In my last post, I shared a bit of my history with the "Mars Spectacular," and I encourage you to take a look at that if you have not already.

But, alas, I can't just stop at refuting this claim by demonstrating that it's been made before. No, I have to refute it with math!

But don't worry, the math needed is almost as basic as math gets . . . once we get past the trigonometry.

For those of you who reel at the sight of sight of trigonometric functions, calm down! I'll just give you the answer: for things that are very small or very far away (like in space), an object's apparent angular size is essentially its actual size (height, width, diameter, what-have-you) divided by its distance from you. There. That's all it is. Now you can skip all the way down to the bottom.

For the rest of you, hold onto your asymptotes, 'cause here we go!

The Angular Size of Things

Surely you have noticed throughout your life that things look smaller the farther away they are, or conversely, things look bigger if they are close. This fact can be described via trigonometry, and don't worry—it's pretty easy to understand.

Imagine that you are standing on the ground, looking at a very tall pole. Or a tower, or a skyscraper, what-have-you. The nature of the structure does not matter. What's essential is that it is way taller than you are. To demonstrate, I made a picture.

In that image, you are the small dot on the ground, the object of interest is located at a distance \(d\) away from you and has a hight of \(H\), and the angle by which you must tilt your head up to see the top of the object is \(\alpha\).

There exists a trigonometric relationship between these three quantities, \begin{equation}\tan{\alpha}=\frac{H}{d},\end{equation}

(Remember, kids: "tangent equals opposite over adjacent!")

which can be inverted to find \(\alpha\). \begin{equation}\alpha=\arctan\left(\frac{H}{d}\right).\label{250818:angle}\end{equation}

What does that mean? Well, as is so often the case, a graph is worth a thousand equations.

But what does that mean? The blue curve represents equation \eqref{250818:angle}. The vertical axis is the angle by which you must crane your neck to see the top of the object, and the horizontal axis is the distance between you and said object, in units of its height. The shorter this distance is, the higher up you must look to see the top. For example, when you are standing at the base of the object so that the distance between you and it is zero, you must tilt your head up by a full 90°. When the distance between you and the object is the same as its height, such that \(d/H=1\), you must tilt your head at a 45° angle. If the object is many times farther away than it is tall, then you don't have to look up much, if at all.

As the object gets farther away (or shorter), the angle you see between its top and the ground gets smaller. Eventually, this angle becomes so small that the object appears to be nothing more than a single point.

(That is why, for example, stars appear to be points of light, though they may be millions of miles wide—they are trillions of miles away!)

So what does any of this have to do with the Moon and Mars? Well, like anything we see with our eyes, the apparent sizes of those objects depend upon their distances from us, and, as is hopefully evident from the following picture, this dependence is very similar to that which we investigated above.

In this case, the solid horizontal line is the distance \(d\) between you and the center of any round object, be it the Moon, Mars, a grapefruit, etc., the solid vertical line is the diameter of that object, \(D\), from which one can derive its radius, \(R=D/2\), and the angle that you see between the object's center and its top, bottom, or any point on its apparent perimeter is the object's angular radius, \(\rho\) (a lower-case Greek letter "rho," which, despite looking like "p," is actually a progenitor of the Latin letter "r"). From the similarities of the geometries of figures 1 and 5, it should come as no surprise that the relationship between \(\rho\), \(d\), and \(R\) is \begin{equation} \rho=\arctan\left(\frac{R}{d}\right). \label{250818:angularradius} \end{equation} This is, in fact, the same equation that we saw in \eqref{250818:angle} above!

So, if \(\rho\) is the angular radius of a round object, what is its angular diameter, \(\delta\) (lower-case Greek "delta," forerunner of the letter "d")? \begin{equation} \delta=2\rho=2\arctan\left(\frac{R}{d}\right). \end{equation}

(As a brief aside for interested readers, note that \(\delta≠\arctan\left(\frac{D}{d}\right)=\arctan\left(\frac{2R}{d}\right)\), as can be seen by the following graph.)

Finally, we have relationship between the apparent size of an object, \(\delta\), its actual size, \(R\), and its distance, \(d\), from you, the observer. \begin{equation} \delta = 2 \arctan\left(\frac{R}{d}\right). \label{250818:angulardiameter}\end{equation}

Let's put it to good use.

Oh, and, so we're all on the same page here, because the tangent of a very small angle is essentially just that angle (\(\tan{\theta}\approx\theta\) for \(\theta\ll 1\)), the inverse tangent of a very small value is essentially just that value (\(\arctan{x}\approx x\) for \(x\ll 1\)). Consequently, if our object of interest is very small or very far away, so that \(R/d \ll 1\), then \begin{equation} \delta = 2\arctan\left(\frac{R}{d}\right) \approx 2 \frac{R}{d} = \frac{D}{d}. \end{equation} In other words, the object's angular diameter is essentially its actual diameter divided by its distance from you: \begin{equation} \delta \approx \frac{D}{d}. \label{250818:smallangulardiameter} \end{equation}.

