# Moon 16,000 km from Earth?

Gannet

And there is a statement
Note that, in those times, the Moon, still burning, was only 16,000 km from the Earth (compared to 384,000 km today)...

I find this statement hard to believe, if so, I would envisioned that the molten Earth and the Moon would be more ellipsoid in shape instead of spheroid.

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spacester
I've seen that before, and I looked at the math once a long time ago and it seemed correct. It is more than a little difficult to visualize, but science cares not about that.

I don't understand what you are getting at with the ellipsoid thing. The close orbit would not preclude the planets from being non-spherical, but it would seem that gravity would pull things into spheres right away.

Ah, are you thinking about extreme tidal forces causing the out-of-spherical shape? Interesting question.

Gannet

Yes it is the effect of tidal forces that I am envisioning.

Furthermore, I envisioned that when the Moon gets into geosynchronous orbit, the tidal forces acting on the same location for prolong period of time would have a detrimental effect on the weak structured Earth at that time.

Gold Member

Yes it is the effect of tidal forces that I am envisioning.

Furthermore, I envisioned that when the Moon gets into geosynchronous orbit, the tidal forces acting on the same location for prolong period of time would have a detrimental effect on the weak structured Earth at that time.

Given that the Earth was initially spinning much faster, I think the Moon would always have been outside a geosynchronous orbit.

spacester
eh? Of course, the rate of spin has nothing to do with the strength of the gravitational field.

Although it does pulsate slightly due to uneven mass distribution, this would be a minor perturbation of the orbit, not at all a determining factor.

Gannet
Given that the Earth was initially spinning much faster, I think the Moon would always have been outside a geosynchronous orbit.

Thanks Jonathan, you caught me imagining without a calculator.

http://www.scientificpsychic.com/etc/timeline/timeline.html" [Broken]

which states the following:
Hadean Eon (4567 to 3800 mya)
- Earth day is 6 hours long

Archaean Eon (3800 to 2500 mya)
- Earth day is 15 hours long

I am going to do some calculations and I'll get back.

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litup
Thanks Jonathan, you caught me imagining without a calculator.

http://www.scientificpsychic.com/etc/timeline/timeline.html" [Broken]

which states the following:

I am going to do some calculations and I'll get back.

I'm waiting:)
Seriously, if the moon was 16K kilometers from Earth, wouldn't that have put it inside the Roche limit? Here is a chart from one Wiki:

Body Satellite Roche limit (rigid) Roche limit (fluid)
Distance (km) R Distance (km) R
Earth Moon 9,496 1.49 18,261 2.86
Earth average Comet 17,880 2.80 34,390 5.39
Sun Earth 554,400 0.80 1,066,300 1.53
Sun Jupiter 890,700 1.28 1,713,000 2.46
Sun Moon 655,300 0.94 1,260,300 1.81
Sun average Comet 1,234,000 1.78 2,374,000 3.42

It seems to be saying if the moon came within 18,000 Km of Earth it would be toast. Does that sound right? If it actually was at 16,000 Km I would think it would have crumpled up into small pieces. No, I guess it has to come within 9,000 km. I see the 18k # is for a fluid body. Sorry.

I guess that would indicate if a Mars sized body crashing into Earth was the cause of making the moon, the detritus came together past 16,000 Km. I assume if the stuff tried to come together at say 1000 Km up, it never would have coalesced into the moon we know today.

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captn
...the rate of spin has nothing to do with the strength of the gravitational field.

True, but it does have to do with where force is applied.

Consider Earth: if the moon were at the geosynchronous distance, the pull of the moon would always be on the same 'spot' on earth. Beyond (wrt earth) the geosync distance, the Earth would spin faster than the moon orbits and thus the moon's affect on the shape of the Earth would wash over the earth-- just like current-day tides.

Moon: when the moon was molten, it was spinning faster than its revolution around the Earth so its bulge would swirl around the moon (similar to the discussion wrt earth).

For both the Earth and moon, gravity may be stronger than the centrifugal force that tries to make then bulge towards each other. Of course, since there are currently tides, the shapes of the Earth and moon are not strictly spherical, but the deviation from sphericity is irrelevant for most non-science/non-oceanic work.

For the moon, this non-sphericity (bulge) is what slowed the moon's rotation and caused the moon's day to lock step with its period of revolution about earth.

