Another day, another awesome google doodle, this time of a lunar eclipse.

You can tell Rhett Allain has infected me, because when I saw this, the first question that popped into my mind was, “Did they get the scale right on this animation?” Hmm, lets see.

This seems like a job for Geogebra.

The first thing I can do is overlay circles to the outline of the moon and the shadow to find their relative sizes. Geogebra is awesome enough that it can draw a circle right through three points I mark on the image. Here’s what this looks like:

Using Geogrbra to measure the ratio of the radius of the moon to the radius of the earth's shadow.

With no work, Geogebra reports the equations of the circles in the standard form:
$(x-x_1)^2+(y-y_1)^2=R^2$,
where $(x_1,y_1)$ is the coordinate of the center of the circle, and R is the radius.

So, based on this, we have
$\frac{R_{\textrm{shadow}}^2}{R_{\textrm{moon}}^2}=\frac{9.64}{1.23}\implies\frac{R_{\textrm{shadow}}}{R_{\textrm{moon}}}=2.8$

Based on this doodle, the radius of the earth’s shadow should be 2.8 times larger than the radius of the moon at the location of the moon. Is this true?

It probably pays to figure out a bit about what causes a lunar eclipse in the first place. Let’s turn to Wikipedia for a nice diagram:

So as you probably remember from elementary school, a lunar eclipse is caused when the Earth passes between the Sun and the moon and casts a shadow on the moon.

There’s only one slight problem with this digram. The scale is horribly misleading. Based on this diagram, the Sun is only about 4 times farther away from the Earth than the Moon, and only a few times larger. This is because it’s completely impractical to draw a scale drawing of this situation. Every drawing you’ve ever seen of the solar system is wrong. If the planets are drawn in the proper scale, i.e. Jupiter’s radius is 10 times larger than the Earth’s radius, like it should be, then there is no way that distances can be drawn to the same scale—the planets would have to be so far apart that they wouldn’t fit on a page. Here’s a great online model, that actually is drawn to scale to help you to see just how vast our solar system is. Neat tip for frustrating the anal-rententive young scientist—ask them to produce a perfect scale model of the solar system or an atom, and tell them they’re doing it wrong every few minutes or so.

But let’s look up some of the data we need to do these comparisons:

Radius of Sun: $6.95 \times 10^5\textrm{km}$
Radius of Earth: $6.39 \times 10^3\textrm{km}$
Radius of Moon: $1.74 \times 10^3\textrm{km}$

Orbital Radius of Earth: $1.52 \times 10^8\textrm{km}$
Orbital Radius of Moon: $3.76 \times 10^5\textrm{km}$

Let’s start by making a drawing (not to scale) of the earth and the Sun in Geogebra. We will also project the extreme most rays from the sun that are blocked by the Earth in order to get an estimate of the size of the darkest part of the shadow (called the umbra).

Here you can see we have two similar triangles, denoted by the red triangle, and the blue dashed triangle. Since these are similar triangles, the ratios of any two sides should be equal. Let us call the distance between the earth and where the shadow reaches a point $R_{\textrm{converge}}$.

$\frac{R_{\textrm{sun}}}{R_{\textrm{earth}}}=\frac{R_{\textrm{orbit}}+R_{\textrm{converge}}}{R_{\textrm{converge}}}$

We can solve this for $R_{\textrm{converge}}$:

$R_{\textrm{converge}}=\frac{R_{\textrm{earth}}R_{\textrm{orbit}}}{R_{\textrm{sun}}-R_{\textrm{earth}}}=1.38\times10^6\;\textrm{km}$

Now, let’s look at another diagram, this time of the Earth and the Moon.

Here again, we have two similar right triangles. One has sides of the $R_{\textrm{earth}}$ and $R_{\textrm{converge}}$ and the other has sides $R_{\textrm{shadow}}$ and $R_{\textrm{converge}}-R_{\textrm{moon orbit}}$.

