I am moving to live on the Moon, I have a solar panel that only makes power when the sun shines on it. I will need to charge batteries to give me power during the lunar night. But I can't figure out how long the day and the night are.

I googled and found several different answers.

My homestead has a good solar location, no mountains to cast shadows, and averages 0 (zero) cloudy days per year.

  • 2
    $\begingroup$ 29 days, 12 hours and 44 minutes, that is 29.53 days, that is only 1 per mille more than 29.5 days. $\endgroup$ – Uwe Nov 8 '19 at 18:22
  • 3
    $\begingroup$ It would be interesting if someone made a heatmap of the moon with different day lengths in different colors. iirc, there's places on the moon that are in permanent night and some that are in permanent light $\endgroup$ – Dragongeek Nov 8 '19 at 20:45
  • 1
    $\begingroup$ (seems) It Never Rains in Southern Mare Imbrium $\endgroup$ – uhoh Nov 9 '19 at 23:37
  • $\begingroup$ why not few mirrors to reflect light to the moon 24x7x365? $\endgroup$ – Thufir May 4 '20 at 22:15
  • 1
    $\begingroup$ @Dragongeek The moon has the same day length everywhere. The places of permanent light and night are found on the poles on the top of mountains and in crevasses respectively. $\endgroup$ – Ingolifs Jul 10 '20 at 0:48

This may belong to Astronomy SE, but the $29.5$ Earth day figure, or more accurately the time in the third reference, is what you should be planning on when you or at least your instruments go to the Moon. This represents the actual cycle between daylight and darkness, the solar day. When one clicks on the references cited in the question, the first and third figures are specifically called the solar day; the others are not.

It also represents the "daily" cycle in temperature -- a cycle which, of course, has wide variation because there is practically no atmosphere to transfer heat around the Moon's surface. Space.com renders this range from $+127°C$ down to $-173°C$.

Edit in response to a comment:

The 27.3 day figure given in the fourth reference in the question is the sidereal day, using the distant stars as reference instead of the Sun. Any object in orbit around the Sun shows a difference between the two day measures because the object is curving around the Sun as it is also rotating. For a combination of prograde rotation and prograde orbital motion around the Sun, as with most Solar System bodies, the relationship is

$\dfrac{1}{\text{Mean solar day}}=\dfrac{1}{\text{Sidereal day}}-\dfrac{1}{\text{Orbital period around the Sun}}$

If we are given the sidereal day as $27.3$ Earth days and the solar orbital period of $365.2$ Earth days (the Moon tags along with the Earth) the equation ends up producing the $29.5$ Earth days quoted above. We can also reverse this process: If we have measured a solar day of $29.5$ Earth days and a solar orbit is $365.2$ days, solving for the sidereal day gives us our $27.3$ day period.

In any event, systems depending on Sunlight input such as solar cells or temperature sensitive astronauts and electronics operate around the solar day. Hence about $29.5$ Earth days on the Moon.

  • 1
    $\begingroup$ The 28.5 day figure may be just a typo, key 8 instead of 9. $\endgroup$ – Uwe Nov 9 '19 at 11:33
  • 1
    $\begingroup$ @uhoh let me know if this works for you. I am really still somewhat befuddled. $\endgroup$ – Oscar Lanzi Nov 10 '19 at 1:01
  • $\begingroup$ @JamesJenkins OscarLanzi has added a further explanation, have another look? $\endgroup$ – uhoh Nov 10 '19 at 13:27
  • $\begingroup$ of possible interest: Is there a timekeeping word for the orbit of a moon? $\endgroup$ – uhoh Feb 16 '20 at 1:17

Maybe not directly related to the question, anyway this site allows calculating "what time it is" in a specific location on the Moon: http://win98.altervista.org/space/exploration/moon/moontime.html

In this page, the moon day duration (29.53 days) is divided into 24 moon-hours; when sunrise terminator reaches specified point, local time will be 06:00; when sunset terminator reaches the location local time is 18:00.

But for a more precise indication of sun position of the Sun in the Moon sky, you can use this page to get the local Azimuth and Elevation of Sun using NASA Horizons data:


You must specify:

  • COMMAND: Sun
  • CENTER: coord@301 (specific location on Moon surface)
  • SITE_COORD: longitude, latitude and altitude of site on the Moon
  • TABLE_TYPE: Observer

For Vikram lander landing site (-70.90267, 22.78110, 0) this will result in this link, which gives:

  • sunrise: 2019-Nov-03
  • noon: 2019-Nov-10, Elevation = 17.8104°
  • sunset: 2019-Nov-17

You can also get just rise/noon/set times by specifying YES for field R_T_S_ONLY and "1m" for STEP_SIZE, getting this link and these results:

 Date__(UT)__HR:MN, , ,Azi_(a-app), Elev_(a-app),
 2019-Nov-03 05:00,*,r, 86.6565, -0.2681,
 2019-Nov-10 09:00,*,t,359.7476, 17.8104,
 2019-Nov-17 13:44, ,s,272.5870, -0.2713,

Note that sun is very low above the horizon even at noon because this specific location is close to south pole (70° latitude South).

  • $\begingroup$ New related question An eternal lunar day? $\endgroup$ – James Jenkins Nov 12 '19 at 12:29
  • 2
    $\begingroup$ There was one of Murphy's rules: Measurements are given using very strange units, for instance speed in furlongs per fortnight. $\endgroup$ – Uwe Nov 12 '19 at 14:54
  • 1
    $\begingroup$ Rods and furlongs: at least he did not use miles, or we'd wonder if it was in geographical miles, nautical miles, British or French miles. (British miles were different from geographical miles till the early-mid 19th century. Don't know when that faded out.) $\endgroup$ – SpaceInMyHead Nov 15 '19 at 22:06
  • $\begingroup$ If strange units are what you want, nothing will ever beat pounds per base box in tinplating. $\endgroup$ – Oscar Lanzi Jul 9 '20 at 20:31
  • $\begingroup$ Degrees MacMichael (°McM) ? (a unit of viscosity) $\endgroup$ – Organic Marble Jul 9 '20 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.