Mars vs. The Moon

Now that we have equation \(\eqref{250818:angulardiameter}\) in our pocket (or equation \(\eqref{250818:smallangulardiameter}\), whichever the case may be), let us use it to determine how big Mars and the Moon look in Earth's sky. For reference data, I am using the Wolfram Knowledgebase, which can be called upon from Wolfram Mathematica—the tool that I use to carry out most of my calculations, run most of my simulations, and create most of my graphs. According to Mathematica, the radii/diameters of the Moon and Mars are as follows. \begin{equation} \begin{array}{r|rr}
 & \text{Radius} & \text{Diameter} \\
 \hline
 \text{Moon} & 1079.6 \ \mathrm{mi} & 2159.1 \ \mathrm{mi} \\
 \text{Mars} & 2106.1 \ \mathrm{mi} & 4212.3 \ \mathrm{mi}
\end{array} \label{250818:linetable} \end{equation}

Yes, I am using miles. I am, after all, born and raised in the USA and that's what's been integrated into my brain.

I am kidding, of course, but funnily enough, when I call up these data in Mathematica without specifying a unit, miles are given by default. (I wonder if it does that for everyone everywhere?) Furthermore, I find it easier to roughly convert from miles to kilometers in my head by simply adding half of the original value. So, if you are pining for kilometers, try doing that. I get \(\sim1500 \ \mathrm{km}\) for the radius of the Moon, and \(\sim3000 \ \mathrm{km}\) for that of Mars.

Well, how about that. It turns out Mars is about twice as wide as the Moon! Keep that information in the back of your mind somewhere. You never know when it will be useful.

Now, to determine the angular sizes of these objects, we need to know their distances from Earth. However, those values are not fixed! This should be obvious for Earth and Mars, considering the fact that sometimes the two planets are on the same side of the Solar System, while at other times they are on opposite sides of the Sun. Because of that, the distance between them can range anywhere from \(\sim50 \ \mathrm{million \ mi}\) to \(\sim250 \ \mathrm{million \ mi}\). However, it might come as a surprise to some readers that the distance between the Earth and the Moon changes, too, since the Moon's orbit is not perfectly circular, nor is it centered exactly on the Earth. We could, I suppose, use the average distances between Earth and these bodies, but Mathematica can do better. If we specify a date and a time, the program will compute what these values are at that moment with very high precision! So, since the date of interest is August 27th, 2018, let us feed that into the database. As for the time of day, the meme on Facebook says 12:30, but it does not specify AM or PM, nor does it give a time zone. The original email says 12:30 AM (local time), which is approximately when an object at opposition will be highest in the sky for any time zone, but that still gives us twenty-four moments in time from which to choose. We need just one. So, I’m going by a clock on the Prime Meridian, and the time of interest is 12:30 AM GMT. At that moment, the distance from the center of the Earth to the centers of the Moon and Mars is \begin{equation} \begin{array}{r|r}
 & \text{Distance From Earth} \\
 \hline
 \text{Moon} & 249\,349 \ \mathrm{mi} \\
 \text{Mars} & 40\,111\,005 \ \mathrm{mi}
\end{array} \label{250818:distancetable} \end{equation}

Wouldn't you know it—Mars is 162 times farther away from Earth than the Moon is. And, if you are using equation \(\eqref{250818:smallangulardiameter}\) and you remember that Mars is twice as wide as the Moon, you can quickly work out that it will appear 81 times smaller in Earth's sky. There, problem solved.

But let's be more precise, shall we?

Feeding our sizes and distances into equation \(\eqref{250818:angulardiameter}\) (and then converting from radians to degrees), we find the angular diameter of the Moon to be \begin{equation*} \delta_{\mathrm{Moon}} = 2 \arctan\left(\frac{1079.6}{249\,349}\right) = 0.496°, \end{equation*} while that of Mars is \begin{equation*} \delta_{\mathrm{Mars}} = 2 \arctan\left(\frac{2106.1}{40\,111\,005}\right) = 0.006°. \end{equation*} In other words, at 12:30 AM GMT on Monday, August 27th, 2018, the Moon is actually about 83 times larger than Mars in Earth's sky.

Meme refuted.

But I can't just stop there! No, I have to go one step farther.

How far would Mars have to be from Earth so that it did look to be the same size as the Moon?

For those of you using \(\eqref{250818:smallangulardiameter}\), that's easy: if Mars is twice the Moon's actual size, then it must have twice the distance from Earth to look the same size to our eyes. That's only half a million miles, which is nothing for planets whose closest approach is usually one hundred times larger.

What would that do to the Solar System?

I don't know, but I'm gonna try to find out. And when I do, you'll see it here.

So stay tuned.

Oh, and as a bonus for reading all the way to the end, here's a picture of the Moon and Mars, scaled to the relative proportions found above. Try to figure out which one's which.