We can now write:

$\frac{R_{\textrm{shadow}}}{R_{\textrm{earth}}}=\frac{R_{\textrm{converge}}-R_{\textrm{moon orbit}}}{R_{\textrm{converge}}}$

Which can be solved for $R_{\textrm{shadow}}$:
$R_{\textrm{shadow}}=\left(\frac{R_{\textrm{converge}}-R_{\textrm{moon orbit}}}{R_{\textrm{converge}}}\right)R_{\textrm{earth}}=4.6\times\;10^3\textrm{km}$

Now we have all the information we need to compute $\frac{R_{\textrm{shadow}}}{R_{\textrm{moon}}}$
$\frac{R_{\textrm{shadow}}}{R_{\textrm{moon}}}=\frac{4.6\times\;10^3\textrm{km}}{1.737\times\;10^3\textrm{km}}=2.63$

Hmm, that’s pretty close, but not quite dead on. Of course, there are lots of complicating factors, like the fact that the moon usually doesn’t pass directly through the center of the earth’s shadow, though in the case of today’s eclipse, it comes very close.

Let’s check our work a bit by looking at this detailed filled plot from NASA:

The diagram tells us that the angular radius of the Earth’s Umbra is 0.7236 degrees. This represents the fraction of the sky taken up by the earth’s shadow, which is a standard measurement in astronomy. A quick hop over to Wikipedia tells us that the Moon varies in angular diameter between 29.3 and 34.1 minutes of arc. We’ll assume an average value of 31.6 arc-minutes. Now if we keep in mind that there are 60 minutes of arc in 1 degree, and convert the umbra radius to a diameter, we get:
$\frac{\delta_{\textrm{shadow}}}{\delta_{\textrm{moon}}}=\frac{2\cdot0.7263\;^\circ}{31.6\;'}\cdot\frac{60\;'}{1\;^\circ}=2.74$

Nice. So it looks like everything is in pretty good agreement, and I can sleep comfortably knowing that not only are the are the good people at Google doing amazing thing allow me to search the web using my voice, their designers also have a very good understanding of our solar system.

To get a sense of just how non-trivial it is to draw things like this correctly—how our mental models often conflict dramatically with reality, I suggest you check out the following awesome video from Derek Muller, where he helps people on the street see just how far away the moon really is.

1. June 16, 2011 7:26 am

I love GeoGebra so very, very much.

2. June 16, 2011 8:51 am

You are out of control. I love it.

• June 16, 2011 5:20 pm

I’d like to coin a new phrase: “Pulling a Rhett” n. Becoming consumed by the analysis of the physics of some small bit of popular culture, often at the expense of work, sleep and other essentials. Truly, it’s like crack. try it once, and you can’t stop yourself.

3. June 16, 2011 5:34 pm

Freak.

June 18, 2011 12:33 pm

My understanding of the lunar eclipse doodle is that the moon image is a compilation of a live video feed (from Cairo I believe). So I would expect it to be correct. That said, that’s a nifty use of Geogebra to explore and confirm.

June 22, 2011 2:32 pm

Geogebra is new to me but you explained it very well! There were a lot of steps to take in figuring this Google eclipse logo out and I think you did a really good job. There aren’t very many people who would figure this out, so you are one in a few! Nice work.

-Emily

6. October 21, 2011 5:58 pm

Just came across this. The images in the Google eclipse doodle are EXACTLY correct. Why? Beacuse they were a live feed from SLOOH.com which deployed a mobile setup to Dubai to capture the event. I know this for sure, because it was me in Dubai taking the images! http://slooh.net/public/Dubai_Setup.jpg

• October 21, 2011 7:29 pm

Awesome. I love that when I write about google doodles, someone intimately involved in their creation always writes back. I didn’t know about SLOOH, either, so thanks for alerting me to that website.

• October 21, 2011 7:29 pm

Awesome. I love that when I write about google doodles, someone intimately involved in their creation always writes back. I didn’t know about SLOOH, either, so thanks for alerting me to that website.