I need to calculate the mean anomaly of an elliptic orbit at a specific point in time.
I found two different formulas for $M(t)$ and I'd like to know how they correlate and if they result in different ouputs.

The first formular is:
$M(t) = \frac{2 \: \pi}{T}\:(t-t_o)$
and its taken from this book.

The second formular is:
$M(t) = M_0 + \Delta t \sqrt{\frac{\mu}{a^3}}$
taken from this document.

$T$... Orbital Period
$t_o$... Starting point in time (epoch)
$t$... Point in time
$\Delta t$... Elasped time: $t-t_0$
$\mu$... Standard gravitational parameter $\mu=GM$
$a$... Semi-major axis
$M_0$... Mean anomaly at epoch


1 Answer 1


Kepler's Third Law including the constants of proportionality is:

$$GM T^2 = 4\pi^2 a^3$$

Substituting $\mu = GM$, this can be rearranged to give:

$$\frac{2\pi}{T} = \sqrt{\frac{\mu}{a^3}}$$

Which lets you rewrite your first formula as your second one and vice-versa.

Note that in the first formula the reference epoch $t_0$ is assumed to be the time of periapsis, which corresponds to the case $M_0 = 0$. Your definition of $\Delta t$ should also read $\Delta t = t - t_0$, i.e. time since the reference epoch.

  • $\begingroup$ what about $M_0$? Does the first formula assume that $M_0$ equals zero? $\endgroup$
    – Laila
    Commented Oct 3, 2020 at 13:10
  • 4
    $\begingroup$ @Laila The first formula assumes $t_0$ is the time of periapsis passage, where the mean anomaly is zero. $\endgroup$ Commented Oct 3, 2020 at 13:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .