I've read that the Moon is moving away from the Earth by 1-3cm per year.

Is this enough to make the Earth lose the Moon before the Sun goes supernova?

I'm asking because I would like to do the calculations for Earth's magnetic pull on this subject. It appears to me that it should be a non-linear function (there are also other factors, like if the Moon could be captured by another planet or having asteroids messing with its orbit on Earth).


2 Answers 2


As HDE 226868 noted in his answer, the Sun is not going to go supernova. That's something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It's not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red giant, but in a shell around a sphere of waste helium. The Sun will start burning helium when it reaches the tip of the red giant phase. At that point it will shrink a bit; a slight reprieve. It will expand to a red giant once again on the asymptotic red giant branch when it burns all the helium at the very core. It will then burn helium in a shell surrounding a sphere of waste carbon and oxygen. Larger stars proceed beyond helium burning. Our Sun is too small. Helium burning is where things stop.

The Sun has two chances as a red giant to consume the Earth. Some scientists say the Sun will consume the Earth, others that it won't. It's all a bit academic because the Earth will be dead long, long before the Sun turns into a red giant. I'll have more to say on this in the third part of my answer.

The current lunar recession rate is 3.82 cm/year, which is outside your one to three centimeters per year window. This rate is anomalously high. In fact, it is extremely high considering that dynamics says that $$\frac {da}{dt} = (\text{some boring constant})\frac k Q \frac 1 {a^{11/2}}$$ Here, $a$ is the semi major axis length of the Moon's orbit, $k$ is the Earth-Moon tidal Love number, and $Q$ is the tidal dissipation quality factor. Qualitatively, a higher Love number means higher tides, and a higher quality factor means less tidal friction.

That inverse $a^{5.5}$ factor indicates something seriously funky must be happening to make the tidal recession rate so very high right now, and this is exactly the case. There are two huge north-south barriers to the flow of the tides right now, the Americas in the western hemisphere and Afro-Eurasia in the eastern hemisphere. This alone increases $k/Q$ by a considerable amount. The oceans are also nicely shaped so as to cause some nice resonances that increase $k/Q$ even further.

If something even funkier happens and the Moon recedes at any average rate of four centimeters per year over the next billion years, the Moon will be at a distance of 425,000 km from the Earth (center to center). That's less than 1/3 of the Earth's Hill sphere. Nearly circular prograde orbits at 1/3 or less of the Hill sphere radius should be stable. Even with that over-the-top recession rate, the Moon will not escape in the next billion years.

What about after a billion years? I chose a billion years because that's about when the Moon's recession should more or less come to a standstill. If the Earth hasn't already died before this billion year mark, this is when the Earth dies.

Dwarf stars such as our Sun get progressively more luminous throughout their life on the main sequence. The Sun will be about 10% more luminous than it is now a billion years into the future. That should be enough to trigger a moist greenhouse, which in turn will trigger a runaway greenhouse. The Earth will become Venus II. All of the Earth's oceans will evaporate. Water vapor will reach well up into what is now the stratosphere. Ultraviolet radiation will photodissociate that water vapor into hydrogen and oxygen. The hydrogen will escape. Eventually the Earth will not only be bare of liquid water on the surface, it will be bare of water vapor in the atmosphere.

Almost all of the Moon's recession is a consequence of ocean tides. Without oceans, that tunar recession will more or less come to a standstill.

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    $\begingroup$ Ahh, beat me to the calculations. Really nice answer. $\endgroup$
    – HDE 226868
    Nov 20, 2014 at 15:52
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    $\begingroup$ There is surely a maximum Earth-Moon distance corresponding to a mutual tidal lock. This must be easy to work out. $\endgroup$
    – Walter
    May 30, 2017 at 7:20

The Sun does not have nearly enough mass to become a supernova. Instead, it will swell to become a red giant, enveloping Mercury, Venus, and possibly Earth. After that, it will shed its outer layers as a planetary nebula, and settle down to become a white dwarf. Wikipedia, apparently, says the exact same things I had though of:

The Sun does not have enough mass to explode as a supernova. Instead it will exit the main sequence in approximately 5.4 billion years and start to turn into a red giant. It is calculated that the Sun will become sufficiently large to engulf the current orbits of the Solar System's inner planets, possibly including Earth.

Still, that does put a deadline for the Earth and the Moon to evolve to the point you're looking for. Wikipedia addresses the possibility that the Earth and Moon survive:

Today, the Moon is tidally locked to the Earth; one of its revolutions around the Earth (currently about 29 days) is equal to one of its rotations about its axis, so it always shows one face to the Earth. The Moon will continue to recede from Earth, and Earth's spin will continue to slow gradually. In about 50 billion years, if they survive the Sun's expansion, the Earth and Moon will become tidally locked to each other; each will be caught up in what is called a "spin–orbit resonance" in which the Moon will circle the Earth in about 47 days and both Moon and Earth will rotate around their axes in the same time, each only visible from one hemisphere of the other.

So the Moon will certainly still be around by the time the Sun becomes a red giant, and for many billions of years after that.

Here's the math: $$\frac{1 \text{ cm}}{1 \text{ year}} \times \frac{1 \text{ m}}{100 \text{ cm}} \times \frac{1 \text{ km}}{1000 \text{ m}} \times 5,400,000,000 \text{ years}=54,000 \text{ km}$$

That's about one-seventh the current distance to the Moon - at its closest pass.

  • 1
    $\begingroup$ Hello. Great answer. Can you improve it by adding the related math? $\endgroup$
    – hawaii
    Nov 18, 2014 at 23:27
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    $\begingroup$ Sure thing. Is there anything more you want? $\endgroup$
    – HDE 226868
    Nov 18, 2014 at 23:36
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    $\begingroup$ The math related to the gravitational pull of Earth, regarding that added distance. Im not sure that it will keep changing by 1cm/year if the Moon keeps getting farther away from Earth. But since you already answered my question, I will check your answer as correct (But I would like to see that calculation, just to be sure, as it is in my original question as well). $\endgroup$
    – hawaii
    Nov 18, 2014 at 23:43
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    $\begingroup$ @Hawaii I'm working on finding some credible calculations, but I can't promise anything today. $\endgroup$
    – HDE 226868
    Nov 19, 2014 at 0:42
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    $\begingroup$ Its ok. I would work on them myself, but I dont have enough skills for that. I think I would take up to a week to get everything right. If you can provide the calculations later, it will be awesome! $\endgroup$
    – hawaii
    Nov 19, 2014 at 0:44

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