How big a coincidence is the Sun and Moon having almost equal apparent sizes?

Question

It's consensus that the very similar apparent sizes of the Moon and the Sun as seen from Earth is a coincidence (as already answered in this site).

This provides us with almost exact total solar eclipses and other interesting sights, such as this one shared in Fermat's Library's tweet:

Here's a great photo of the ISS by J. Mccarthy.

The reason why the ISS keeps its relative size against both the Sun and the Moon is due to a remarkable coincidence: the diameter of the Moon is 400x smaller than the diameter of the Sun, but it is also 400x closer to us!

This state of affairs is not permanent, of course - as the moon keeps retreating from us - but total solar eclipses are estimated to keep occurring for the next 600 million to 1 billion years from now.

We're currently unable to answer how big this coincidence is, since we don't have enough statistical data on planet-moon systems yet, however we do have models of solar system formation.

So my question is, irrespective of the presence of intelligent beings or the possibility of actually observe total eclipses:

• According to our knowledge of planet formation, how unlikely is this coincidence?

Answer 1

Moons in the solar system range in size from about 0 to a little larger than our moon. The distance must be close enough that other planets won't pull it loose and I suppose far enough that it won't hit Earth when at closest. You could figure out a number from this.

But our moon is unusual in that it is much larger fraction of the planet than other moons. Mars has tiny moons that don't make total eclipses. Something the size of Mars hit proto-Earth and launched a huge amount of debris into space. Some of it must have escaped and some returned to Earth. What are the odds that something so big would be left in orbit?

As I understand it, many systems form with a lot of small objects everywhere, and large gas giants in the outer reaches. The giants throw small objects everywhere. If an object is ejected, it isn't thrown again. Over time, objects get ejected until none are left. On average small objects migrate out and gas giants migrate in. Many systems have a hot Jupiter close the the Sun, an Oort cloud, and nothing else. This process started in our system, and is responsible for the late heavy bombardment. But it stopped short of completion. What are the odds of our system running out of small objects before Jupiter got to the inner planets?

From looking at our system, distant gas giants may well have moons big enough to cover a tiny apparent Sun. Do they count? Suppose Venus had a big moon. Would it count? Their atmospheres are so thick that you couldn't see the eclipse from the surface. If they have a surface. For that matter, most of the surface of the Earth is covered with enough water that you couldn't see an eclipse. Supposing they do count, how likely is it that there will be distant gas giants with large moons?

We need more data before we can answer these questions.

Answer 2

A rough and ready approximation.

This is basically a mathematical/probabilistic approach.
This approach makes several assumptions. If you change the assumptions you will get a different result, but I think these assumptions are reasonable.

1. The sun always subtends the same angle with respect to the earth. Even thought the earth’s orbit is elliptical we assume it is circular and take this angle to be a given. (This is really accounted for in assumption 8)

2. The largest possible orbit for a moon is that of our moon, max distance 252,000miles. link It is generally accepted that the size of the moon relative to the earth is exceptionally large, so the probability of having a larger moon is small.

3. The smallest orbit is taken to be 1200 miles (LEO).

4. The total probability of earth having a moon is 1. Although Mercury and Venus do not have moons, it is most likely that Venus once did have a moon. see Other planets have multiple moons and it seems that further our even pebbles can have moons. The asteroid Elektra 130 has three. Elektra 130

5. The largest diameter of a moon is 2200 miles and the smallest is 10 miles. These are the sizes required to cause total eclipses at the maximum and minimum orbits.

6. There is an equal probability for the moon to have any size. Total probability =1. As we don’t know the actual probability of a moon having a particular size we apply the principle of equal a priory probability link

7. There is an equal probability of a moon having any orbit within the range given. See previous assumption.

8. A moon occupies an orbit between 90% and 100% of its maximum distance from the earth. As the orbit of the moon varies from about 225,300 miles to 251900 miles which we will call a 10% variation, it basically occupies this volume of space.

9. To find the probability of a moon having a particular orbit we have to assign a range to each orbit. For example if the average speed of cars passing a point is 60mph you can find the probability of a car have a speed between 59 and 61mph. If you try to find the probability of a car having a speed of exactly 60mph that probability is 0. The range we will assign is that from the last assumption, that is it lies between 90% and 100% of its maximum orbit. A simple calculation reveals 51 possible orbits allowed.

10. For each possible orbit only one size of moon is allowed - that is that which will just cause a total eclipse. This means there are 51 possible moon sizes. A moon in other than its optimum orbit will not match this requirement exactly.

11. All moon orbits lie on the ecliptic

12. All moon are approximately spherical

Then choose any moon size. Given that choice of moon, the probability that this will be is its proper orbit to cause a total eclipse is simply 1 in 51 or about 2%. (Think of it this way, you have two boxes each with ten different colored balls inside. Pick one ball from the first box, what's the probability that you will choose the same color ball from the second box - just one in ten. It doesn't make any difference which colored ball you pick out of the first box. So the probability that the chosen moon will match its orbit is 1/51)
Therefore we can say that the probability of the earth having a moon that would cause a total eclipse is about 2%. Given all the assumptions made let us say the probability of earth having a moon that would cause a total eclipse is between 1% and 4%.

