# How long does one orbit around the Moon take in a typical parking orbit for a stage a lander is going to meet?

I have been pondering the relative advantages of placing a base near the Moon's equator, or at the poles. The poles have been defended as having a superior business case because of deposits of volatiles, and the peaks of (mostly) eternal light. I lean more towards the equator because:

• you can place it between a mare and a highland and thus have access to the resources of both

• you might find a mighty handy lava tube around there

• and if several fueled stages or fuel depots (let's say fuel depot, shall we?) have been parked around the equator for handy rendezvous, there would be launch opportunities several times a day.

But how many times a day, supposing that the base is right on the equator? Which is to say, how long would it take for the fuel depot to make a full orbit?

And how long would the gaps be when the depot's orbital inclination means it isn't passing overhead for a while, if a base was, say, 20°N, where the Apollo 17 mission landed, which seems like a nice place to put a base according to the parameters mentioned above?

How many stages or depots would be needed to be in orbit in order for one to be sufficiently well aligned for rendezvous, once it passes overhead, at any time?

• When picking an actual orbit around the Moon, you also have to consider the issue of mascons. Only a few orbital inclinations around the Moon are stable. – user Apr 23 '19 at 15:21

For the first part of your question, and assuming an equatorial orbit, it depends on the orbital altitude, but maths are simple enough to start with:

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

Where $$a$$ is semi-major axis (in our case Moon's equatorial radius of 1,738.14 km, plus mean orbital altitude above it), and $$\mu$$ is standard gravitational parameter of the central body (in our case 4,903.86 km³/s², someone mistyped something in that Wikipedia page) that is a product of gravitational constant $$G$$ (6.674×10¹¹ N⋅m²/kg²) and central body's mass $$M$$ (7.3477×10²² kg).

So for an orbital altitude of 200 km above the Moon, our orbital period $$T$$ is 127 minutes and 34 seconds. But, during this time, the Moon also rotates 1.08°, so we have to do this a bit differently. We have to find our satellite's sidereal time. Since we know it's not off by much to our satellite's orbital time $$T$$, we can simplify this and say that the Moon's rotation is increasing our sidereal time for prograde orbits, and decreasing it for retrograde orbits (we know this because it rotated on its axis for less than 180° during one of our satellite's orbits). So we can say that:

• Our satellite's orbital speed ($$\sqrt{\mu/r}$$) is 1,590.53671 m/s, so its angular velocity $$\omega = \Delta\phi/\Delta t$$ is 0.000820651095 rad/s
• Moon's orbital period is 29.530589 days, or in $$\omega$$, 2.4626008 × 10-6 rad/s and we'll add it to our satellite's orbital speed for retrograde orbit, and subtract it from it for prograde orbit

So, to complete one sidereal time of a 200 km orbit around the Moon, and for an equatorial station where this sidereal time is observed, and assuming circular orbit, we have:

In case you were wondering so far why I did it the long way around, I think you started appreciating it. Or, you better have, because it gets even more complicated for inclined and/or elliptical orbits. I won't do exact maths for it, because we also have to consider orbital node (or more specifically its phase, i.e. where in the orbital position is the satellite when its ground track aligns with our off-equatorial longitude) and time of insertion into orbit (our epoch), as well as decide on what range is sufficiently close for our case (likely different margins for longitudinal drift and phase shift), but assuming no orbital precession (nodal due to non-oblateness of the central body, apsidal due to eccentricity of non-circular orbits, and so on), equally spread constellation, and circular horizon within which we still count an overshot as our conjunction, you should be able to relatively simply™ integrate it (you'd probably want to use Lagrange interpolating polynomial). Someone else might be more merciful and post a wall of formulae for you, but I won't be so cruel now (also on myself, because it's been a while since they made me do it). Have fun!

This is just a supplement to the accepted answer.

For a low orbits where the semi-major axis is close to the radius of the central body, the period is related to the average density of the body and unrelated to it's size.

So a low orbit around a spherical asteroid (which there usually aren't) made of a mixture of rock and iron (which there usually aren't) will be roughly 90 minutes just like LEO, even if it's only 1 km in diameter.

Starting with

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

and

$$\rho = \frac{m}{\frac{4}{3} \pi R^3}$$

$$\frac{1}{m} = \frac{1}{\frac{4}{3} \rho \pi R^3}$$

where $$\rho$$ is the density and $$R$$ is the body's radius. Then:

$$T = 2\pi\sqrt{\frac{a^3}{G\frac{4}{3} \rho \pi R^3}}$$

If you first set $$a=R$$ you get

$$T = 2\pi\sqrt{\frac{3}{4G\pi \rho}}$$

and finally

# $$T = \sqrt{\frac{3 \pi}{G\rho}}$$

So they are related by inverse square root.

According to Wikipedia, the mean density of the moon (3.344 g/cm^3) is 0.61 of Earth, so a low moon orbit should be $$1/\sqrt{0.61}$$ or 28% longer than that of Earth, based only on density.

Testing:

If you scale the altitudes to the radii (1738 and 6378 km) as well, say 109 km Moon orbit and a 400 km Earth orbit, it should agree exactly. Let's try it!

Wikipedia gives the standard gravitational parameter $$GM$$ for the Moon and Earth as 4.905E+12 and 3.986E+14 m3/s2.

Moon:

$$2\pi\sqrt{\frac{(1738000+109000)^3}{4.905 \times 10^{12}}} = 118.7 minutes$$

$$2\pi\sqrt{\frac{(6378000+400000)^3}{3.986 \times 10^{14}}} = 92.6 minutes$$

and voila! The period of the Moon's low 109 km orbit is 28% longer than the Earth's 400 km orbit!