I know about the equations to describe the orbit of a moon around a planet. I know the moon's semi-major axis and eccentricity, and the same for its host world with the star they orbit.

Is there any equation that tells me how much of the moon is illuminated at night, and possibly how bright, as seen from the planet?


3 Answers 3


The Moon phases can be defined by the phase angle between the Sun, Moon and Earth; for example, at 0°, the Moon is defined as full, and at 180° it is defined as new. If you want to know how bright the Moon is at a given angle, we would use the phase angle to find the apparent and absolute magnitudes of the Moon.

Absolute magnitude, when referring to illuminated objects (objects that do not produce their own visible light), simply means their apparent magnitude if viewed from 1 AU away. This means it is almost entirely dependent on the object's phase angle. Right now, you're asking about how bright the Moon would seem to a person on Earth, so we'll find the apparent magnitude. The formula to find an illuminated object's apparent magnitude (in the Solar System), if we know its absolute magnitude $H$, is:

$$m = H + 2.5 \log_{10}{\left(\frac{d_{BS}^2 d_{BO}^2}{p(\chi) d_0^4}\right)}\!\,$$

Where $d_0$ is 1 AU, $\chi$ is the phase angle (in radians) and $p(\chi)$ is the phase integral (integration of reflected light). $d_{BO}$ is the distance between the observer and the body, $d_{BS}$ is the distance between the Sun and the body, and $d_{OS}$ is the distance between the observer and the Sun. This formula probably looks pretty scary, but it can be simplified with some approximations. First, we can approximate the phase integral as this: $$p(\chi) = \frac{2}{3} \left( \left(1 - \frac{\chi}{\pi}\right) \cos{\chi} + \frac{1}{\pi} \sin{\chi} \right)$$ Where $\chi$ is the phase angle, in radians. In the Moon's case, we can set $H_{Moon} = +0.25$ (this is the absolute magnitude during a full moon), $d_{OS} = d_{BS} = 1$ AU and $d_{BO} = 0.00257$ AU. Now we get the formula:

$$m_{Moon} = 0.25 + 2.5 \log_{10}{\left(\frac{0.00257^2}{p(\chi)}\right)}$$

So now, we've got a formula that approximates the apparent magnitude of the Moon at any given phase angle. However, even though this gives a close approximation, it is not 100% accurate. Astronomers use empirically derived relationships to predict apparent magnitudes when accuracy is required.

Here's a quick script I wrote to calculate the apparent magnitude, given any phase angle: https://jsfiddle.net/fNPvf/33429/


Here's a practical approach - the algorithm and the equations are packaged as a software library.

Install PyEphem:


Run it:

$ python
Python 2.7.12 (default, Jun 29 2016, 14:05:02) 
[GCC 4.2.1 Compatible Apple LLVM 7.3.0 (clang-703.0.31)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import ephem
>>> moon = ephem.Moon(ephem.now())
>>> print moon.phase
>>> print(ephem.next_new_moon(ephem.now()))
2016/9/1 09:03:05
>>> print(ephem.next_full_moon(ephem.now()))
2016/9/16 19:05:05

'phase' is between 0 (new moon) and 100 (full moon).

More details:


  • $\begingroup$ Wow - I didn't realize PyEphem was so easy to use! Thanks for posting the script - I'll give it a test drive. $\endgroup$
    – uhoh
    Commented Aug 27, 2016 at 4:18

Short Answer:

A program for calculating the exact time of full Moon is a lot less expensive than the astronomical instruments necessary to observe when the Moon is exactly full.

Long Answer:

I can tell you that naked eye observation is pretty useless for telling whether the Moon is exactly full.

When I was a child, I noticed the Moon in the sky one day, and thus realized that the Moon is sometimes visible during the day, and I have seen the Moon the day many time since then.

One afternoon, in bright sunlight, I was on the boardwalk at Cape May, New Jersey, near Convention Hall. And I saw what looked like the full Moon low above the Atlantic Ocean. In the blue daylight sky. But if the full Moon has to be 180 degrees from the Sun as seen from Earth, the Sun should have been below the opposite horizon and the sky should have been darker.

When the Moon is full, shouldn't the most of the full Moon and the Sun one could hope to see at the same time be to see half the Moon and half the Sun at the opposite horizon?

So I turned around and I saw the full orb of the Sun low in the sky. How is that possible?

One) The Earth is spherical so the horizon is a little lower than half a sphere, and the sky is more than half a sphere. So the Sun and the Moon could be 180 degrees apart and still both be above the horizon.

Two) The Earth's atmosphere refracts light and makes object near the horizon appear higher than they are, so the Sun and the Moon could have actually been partially blocked by the horizon but appeared higher in the sky.

Three) And maybe the Moon looked full to my naked eye, but was not totally full, in which case the Sun and the Moon would be less than 180 degrees apart.

Years later, I was at the train station at Chalfont, Pennsylvania one morning, and I noticed what appeared to be a full Moon in the bright blue sky. The train station is on low ground, with hills, buildings, and trees around it. I found the Sun in the sky, and then I pointed one arm at the Sun and one arm at the Moon. And my two arms were pointing noticeably less than 180 degrees apart.

Therefore the Moon could not have been totally full at that time, when it was noticeably less than 180 degrees from the direction to the Sun. Therefore, the Moon can look full to my naked eye, and presumably to anyone else's naked eye, for a few days around the time of the full Moon.

The Moon can look full, as well as you can see with the unaided eye, when it is less than full.

So to find the exact day, or hour, or minute, or second, when the Moon is exactly full, a telescope and possibly some elaborate astronomical tools to measure its angular diameter would be necessary. The more precisely someone desires to know the instant of full Moon, the more their astronomical equipment and its expense would have to equal that of a professional astronomical observatory.

So using a lunar phase calculating program would be much less expensive than the equipment necessary to observe the moment of full Moon.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .