# How closely could the Moon orbit the Earth

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without having devistating consequences for Earth?

My kids (6 and 4) came up with the question last night at bedtime and it is an interesting one.

Clearly, none of the moon can contact the roughly 65 miles above Earth that includes the atmosphere, and given its size that puts a bare minimum at about 700 miles, but I suspect that it would still be a huge problem that close, and it has to be stable. But, a fairly close orbit could be stable with the right momentum.

With just 1.2% of Earth's mass, I'd think you could get fairly close. Of course, the lunar cycle would get shorter, and the tides would get stronger since the gravitational effect as a shorter distance would be greater. The Moon would also look huge in the sky.

Right now the Moon is at 225 to 252 thousand miles at any given time, and is about 1,080 miles in radius. Could it have a stable orbit that would wreck total havoc on Earth at 22-25 thousand miles, for example?

Zelos
it cant be so close to the earth so the atmoshperic friction comes into action it doesnt matter exept it how close it is aslongest it travel fast enough. and it travel at a distance of 384400 km. ;) when u talk about science use SI units.

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ohwilleke said:
without having devistating consequences for Earth?
Devistating is a subjective word.

If we lived in a world with no moon, and then one day the Moon popped into existance in its current orbit, we'd probably consider that devistatning as we never experienced tides that are the size the Moon generates.

If the Moon were just a few thousand kilometers away it would generate HUGE tides. The tidal force equation can tell you how strong they would be:

$$F_{tidal} =\frac{2GMmr}{d^3}$$

Where G = 6.67e-11, M (mass of Earth) = 5.98e24, m (mass of moon) = 7.349E+22, d = the distance that seperates them, and r (radius of Earth) = 6371000.

Notice that d gets cubed. So moving distance is a very sensitive variable. That's why the tides generated by the Sun are small compared to tides generated by the Moon, despite the Sun being 30 million times as massive as the Moon. It's 400 times farther away and that greatly neutralizes its ability to pull tides.

Distances are measured from center to center, so the Moon could not possibly be any closer than the radius of the Moon + the radius of the Earth + a little extra for Earth's atmosphere(~8000 km). But watch out for the Roche limit. (It'd be easier for you to google it than for me to explain it ).

The Moon must be far enough away that its orbit takes at least 24 hours. Otherwise, the Moon will sprial in and hit Earth rather than spiral outward.

But if the Moon orbited the Earth in exactly the same time as its orbital period, and the Earth and Moon were tidally locked, then you would perceive no tides even though the Moon generated tides would be huge.

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derekmohammed
I am sorry but I have never seen this version of the tidal wave equation before. Could you explain (briefly, don't want to be a pest...) its derivation?
Thanks
Derek

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ohwilleke said:
without having devistating consequences for Earth?

My kids (6 and 4) came up with the question last night at bedtime and it is an interesting one.

If you consider devastating consequences for the moon, there's a well-known answer, the Roche limit (I see Tony has mentioned this already).

http://en.wikipedia.org/wiki/Roche_limit

Numerically, this is somewhere between 1.26 R $\sqrt[3]{\rho_m/\rho_e}$ and 2.423 R $\sqrt[3]{\rho_m/\rho_e}$

call it 6800 km to 13000 km

This is well under a 24 hour period for the orbit, though. The low value might also be within the atmosphere (so it's probably too low). Tony is right in saying that a moon in this low of an orbit would tend to spiral inwards rather than outwards - unless you shorten the length of the Earth's day to match.

This is starting to get complicated - bear with me as I try to figure it out.

While the moon's orbit will decay, the Earth's spin will increase (and it's day will shorten). So being in an orbit with a lower than 24 hour period doesn't necessarily mean the moon moves inside the Roche limit and breaks up, what will tend to happen if the system has enough angular momentum is that the moon will spiral in a bit, the Earth will speed up, and the system will achieve stability when the moon is tide-locked to the Earth.

I think that the bottom line is that as long as the Moon stays outside the Roche limit, everything will probably be OK.

What will tend to happen is that the Earth-moon will tidal lock very rapidly. This has nearly happened already - and with the process depending on the sixth power of the separation, if the moon was any closer, it already would have happened already.

I would guess that the tidal lock process would happen even before the Earth cooled down. (I ought to work out how much energy is being dumped into the Earth by the tidal lock process, but I don't recall the exact coefficients of the process. I've seen them posted once. The concern is how much this affects the temperature of the planet).

[add - basically, you look at the total amount of energy stored in the tides, and assume some fraction of it is dissipated every orbital cycle via friction. This gives the energy dissipation rate and also the lock time. This is the origin of the 6th power law, the tidal height is proportional to 1/r^3, and the tidal energy is proportional to the square of the tidal height.]

But I think that the tidal lock would happen quite rapidly, and all the fireworks would be over before life started to evolve. (I think current theories of moon formation are based on a large impactor very early in the Earth's history).

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pervect said:
I think that the bottom line is that as long as the Moon stays outside the Roche limit, everything will probably be OK.

I may be mistaken, but I don't think of the moon is nearly enough structural strength to withstand title forces at such close proximity. Wouldn't it be torn to shreds, with many of the pieces falling to Earth?

