# What is the apparent size of the Earth and Sun from Moon

Wanted to understand how lunar eclipse will look like from the Moon. What is the apparent size of the Earth and Sun from Moon. How to calculate it. How much Earth will cover Sun during lunar eclipse from the moon.

• I am not quite good enough to come up with the calculation, but I do know that the angular diameter of the earth is quite a bit larger than the sun, from the moon. However many times bigger the earth is than the moon, pretty much Apr 10 at 19:27
• The earth's shadow is much larger than the moon, that's why total lunar eclipses are more common and last longer than solar eclipses. And observing the earth's shadow moving across the moon's face during lunar eclipses is one of the ways the ancient greeks (among other people) determined that the earth is a sphere (because the shadow is always curved, never showing a lip or edge, and the only shape that can cast a rounded shadow from every angle is a sphere). Apr 11 at 13:52
• astronomy.stackexchange.com/questions/7736/… Apr 11 at 20:09
• Regarding how it will look, there's this question. Doesn't show the relative sizes, but does demonstrate that, although the solar disk is much smaller than that of the Earth, some shininess from the Sun is still present. Apr 12 at 15:22

The earth's diameter is about 4 times the moon's, therefore, it will look about 4 times as big from the moon as the moon looks from the earth. The moon is not appreciably further from the sun than the earth is (only about a maximum of 0.25% further during a lunar eclipse), so the sun's angular diameter viewed from the moon is basically the same as when viewed from the earth.

When viewed from the moon, the earth has about 4x the angular diameter of the sun. From the moon's perspective, it might be better to say the sun is occluded rather than eclipsed, although there is no firm boundary on how close the relative apparent diameters need to be to be called an eclipse.

• For half of each month, the moon is closer to the Sun than Earth is. Apr 11 at 7:14
• @DaleM The OP asks about lunar eclipses, during which the moon must be further from the sun. Not that it matters much either way, it's never appreciably further or closer to the Sun than the Earth is. Apr 11 at 13:34
• Perception of size tends to be closer to projected solid angle rather than angular diameter, but can range anywhere from that to an estimate of apparent volume. Some recent immigrants to the lunar settlement will say the earth is 4× bigger, while others will say it's 64× bigger, but most will say it's closer to 16× bigger. Apr 11 at 20:36

The apparent size of a celestial body is given by its angular diameter or angular radius. In this diagram, both the circles have the same angular diameter, as seen from the point at the origin.

For a body of radius $$r$$, with its centre at distance $$s$$ from the observer, the angular radius $$\theta$$ is given by: $$\sin(\theta) = r / s$$

The Moon's orbit is moderately eccentric, so the Earth-Moon distance varies over the month. But due to the influence of the Sun the Earth-Moon distance varies over the year, too.

We can get the angular diameter from JPL Horizons. Horizons gives angular diameter in arc-seconds. (There are 3600 arc-seconds in a degree).

Here are some angular diameter plots for last year using Horizons data. The time step is one day. To keep the plots simple, the observer is located at the centre of the body, although Horizons can easily produce data for surface observers.

Here's the angular diameter of the Earth, as seen from the Moon.

Here's the angular diameter of the Sun, as seen from the Moon.

Here's the angular diameter of the Sun, as seen from the Earth.

So, on average, to a Moon observer, the Earth looks ~3.575 times bigger than the Sun, but there's a lot of variation.

Here's the angular diameter Sage / Python script running on the SageMathCell server.

Here are the angular diameters of the Sun and Earth, as seen from the Apollo 11 LRRR (Laser Ranging Retroreflector).

Here's the ratio of those angular diameters.

Here's the script that generated those plots. You can use any major body as the center, and any number of major or small bodies for the targets, separated by commas.

Colors in the palette list may be specified by name or in RGB hex notation, eg #f00 is red. You may supply more colors than targets. If there are fewer colors than targets, the targets without a matching color are ignored.

The X axis checkbox lets you select whether the graph is forced to have an X axis.

The ratio checkbox activates ratio mode. In this mode, the diameters of the second and any subsequent targets are divided by the corresponding diameter of the first target.

Please see the Horizons manual for full details on how to specify bodies, and site locations on observing centers.

Briefly, major bodies include, the Sun, planets, moons, and spacecraft. Small bodies are asteroids and comets.

You can do body ID lookups by entering the body name into the targets field. Horizons will respond with a list of matching sites. You can also do substring searches by appending *. Append ; to force a small-body lookup. Eg Juno* will find spacecraft named Juno, but Juno*; will find the asteroid named Juno. You may also use a partial ID number in a substring search.

Here's a script that does body name lookup, and prints some data about the body.

If you use a center like @301 the angular diameters are computed for an observer at the center of the body. You can get the diameters for a surface point by prepending a site ID number, or the letter c, which specifies a location of latitude 0, longitude 0, at an altitude of 0 km relative to the body surface. For gas & ice giants, that's the usual nominal surface, where the atmospheric pressure equals 1 bar.

Site ID numbers are available for the Moon, Earth, and Mars. Note that Earth has a lot of site IDs. Here's a site ID script.

• Those graphs would be much improved if you included zero on the y axis. As it stands, it gives a zoom in on detail with no visual indication of how important that detail is (which is tiny). All of the information is there, but the visual display isn't proportional. (Having all of the information isn't enough for it to be a good graph: you could give the initial value and graph the derivative and get another graph that focused on the details of the wobble even further) Apr 12 at 14:09
• Well, 13% 3% and 3% total wiggle - my point is, without zeroing, all 3 graphs show similar "wiggle", but one of them has more than 4 times as much! The Earth does vary in size significantly from the moon, but the sun from the earth and moon is almost entirely fixed in size (and the sun from the moon and sun from earth are nearly indistinguishable) Apr 12 at 14:32
• Log scale the ratio? That puts 1:1 at 0, and 100:1 and 1:100 equidistant from the baseline. (And yes, I know this is just nits and the answer is already great!) Apr 13 at 16:44
• @Yakk Thanks. :) log scale is very easy, it's just a keyword arg, but of course I'd have to disable the X axis option. :D OTOH, the GUI is already a bit too complicated... Apr 13 at 17:01

Others have answered the apparent size of the Sun and the Moon, but I'd just like to chime in that even though Earth will cover the whole Sun during a total lunar eclipse, it still wouldn't look boring from the Moon. Because of our atmosphere, you'd see a glowing red ring around Earth.

Considering that it is this red light is bright enough to color the entire lunar surface red during an eclipse, the glow would probably look pretty bright, if seen from the Moon.

Wanted to understand how lunar eclipse will look like from the Moon.

Probably like this:

This picture was taken by the Apollo 12 astronauts as the Earth moved between them and the Sun, thus "eclipsing" the Sun.
Since what happens in a lunar eclipse, from the point of view of the Moon, is exactly the same — i.e. the Earth positions itself between the Moon and the Sun, the view from the Moon would be similar to this.