# How do different formulas for calculating the mean anomaly of an elliptic orbit correlate?

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.

Inputs:
$$T$$... 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

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.

• what about $M_0$? Does the first formula assume that $M_0$ equals zero? – Laila Oct 3 '20 at 13:10
• @Laila The first formula assumes $t_0$ is the time of periapsis passage, where the mean anomaly is zero. – David Hammen Oct 3 '20 at 13:47