As has been stated earlier, "devastating consequences" is a rather subjective term. However, I seem to recall hearing that if the Moon were just 2% closer to the Earth than it is, nearly the entire face of the planet would be washed completely over twice daily by the tides. The unaffected areas would mostly be some very high mountain tops which are considered uninhabitable anyway.

Not sure how much affect the title flexing of the crust would have on tectonic activity, or climate.

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Zelos said:
it cant be so close to the earth so the atmoshperic friction comes into action it doesnt matter exept it how close it is aslongest it travel fast enough. and it travel at a distance of 384400 km. ;) when u talk about science use SI units.

My source was in miles and I didn't have time to do the quick conversion.

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LURCH said:
I may be mistaken, but I don't think of the moon is nearly enough structural strength to withstand title forces at such close proximity. Wouldn't it be torn to shreds, with many of the pieces falling to Earth?

As has been stated earlier, "devastating consequences" is a rather subjective term. However, I seem to recall hearing that if the Moon were just 2% closer to the Earth than it is, nearly the entire face of the planet would be washed completely over twice daily by the tides. The unaffected areas would mostly be some very high mountain tops which are considered uninhabitable anyway.

Not sure how much affect the title flexing of the crust would have on tectonic activity, or climate.

Given the formula's stated above, that can't be the case. 1/.98^3=is about a 6% increase.

At a 10th of the distance, the distance I suggested as an example, the tides would be up by a factor of 1,000 (subject to the caveat of rapid locking).

The average tidal rise and fall in various North American cities typically ranges from about .3 meters (Cristobal, Panama) to 3.7 meters (Seattle). The maximum tidal variation in the Bay of Fundy, in Nova Scotia is about 13.3 meters. Certainly, a 13 km tide would be pretty freaky.

So, absent locking, that would be pretty devistating.

But, a levee could certainly handle 100m tides and they wouldn't go all that far inland. To get 100m tides in Seattle, and 50m tides in New York, you are talking about a Moon which is at about 30% of its current distance from Earth.

Thanks all for the help.

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derekmohammed said:
I am sorry but I have never seen this version of the tidal wave equation before. Could you explain (briefly, don't want to be a pest...) its derivation?
Thanks
Derek
The tidal force that the Moon raises on the Earth is the difference between the force it exerts on the side of Earth closest to it, and the force it exerts on the side of Earth farthest from it.

That's why distance is so critical. Not only does gravitational force decrease exponentially with distance (inverse square law):

$$F_{gravity}=\frac{GMm}{d^2}$$

But the ratio between the distance of the near side of an object and the far side of an object also diminishes with distance.

Just a trivial fact. If your car is up on a hydrolic lift for an oil change, and you stand under your car, your car exerts a larger tidal force on your body than the Moon does. Remember that next time someone tells you that your body is 90% water, therefore the full moon influences you just like it influneces Earth.

Here's a link discussing how the tidal force equation is derived:

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tony873004 said:
That's why distance is so critical. Not only does gravitational force decrease exponentially with distance (inverse square law):

$$F_{gravity}=\frac{GMm}{d^2}$$

But the ratio between the distance of the near side of an object and the far side of an object also diminishes with distance.

Just some small things (sorry to nitpick). An inverse square law is not an exponential decrease, it's a power law. Also, the last sentence isn't quite right. What really decreases with distance is the ratio of the size of the object being effected (in this case, the earth) and the distance to the object (earth-moon distance).

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SpaceTiger said:
Just some small things (sorry to nitpick). An inverse square law is not an exponential decrease, it's a power law. Also, the last sentence isn't quite right. What really decreases with distance is the ratio of the size of the object being effected (in this case, the earth) and the distance to the object (earth-moon distance).
I don't mind the nitpickin' :yuck:

You're right. I've got it clear in my mind but didn't take enough time phrase it properly. The last sentence, the way I phrased it would be a constant. Thanks for cleaning it up ST.

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ohwilleke said:
Given the formula's stated above, that can't be the case. 1/.98^3=is about a 6% increase.

At a 10th of the distance, the distance I suggested as an example, the tides would be up by a factor of 1,000 (subject to the caveat of rapid locking).

Yep. Maybe I'm belaboring the point a bit, but if the moon was 10x closer, and the tides 1000x bigger, tidal locking would happen 1,000,000 times faster. So, in my version of the problem, I imagined a tidally locked moon that was 10x closer.

The average tidal rise and fall in various North American cities typically ranges from about .3 meters (Cristobal, Panama) to 3.7 meters (Seattle). The maximum tidal variation in the Bay of Fundy, in Nova Scotia is about 13.3 meters. Certainly, a 13 km tide would be pretty freaky.

Quite a long time ago I thought about this problem a little bit. You can see the archived discussion at the link below (from rec.arts.sf.science).

