Brian Stonelake

T he sun is bigger than the moon. Signifi-cantly. This fact probably doesn’t surprise you, but hear me out. The sun is so much bigger than the moon that it’s hard to picture both, to scale. Figure 1 shows the

relative size of the sun and Earth and the Earth and the moon.

Everyone knows that the sun is big and the moon is (relatively) small. But perhaps that’s not as obvious as you think. My dog, which is pretty smart for a dog, probably thinks they’re similarly sized, as their apparent size from Earth is similar.

Apparent size is a function of size and distance. All else being equal, big things look big and small things look small. Similarly, close things look big and far things look small. Where things get interesting is when big things are far and small things are close. How many snapshots have you seen of a tourist pinching the Eifel Tower or the Taj Mahal between thumb and forefinger?

Your patience is waning (pardon the pun). What’s my point? Our sun is big (1,391,400 km diameter) and far (149,597,870 km from Earth) while the moon is small (3,474 km diameter) and close (384,400 km from Earth). If you sketch some triangles and apply some trig, you can verify that the angular diameters are both approximately the same—roughly half a

degree. U.S. readers saw this for themselves on August 21 when they watched the moon perfectly obscure the sun. This also explains my dog’s confusion.

Now imagine that you, thanks to Elon Musk and his SpaceX efforts, are gallivanting around in space. Where could you go to make the apparent size of the Earth and the moon equal? This question, which came from a colleague’s friend’s child, has recently kept me from my grading.

You could stop at some point on the line connecting the centers of these two celestial bodies—close to the moon and far from the Earth—to get this perfect van-tage point. But where else could you go? What would the collection of all perfect vantage points look like in 3-space? (Take a guess . . . Mine was way off.)

Perfect Vantage PointsFirst, let’s limit our perspective to 2-space and Euclid-eanize our situation. Let’s place the Earth at the origin and let the moon hang out on the y-axis at the point (0,a), where km.

Any perfect vantage point is the vertex of two similar isosceles triangles, the bases of which are the diameters of the Earth and moon (as in figure 2). To avoid big numbers (for now) let’s call those diameters d1 and d2, respectively, and the distance from your vantage point

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Optimally Observing Orbiting Orbs

Figure 1. Relative sizes of celestial bodies.

to the center of the Earth w. Because the triangles are similar, the distance from your vantage point to the center of the moon is

We no longer need these isosceles triangles, so I’ll redraw figure 2, letting (x,y) represent the coordinates of an ideal vantage point. If you have Pythagorean proclivities, you may notice that the two right triangles in figure 3 imply the equations and

Scaling the first equation by allows us to set the equations equal to each other, yielding

Setting this equation to 0, expanding, and factoring gives

For mathematical aesthetics (mathsthetics?), we can substitute to clean things up a little:

By completing the square, we obtain

This, of course, is the equation of a circle—specifically, the circle centered at and radius Recall that b was just a mathsthetical substitution, so really our circle has center and radius (check my algebra, I dare you!).

So, from anywhere on this circle, the Earth and the moon will appear to be the same size. But our analysis was in two-dimensional space, and we should be consid-ering 3-space. Thus, our solution set is the sphere with

the above radius and center. (Take a moment to con-vince yourself; I’ll wait.)

The values of our constants in kilometers (because even though I’m American, I recognize that the metric system is superior) are approximately

and Plugging these values into our formula, we see that our sphere has a radius of 113,219 km and is centered 415,268 km away on the ray originating from Earth’s center, passing through the moon.

Unless you’re Neil deGrasse Tyson (in which case, ’sup Neil, big fan!) those numbers probably mean little to you, so let me provide some context. This point is roughly 8 percent farther than the moon (from Earth). The radius is roughly 30 percent of the distance between these celestial bodies (see figure 4).

The TrifectaDoes that answer your question, colleague’s friend’s son? Let’s suppose it doesn’t. Maybe you want a third celes-

28 November 2017 : : Math Horizons : : www.maa.org/mathhorizons

α

αd2

a w

d1

x

y

Figure 2. Similar Earth and moon triangles.

(x,y)

y

x

x

Moon

Earth

You(0,a)

(0,0)

w

Figure 3. Earth and moon right triangles.

Moon

Earth

Moon’sorbit

Figure 4. The perfect Earth-moon vantage points sit on a solution sphere about the moon.

tial body to also appear to be equally sized. Is this possible? Let’s try to get the sun in on this fun!

We need to find a point on our Earth-moon solution sphere where Earth and the sun appear equally sized. Looking for this point is easier than it sounds, as we already have an equation to use. Just take to be the distance between Earth and the sun, to be the diameter of the sun and to be the diameter of Earth. If the result-ing Earth-sun sphere intersects our Earth-moon sphere, we win!

Drumroll please. . . . To get to the center of this sphere, start at the center of the sun, and walk toward (and through) the center of the Earth. When you’ve walked 149,610,417 km, stop and admire the center of the Earth-sun solution sphere. Note that this point is only 12,547 km beyond the Earth’s center, so you’re between low-Earth orbit and geostationary orbit. In other words, you are farther from Earth than top-secret mili-tary spy satellites, but not as far as the more important satel-lite that is beaming me tonight’s Warriors game (KD’s fixin’ to drop 40, mark my words).

However, the radius of this solution sphere is huge, measur-ing 1,370,085 km. This is more than 10 times the radius of the Earth-moon solution sphere, and more than triple the distance from Earth to the moon. This is bad. It means that for the sun and Earth to appear equally sized, you have to be very far from Earth. This is pictured in the fictitious figure 5, which I’ll leave to your imagination. It looks remarkably like figure 4, except there is a huge yellow sphere representing the Earth-sun solu-tion set, centered very close to Earth, with a radius 10 times that of the red sphere. Clearly these spheres never intersect, even if you account for ignored minutiae like elliptical orbits and equatorial bulge.

I guess if you really want to see equally sized celestial objects, head to Jupiter and find an ideal vantage point of some of its 67 moons. I’m guessing you can find intersecting spheres. n

Brian Stonelake teaches mathematics and statistics at Southern Oregon University. In addition to pondering cosmic trivialities, he enjoys publishing articles that attempt to justify the recent lack of success of his beloved San Francisco 49ers. Special thanks to Curtis Feist and the inquisitive minds of his friend’s sons Blake (14) and Nathan (10).

http://dx.doi.org/10.4169/mathhorizons.25.2.27

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Crossword solution: Torus

Crossword solution: Klein Bottle

Crossword solution: Sphere