EDIT: Clarification
The moon varies in its distance from the earth from a max of 251900 miles to a min of 225300 miles. This is 10.56% less than its maximum orbit. So this space from 225300 to 251900 miles is “occupied” by the moon. At the moon’s furthest distance from earth a solar eclipse is not total; there remains a corona of the sun visible.
A moon with a smaller orbit from 202770 miles to 225300 would have to be smaller to subtend the same angle as the sun, and it would only be in this orbit that it would do this. If it were in a larger orbit it would subtend a smaller angle than the sun and never gives us a total eclipse. Similarly if it were in a smaller orbit it would subtend a larger angle than the sun, and would not be a “perfect fit”. For each size of moon there is only one orbit that it can have that allows this to occur.

If we allow each orbit to have the same 10% spread from max distance to min distance, this gives us 51 orbits between 1200 miles (LEO) and 252000 miles. That means there are 51 sizes of moon, one for each orbit, that will subtend the same angle as the sun and give us this perfect fit during eclipses. We have assumed that the probability of the earth having a moon is 100%. If there are 51 possible sizes of moon available, there is about a 2% chance of picking any one, but if you choose any moon at random, the probability of having a moon at all is 100%. You have a moon that you just picked.

For any particular moon there is only one orbit that will match the angle subtended by the sun. If there are 51 equally likely orbits possible, there is one chance in fifty one that this orbit will match the moon. If we number the moons from smallest to largest 1 to 51, and the orbits likewise from 1 to 51, moon #29 say will only match the sun if it is also in orbit #29.
Given that you have a moon, the probability of it being in its proper orbit is 1:51. This is where the 1/51 comes from.
But, you may ask, what if we had made a division into 100 orbits and 100 moon sizes; wouldn’t the chance then be 1/100? Good question. The answer is no, because you would find that in this case there would be 2 orbits for each size moon that would fit our criteria.

However, if you doubled the number of possible orbits to 110, extending them out to about 54 million miles (still using the 10% rule) but kept the possible sizes of the moon the same (51 sizes from about 10 to 2200 miles) then the probability of have a moon the same apparent size as the sun would only be about 1%.

EDIT 2
Following Kyle Oman’s argument that the probability of larger moons decreases as the square of their diameter, let us recalculate. We will also make the assumption that the moon may be in any orbit such that it subtends an angle equal to or larger than that of the sun; that is, that it will more than completely block the sun. Also let us assume that the moons may take 200 possible sizes and 200 possible orbits (one orbit of which any moon just fits our criteria), with the smallest approx 10 miles diameter and the largest 2000 miles diameter. We still assume that any orbit is equally likely.

Let $a_i$ denote the number of moons of diameter i. $a_i=i^2$, with i=1 the largest moon. Note that every moon with diameter i will block the sun in 201-i different orbits. The smallest moon will block the sun in only one orbit, LEO. The probability of a moon 2000 miles in diameter is thus $\frac{1}{200^2}$ that of the smallest moon.

Then the probability that a moon will block the sun is:

$\frac{(a_1)^2*(201-1)+(a_2)^2*(201-2)+...+(a_{200})^2*(201-200)}{200*200*201*401/6}$

$=\frac{1^2*(201-1)+2^2(201-2)+….200^2(201-200)}{200*200*201*401/6}$

$=\frac{(1^2+2^2+….200^2)*201-(1^3+2^3+….200^3)}{200*200*201*401/6}$

$=\frac{(200*201*401/6)*201-200^2*201^2/4)}{200*200*201*401/6}$

$=\frac{1360167}{5373400}$ = 25.3%

(Need to multiply this by the probability that earth has a moon, in our solar system that's close to unity.)

This is the probability that a moon will completely block the sun. As we know, tidal effects on a moon gradually increase its orbit, so that for every moon that completely blocks the sun, as its orbit increases over time, there will come a time when it exactly coincides with the sun (at some point in its orbit). This is much higher than the previous calculation as we have increased the time frame from one moment to the entire time a moon is in orbit.

The probability that we are around to observe this phenomenon is not calculated.

Answer 3

If one does not care about the particular time of observation (as requested by the OP) then the probability that the Moon has the same apparent size as the Sun at some point in time is actually close to 1 (unless the sizes of the Sun or Moon would be massively different). The reason is the fact that the Moon as been receding from the Earth since its formation from a distance of just a few Earth radii to it present distance of 60 Earth radii (4 billion years ago it was still just about 20 Earth radii; see https://www.lpi.usra.edu/exploration/training/illustrations/earthMoon/ ). And it will be receding further due to tidal effects slowing down the Earth's rotation. The rotational kinetic energy of the Earth at the moment is about $2\cdot10^{29} J$ , whereas the potential gravitational energy of the Earth-Moon system is about -$8\cdot10^{28} J$ and thus (from the virial theorem) the orbital kinetic energy $4\cdot10^{28} J$, so a total orbital energy of $-4\cdot10^{28} J$. When the Earth has stopped rotating, the present rotational energy of $2\cdot10^{29} J$ would have been added to the present total orbital energy, which however would mean that the latter has become positive i.e. the Moon is not bound to the Earth anymore and has become a new planet of the Sun (provided of course the latter still exists)

Whatever the eventual outcome of this, the Moon will overall have moved from initially a fraction to eventually several times the present distance, so even with a considerably different distance to the Sun or a different size of the latter, the apparent sizes would match at some point in time. The real coincidence (if it is one then) is that we just happen to be here at the right time to observe this.