The Sun is huge, when compared to the Moon. Despite the huge difference in their size and distance from Earth, is it purely coincidental that they both look almost the same from Earth?

  • $\begingroup$ It's a sheer coincidence that at this point in time, just when humans have modern society, they are identical. Of course, it's a staple of sci-fi plots, and conspiracy theories, that since it is such a ridiculously amazing coincidence they happen to be identical, this must have something to do with a message from aliens, higher beings, or similar! It is a totally outlandish coincidence, whatever you think of it! $\endgroup$
    – Fattie
    Commented Oct 24, 2016 at 10:00
  • $\begingroup$ Better question might be - is there any consequence related to similar apparent sizes, aesthetics aside. $\endgroup$ Commented Aug 27, 2017 at 23:40

2 Answers 2


The coincidence isn't so much that they appear very similar sizes from Earth, but that we are alive to see them at the point in time in which they appear very similar sizes. The moon is slowly moving away from the Earth, and at some point in the future the moon will be unable to totally eclipse the sun and conversely, if you could step far into prehistory, you would be able to see the moon with a much greater angular diameter than you see it now.

Most research I've found on the topic seem to be unavailable through my institute, however I did find one paper, "Outcomes of tidal evolution", which references results from Goldreich's research on the subject.

This qualitative description of the eventual disruption of the Earth-Moon system is confirmed by the results of Goldreich's numerical integration, which showed that the moon will recede to 75 Earth radii, when spin-orbit synchronism will be reached; then the Moon's orbit will decay steadily inward because of the influence of the Sun.

For reference, the Moon is currently at a distance of approximately 60.3 Earth radii. As such, the moon will steadily move away until synchronism would be reached, and from that point begin to recede towards the Earth due to the tidal affects of the Sun on the Earth disturbing the synchronization. It would seem that at some eventual point in the far distant future, it will return to this coincidental position once again.

Counselman III, Charles C. "Outcomes of tidal evolution." The Astrophysical Journal 180 (1973): 307-316.

  • 1
    $\begingroup$ What do they mean by "spin-orbit synchronism"? $\endgroup$
    – Py-ser
    Commented Apr 24, 2014 at 2:43
  • 2
    $\begingroup$ This refers to tidal locking between two bodies. In this case, reaching a point where the Earth and Moon are tidally locked. This is important, as the mechanism for the Moon moving away from the Earth is due to the tidal forces of unsynchronized movement between the two. As they reach synchronization the distance would stabilize, if not for the additional tidal forces from the Sun destabilizing the system. $\endgroup$ Commented Apr 24, 2014 at 2:50
  • 2
    $\begingroup$ The Moon's rotation is already synchronized to its orbit. In the distant future, the Earth's rotation will also synchronize to the Moon's orbit, so the Moon will be visible only from one hemisphere and a day will be a month long (and a month will be even longer than it is now). Pluto and Charon are mutually locked in this way. $\endgroup$ Commented May 1, 2014 at 1:23
  • $\begingroup$ Moon size changes visibly over time: Micro Moon over Super Moon apod.nasa.gov/apod/ap140121.html This is why we see annular solar eclipses sometimes: en.wikipedia.org/wiki/Solar_eclipse#Types $\endgroup$ Commented Aug 9, 2014 at 18:04

I'd argue it isn't a total coincidence, but it isn't artificial either.

Out of possible arrangements allowing for stable orbits, what positions of the moon on the Sun-Moon-Earth line offer near equal angular measures?


$$l_s=150*10^6 km=1AU$$ $$r_s=695,508km$$ $$r_m=1,737km$$ $$l_m=384,400km$$

Where $l_s$ is distance from earth to the sun, $r_s$ is radius of the sun, $r_m$ is radius of moon. $l_m$ is distance from earth to moon, hear assumed to be along the straight line running from earth to sun. $l_m$ will be variable, but empirically on average has the value given above.

For what values of $l_m$ does $r_m/l_m$ , the tangent of the angular radius of the moon, fall within 10% of $r_s/l_s$, the tangent of the angular radius of the sun? i.e.:

$$(0.9)(r_s/l_s)<(r_m/l_m)<(1.1)(r_s/l_s)$$ $$0<\Delta(r_m/l_m)<(0.2)(r_s/l_s)$$ $$0<\frac{r_m}{l_m^2}\Delta l_m<(0.2)(r_s/l_s)$$ $$0<\Delta l_m<(0.2)\frac{r_sl_m^2}{(r_ml_s)}$$

So if $(r_m/l_m\approx r_s/l_s)$, then $l_m$ can vary by $20\%$ of its current value and still maintain an approximately equal angular radius- a Goldilock's Zone.

There are a few back of the envelope methods to gauge how far the moon can get from earth to remain in a stable orbit.

How far can the moon get without escaping from the earth? The Hill Sphere (h/t uhoh!)of the Earth has a radius of about $(0.01)l_s$. Further away from there and the sun perturbs the moon away from earth, establishing a maximum distance for our allowable range. The Roche Limit is a lower bound.

Whereas the probability of the angular radii being about equal if allowing the moon to be any place on the straight line joining the earth and the moon is approximately $\frac{(0.2)l_m}{l_s}=0.051253\%$, factoring in that the moon can only be just so far from the earth without decoupling, that probability goes up to at least $\frac{(20.0)l_m}{l_s}=5.1253\%$ by the above arguments.

A more thorough approach uses tighter bounds. There's a Goldilock's Zone for the moon and sun to have the same angular radius. The probability of falling within that Goldilock's Zone is the ratio of the Goldilock's Zone $(0.2l_m)$ to the interval between the minimum and maximum distances from the earth to the moon factoring in orbital stability, the moon's recession, and eventual return closer to earth. From answer above, the moon is 60.3 Earth Radii from Earth and will not get further away than 75 Earth Radii. The Goldilock's Zone spans from about 38 Earth Radii to about 72 earth radii. So between now and maximum recession of the moon, and assuming uniform time spent at each distance, the angular radii will match to within 10% about 75% of the time.

Overall its a low probability that the angular radii match so well. Given the distance the Earth lies from the sun and the radius of the moon which are fairly arbitrary, gravity allows the above mentioned Goldilock's Zone to have significant overlap with realistic lunar trajectories.

Put another way, if the Equal Angular Goldilock's Zone has sufficient overlap with allowable lunar orbits, then the angular match becomes about as likely as having a moon in the first place which itself is a coincidence.

  • 1
    $\begingroup$ It would be a good idea to calculate the size of the Hill sphere (also here) around the Earth, outside of which gravitational effects from the Sun (rather than Venus) will have an effect. I don't know if that's required using your approach, but I think you should rule it in or out. $\endgroup$
    – uhoh
    Commented Oct 31, 2018 at 1:59
  • 1
    $\begingroup$ Thanks @uhoh! Is required. I'll patch it up manana. $\endgroup$
    – R. Romero
    Commented Oct 31, 2018 at 3:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .