As I know on Low Earth Orbit it takes around 90 minutes to orbit the Earth, but this changes with altitude. My question is at what altitude did the Apollo Command Module orbit the moon and how much time did it take?


2 Answers 2


Please note these are average periods, since the CSM did its maneuvering (DOI, descent orbit initiation) to make it easier for the LM to land on the Moon.

\begin{matrix} \text{Mission} & \text{Avg period, mm:ss} \\ \text{Apollo 8} & 121:01 \\ \text{Apollo 10} & 119:27 \\ \text{Apollo 11} & 119:00 \\ \text{Apollo 12} & 118:37 \\ \text{Apollo 14} & 117:31 \\ \text{Apollo 15} & 117:44 \\ \text{Apollo 16} & 117:57 \\ \text{Apollo 17} & 118:10 \end{matrix}

Sources: Apollo by the numbers. (http://history.nasa.gov/SP-4029/Apollo_18-01_General_Background.htm) Individual parameters (exact inclination, for instance) for some missions can also be found (not always reliable) at http://nssdc.gsfc.nasa.gov

The orbit altitudes were on the order of 60 nmi, or 110 km. See the links above for more details.


Interestingly the low circular orbit period about a body is independent of its size. It depends only on the density of the object.

$$T=2\pi\sqrt{a^3\over G m}$$

Let's say that the low circular orbit altitude is 6% of the radius of a sphere with average density $\rho$. Then:

$$T=2\pi\sqrt{3\left(1.06 r\right)^3\over{4\pi G\rho r^3}}$$


So low orbits around rocky bodies will be on the order of two hours, regardless of the size.

For Earth, $\rho=5.51\,\mathrm{g/cm^3}$, which gives a period of 92 minutes. For the Moon with $\rho=3.34\,\mathrm{g/cm^3}$, the period is 118 minutes. Mars' $\rho=3.93\,\mathrm{g/cm^3}$, giving 109 minutes. Tiny Ceres at $\rho=2.08\,\mathrm{g/cm^3}$ gives 150 minutes. (Dawn won't get that low.)

  • $\begingroup$ But, turns out, Dawn is getting darned close! $\endgroup$
    – Mark Adler
    Jul 22, 2018 at 16:20

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