference between the Mean Anomalies, calculate time since ascending node, $t$

If the mean anomaly of the satellite is $M_s$ and the mean anomaly of the ascending node is $M_{\omega}$:

$$M_s >= M_{\omega}$$ $$M_s< M_{\omega}$$
$$t = \frac{M_s - M_{\omega}}{n}$$ $$t = \frac{ 2\pi + M_s - M_{\omega}}{n}$$
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    $\begingroup$ great answer, but the numerator for eccentric anomaly should be e + cos(f), not e*cos(f). $\endgroup$
    – Hagbeard
    Jan 6, 2023 at 19:32

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