I estimated about 500m tides with the moon 10x closer as you can see in the link. But there exists better formulas than the ones I used, they are supplied in the above link via Brian Davis. I'll re-quote them here, I'm not sure if Brian gave his source in the thread referenced, but I'm fairly sure it was

http://www.uapress.arizona.edu/books/BID76.htm

(I don't own this book personally, though it looks nice)

Brian said:
r_1 = ( 1 + (14/9) h gamma ) r radius along primary-satellite line
r_2 = ( 1 - (4/9) h gamma ) r radius perpendicular to r_1 and
rotation axis
r_3 = ( 1 - (10/9) h gamma ) r radius along rotation axis
where,
gamma = equilibrium tidal height
= (3/4) (M/m) (r/a)^3
M = mass raising the tides
m = mass experiencing the tides
a = seperation distance between the two
h = correction factor for self-gravitation and rigidity
= 0.4167 for Earth
= 0.0333 for Moon
= 5/2 for an ideal self-gravitating liquid (zero strength) body

I haven't plugged the numbers into this better formula to see what pops out for estimated tidal heights.

One of the things I thought of back then but didn't think of this go-around was the problem of orbital plane precession and circularization.

Unfortunately from what I can tell, by ignoring the effect of the Sun I did not treat the problem properly.

The moon currently has an 18 year orbital plane precession frequency from what I can determine. See "draconic month" in the URL below. That's not what the formula I was quoting from Goldstein would predict at all! So I'm probably off many, many orders of magnitude in the earlier post I cited - beware.

http://www.hermit.org/Eclipse/why_months.html

Simulating this lunar plane precession would be an interesting task, but it requires the treatment of the gravity field of the non-spherical Earth, so simple gravity simulators that use a "point mass" approximation won't work. One needs to simulate at least the "quadropole moment" terms of the Earth's gravity to see the effect. It looks like the Suns' gravity is important too (bevcause the quadropole formula alone didn't give anywhere near the right answer).

Note that even with the Moon's orbit regularized, there will be some pertubations of it via the sun's gravity (and that of other planets). These pertuabations are potentially important because of the very large size of the tides that the moon generates.

So there are a lot of interesting things to consider, even if the moon is tidallly locked, because the "lock" won't really be a perfect lock.

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LURCH said:
I may be mistaken, but I don't think of the moon is nearly enough structural strength to withstand title forces at such close proximity. Wouldn't it be torn to shreds, with many of the pieces falling to Earth?

If the moon is inside the Roche limit, it will get torn apart by the Earth's gravity. That's the whole point of the calculation I was doing (see the wikipedia article). Consider the two numbers I computed (from the formulas in the Wikipedia) as an upper and lower bound on when one would expect the "tearing apart" process to occur.

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pervect said:
Simulating this lunar plane precession would be an interesting task, but it requires the treatment of the gravity field of the non-spherical Earth, so simple gravity simulators that use a "point mass" approximation won't work.

Actually, they will. Here's a link describing a simulation I did on this 18 year precession of node period:

http://www.orbitsimulator.com/gravity/articles/saros.html

There's an animated GIF towards the bottom of the page that shows the orientation of the Moon's orbit precessing in an 18 year period.

This simulation uses point masses for the Earth, Moon, AND Sun. If you delete the Sun from the simulation, the orbit doesn't precess at all. So the precession is caused exclusively by the Sun's influence.

As your distance increases from an object, the object, no matter what its shape, begins behaving like a point mass.

That's why the Moon only receeds a few inches a year as a result of Earth's non-spherical shape. (The non-spherical part caused by the tides the Moon pulls.)

The interesting thing about my simulation is that it is run way too fast (16384 second time step) to keep the Moon accurate with regards to most properties of its orbit. With a slower time step (~16 seconds - 256 seconds) the Moon's position almost remains accurate enough to predict eclipses 50 years in the future. (Probably non-spherical Earth is a major player in preventing perfect accuracy, as well as GR, and uncertainty in masses and G). But run at the speed you see in the animated GIF, positioning accuracy is thrown out the window. However, the precession of nodes marches forward on schedule.

Another thing to consider when thinking about how much higher a water tide would be in the presence of a stronger gravitational tide is that the water tide is happening on top of a land tide. The Earth is basically a liquid / semi-liquid sphere with a thin solid shell. The tides raised by the Moon are much larger than we observe. We only observe how much higher the water tides rise above the land tides.

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Apparently I'm going to have to study the whole precessional thing more closely. It's possible I'm getting confused between different types of precession. (Precession of the line of nodes vs orbital planar precession). So far, Google finds a number of "cranky" webpages but not much really good stuff on standard lunar-solar precession theory :-(.

In any event, these sorts of small, "fiddly" effects would become much more important if the moon was 10x closer to the Earth, because anything that moves the Moon at all is going to move it's tides as well, which are quite large. This is what makes the problem interesting (and challenging).

I do exepct that omitting the oblatness of the Earth is the biggest factor that limits your (Tony's) simulations. The effect will be most important for close-in objects (it's got a huge effect on low earth orbit satellites, for instance). Relativistic effects are probably pretty minor.

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pervect said:
So there are a lot of interesting things to consider, even if the moon is tidallly locked, because the "lock" won't really be a perfect lock.

In other words, while typical tides might actually appear milder in a locked situation, if that lock was imperfect, an occassional bobble might produce tsunami class tidal waves from time to time.

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