Neil

Staff Emeritus
Gold Member

And there is a statement

I find this statement hard to believe, if so, I would envisioned that the molten Earth and the Moon would be more ellipsoid in shape instead of spheroid.

16000 is 1/24 of the distance of 384000. Since tides decrease by the cube of the distance, the tidal force of moon on Earth would be 13824 times larger.

At present, the deformation of the Earth due to Lunar tidehas a maximum of 55 cm.

The Earth being "molten" at that time really doesn't matter. The Earth today is mostly molten with just a thin crust and that thin crust has little effect on the lunar tidal deformation of the Earth.

Now we can estimate that the highest Earth tidal bulge would be around 7600m or about 1200 m less than the height of Mt Everest.

Given that scaled to the size of a billiard ball, the Earth, with all its mountains, would be smoother than a billiard ball, you wouldn't see any visible deviance from a spheroid shape for the Earth.

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litup
16000 is 1/24 of the distance of 384000. Since tides decrease by the cube of the distance, the tidal force of moon on Earth would be 13824 times larger.

At present, the deformation of the Earth due to Lunar tidehas a maximum of 55 cm.

The Earth being "molten" at that time really doesn't matter. The Earth today is mostly molten with just a thin crust and that thin crust has little effect on the lunar tidal deformation of the Earth.

Now we can estimate that the highest Earth tidal bulge would be around 7600m or about 1200 m less than the height of Mt Everest.

Given that scaled to the size of a billiard ball, the Earth, with all its mountains, would be smoother than a billiard ball, you wouldn't see any visible deviance from a spheroid shape for the Earth.

But if the moon came that close and therefore tore itself apart the tidal effect would get a lot less severe I would think due to the moon's mass spreading out in a disk, so the tidal effect would pretty much disappear at that point.

Staff Emeritus
The Earth today is mostly molten with just a thin crust and that thin crust has little effect on the lunar tidal deformation of the Earth.
The Earth is mostly solid. Plastic, but solid.

But if the moon came that close and therefore tore itself apart the tidal effect would get a lot less severe I would think due to the moon's mass spreading out in a disk, so the tidal effect would pretty much disappear at that point.
The leading idea regarding the formation of the Moon is that a Mars-sized object collided with the Earth shortly after the Earth formed. Most of the mass from that collision added to the Earth's mass, but some went into orbit. That orbiting ejecta quickly formed the Moon, but orbiting much closer to the Earth than the current 384,000 km orbit. In short, the Moon didn't tear itself apart. It pulled itself together.

Gannet

Thanks litup I forgot about the Roche Limit which was mentioned in my Astronomy book Jastrow, Robert & Malcolm H. Thompson; Astronomy: Fundamentals and Frontiers, Second Edition; John Wiley & Sons, Inc.; New York; 1974; p355 & 356 (It was current when I took Astronomy as an Art & Science elective) which states:
All planets and stars possesses Roche limits. The Roche limit for the Sun is one million miles from its center, or approximately 500,000 miles above its surface. No planet could form within this distance. The Roche limit for the Earth is 10,000 miles from its surface, well inside the orbit of the Moon.

I found this site http://media4.obspm.fr/exoplanets/pages_outil-roche/calcul-limiteRoche.html" [Broken] which gave an explanation that was easy for me to understand

Therefore, the two masses are separated if the distance D is less than (2^4/3)R(rhoplanet/rhosatelite)^1/3
and assuming crudely that the density of the Earth and moon and radius of the Earth were approximately the same as today I get 18,482 km.

Assumption above is based on http://www.palaeos.com/Hadean/Cryptic.html" [Broken] which states
After that, things were kept lively for another 30 My or so by impacts between the inner planets and a few remaining planetoids, such as the Earth-Theia collision. For the most part, these late encounters added no net mass to the inner planets. The bodies involved were probably still molten. Greenwood et al. (2005).

It seems to me, that the 16,000 km was deduce such that when the Earth captured the Moon it was just outside the Roche limit. If this was so, wouldn't the Lunar Seas on the Moon which after Apollo flowed from 3.1 to 3.8 bya be more like blisters instead of flats? In other words, I would expect a lot more visual indicators on the Moon of the Earth's tidal forces.

Janus thanks for your excellent and clearly understandable comments. If the Earth deforms 55 cm today due to the tidal forces, it makes me wonder how much of an effect the Earth has on the Moon today and then amplify that using the 13,824 factor.

This discussion got me wondering about all the other planetary satellites but that will be another thread

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