The Moon orbits Earth at a semi-major axis of 384400 km, with its periapsis being 363300 km and apoapsis being 405500 km. (All figures from this NASA fact sheet.)

If the Moon orbited Earth at a constant velocity, the average distance would be 384,400 km. Unfortunately, as Kepler found out, celestial objects move faster near periapsis and slower at apoapsis.

This means that the Moon's actual average distance from Earth is slightly bigger than 384400 km, due to it having a lower orbital velocity when farther from Earth.

So, what is the Moon's actual average distance from Earth? Is there a formula (I'm assuming including calculus) that would give a solution to any two-body-system?

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    $\begingroup$ All averages, especially averages of continuous quantities, require a weight function. This question seems to come down to treating the mean distance weighted by time as the "actual" average, rather than one particular average. And that's not even getting into things like arithmetic mean versus other types of mean such as geometric. $\endgroup$ May 4, 2022 at 21:41

3 Answers 3


Mean distance averaged over time for any Keplerian orbit is $a(1+\frac{1}{2}e^2)$, where $a$ is the semi-major axis and $e$ is the eccentricity. Using your NASA fact sheet, I get about 384,979 km for the Moon (which is pretty close to wikipedia's value of 385,000 km). This, as expected is a bit larger than $a$ at 384,400 km.

Note: I wouldn't call this the "actual" average distance because there is more than one way to compute an average.

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    $\begingroup$ Indeed - beware the word "actual" in any context. $\endgroup$
    – J...
    May 4, 2022 at 13:41
  • 1
    $\begingroup$ Indeed - one should also beware of the word "average". This answer gives the mean distance. Is the median distance the same or different? $\endgroup$
    – Dale M
    May 5, 2022 at 2:27
  • $\begingroup$ @DaleM I think the median distance over time will be different if you define the median distance $\tilde{d}$ and the distance between the Earth and Moon as $d$ such that $d<\tilde{d}$ for exactly half the period of the orbit. This is because distance as a function of time is non-linear. How to calculate such a $\tilde{d}$ might be another interesting Astronomy SE question! (hint hint) $\endgroup$
    – Connor Garcia
    May 5, 2022 at 16:05
  • $\begingroup$ Do you have examples of different, meaningful averages for the distance between Moon and Earth? $\endgroup$ May 5, 2022 at 17:28
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    $\begingroup$ @EricDuminil Yes, there are three quite nice examples from wikipedia below: astronomy.stackexchange.com/questions/49258/…. Note that all three are the same value for circular orbits. $\endgroup$
    – Connor Garcia
    May 5, 2022 at 17:57

Lunar theory is complicated. ;) The Moon's orbit around the Earth is only approximately a constant Keplerian ellipse. As I mentioned here, the Moon's orbit around the Earth-Moon barycentre has an eccentricity of ~0.0549 and it's quite dynamic, with relatively short apsidal and nodal precession cycles, primarily due to perturbation by the Sun.

The size of the Moon's orbit varies over the course of the year. Here's a daily plot (courtesy of JPL Horizons) of the Earth-Moon distance for 13 (anomalistic) lunar months.

Earth-Moon distance

The mean distance over that period is roughly 384,975 km. The maximum monthly variation occurs when the Earth-Moon system is near perihelion or aphelion.

That plot was created using this Sage / Python script, running on the SageMathCell server. There are brief instructions on using an earlier version of that script here.

As Connor mentions, there are a few ways to define the average distance. Here's some info from Wikipedia Semi-major and semi-minor axes, recycled from a previous answer:

It is often said that the semi-major axis is the "average" distance between the primary focus of the ellipse and the orbiting body. This is not quite accurate, because it depends on what the average is taken over.

  • averaging the distance over the eccentric anomaly indeed results in the semi-major axis.

  • averaging over the true anomaly (the true orbital angle, measured at the focus) results in the semi-minor axis, $b=a{\sqrt {1-e^{2}}}$

  • averaging over the mean anomaly (the fraction of the orbital period that has elapsed since pericentre, expressed as an angle) gives the time-average, $a\left(1+{\frac {e^{2}}{2}}\right)$.

In the above equations, $e$ is the eccentricity of the ellipse.


Your guess on the requirement for calculus is correct. What you need to do is integrate the moon's distance over one orbit to find an average distance.

Starting from the equation below, the Moon's (or really any orbiting body's) orbital distance, $r$, as a function of azimuthal angle $\theta$ is given by the Orbit Equation.

$$r(\theta) = \frac{L^2}{m^2\mu}\frac{1}{1+e\cos(\theta)}$$

In this equation, the variables have the following definitions:

  • $r$ - The distance the body is from the body it is orbiting.
  • $\theta$ - The angle of the body at a given point in the orbit, as measured from the periapsis. This means $\theta = 0$ at periapsis and $\theta = \pi$ at apoapsis.
  • $L$ - The orbital angular momentum of the body. This is simply given by $mrv$ where $m$ is the mass, $r$ is the distance at any point in the orbit, and $v$ is the orbital velocity at that point. Because the orbital angular momentum is conserved, any point along the orbit will give the same value. The fact sheet you linked to gives both the apoapsis distance and the velocity at that point so you could calculate $L$ from this.
  • $m$ - As above, the mass of the orbiting body. Note that technically the formula for $L$ cancels with the $m$ in this equation, which is why you often see the $L^2/m^2\mu$ portion often written as $\ell^2/\mu$ where $\ell$ is the angular momentum per unit mass (i.e., just $rv$).
  • $\mu$ - The standard gravitational parameter of the system, given by $G(m_1 + m_2)$ where $m_1$ and $m_2$ are the masses of the two bodies involved (the Earth and the Moon in this case) and $G$ is the gravitational constant.
  • $e$ - The eccentricity of the orbit. Also on the fact sheet.

So there's a bit of work here, but you can find all the numbers and plug them in to the equation.

Now, here comes the calculus part. In order to find the average orbital distance, $\bar{r}$, you have to integrate $r(\theta)$ over an entire orbit. So the average becomes

$$\bar{r} = \frac{1}{2\pi}\int_0^{2\pi} r(\theta)\ d\theta = \frac{\ell^2/\mu}{2\pi}\int_0^{2\pi} \frac{1}{1+0.0549\cos(\theta)}\ d\theta$$

I have no interest in doing that nasty integral so I'll just plug it into a calculator for me. I find that this works out to the following:

$$\bar{r} = 1.00151\frac{\ell^2}{\mu}$$

This boils down to just calculating $\ell$ and $\mu$ now and plugging in. Using the fact sheet you linked in your question, I find the following values of $\ell = 3.922\times10^{11}\:\mathrm{m^2/s}$ and $\mu = 4.033\times10^{14}\:\mathrm{m^3/s^{-2}}$. Which means, when I plug those numbers into the equation, I get a mean orbital distance of

$$\bar{r} = 384\ 234\:\mathrm{km}$$

However, all of that was a lot of work to calculate something that could've been arrived at much easier. Remember above, I said that $L$ is conserved so the value is constant at any point in the orbit. Thus you can say $r_1v_1 = r_2v_2$ and because the fact sheet gives you the radius and velocity at apogee and the average velocity, you can easily calculate the average radius. You'll find the same value though.

And here again, we've done more work than we need. We integrated over the true anomaly to find the average orbital distance, but that's just the semi-minor axis which is $383\ 800\:\mathrm{km}$, not too far off from what I calculated. We could've just looked it up from the get go!

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    $\begingroup$ OP wants a time average, not an angular average, as far as I can tell. $\endgroup$
    – benrg
    May 4, 2022 at 2:48

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