MarkBour wrote:
The Moon's orbit is far enough out that it only circles the Earth once every ~28 days, whereas the Earth's own rotation is once every day (of course). (Yeah, yeah, solar, not sidereal, etc. and plenty of approximations.)
Best to round that number down to once every ~27 days in order that Kepler's 3rd Law
would state that the Moon is ~9 [=27
(2/3)] times further out than geostationary orbit.
MarkBour wrote:
The Moon pulls up some of the ocean, right towards it. But the terra firma underneath the ocean is rushing around ahead of the Moon, and drags a lot of the water forward of the Moon's orbit. This (I had recently read) is the reason the Moon is being urged forward, because of the gravitational imbalance this creates, causing a net force vector that urges the Moon to rotate faster around the Earth. It would be quite complicated, I am sure, to prove this, my sense is that it's a very messy calculus problem. But I'm accepting those statements.
A geostationary moon (like Pluto's Charon) would produce a
fixed ("
pulled up right towards it")
minimal energy tidal distortion that would
not dissipate any energy.
A non-geostationary moon would produce a
moving tidal distortion that would of necessity dissipate energy. This moving tidal energy is provided by having the peak tide
lag (in time) somewhat behind the moon.
If the moon is orbiting faster than geostationary then the peak tide also lags somewhat (in space)
but if the moon is slower than geostationary the time lag causes the peak tide to actually precede the moon (in space).
In either case the faster rotating object slows down in order to provide:
- 1) the tidal dissipation energy as well as
2) to impart energy to the slower rotating object.
MarkBour wrote:
as the Moon gets farther away, the entire effect must surely weaken.
When the Moon is twice as far from Earth as it is now, no doubt the tidal bulge itself will be smaller (I've no clue how much smaller, perhaps it's inverse square), and although it will now be even farther forward of the Moon, I'm sure it's going to have a much smaller effect on the Moon (perhaps inverse of distance to the fourth).
Tidal distortion forces drop off as
the inverse cube.
The tidal dissipation energy loss will depend upon
the square of the tidal distortion ... hence as 1/r
6.
The Earth has ~16 times the kinetic energy needed to send the Moon flying off into space; however, most of that energy will get dissipated by the Earth's moving tides.
If we could have ignored that tidal dissipation slowdown (as would be the case with Mars/Deimos) then a simple calculation would give:
d(E)/dt = d(1/r)/dt = -1/r
6 or r = t
(1/5) ... sending the Moon
slowly out into space [r => ∞ as t => ∞].
However, even this simple situation would get complicated if the Moon should ever reach the unstable Lagrangian points L1 & L2 [at 1.5 million kilometers (~4 lunar distances) from Earth for our Moon. However, the Sun would have become a red giant long before that].
MarkBour wrote:
So, although I may again have already taken a couple of wrong turns, I'm assuming this effect damp[en]s out. Perhaps the Moon never stops retreating, but it ought to slow down its retreat, if this is the only cause of it.
I still think I had one component of this right the first time. The other effect here is that the Moon is dragging the Earth's rotation, slowing it down through this tidal mechanism. (I have no idea of the comparison of the two effects: whether the Moon will slow down our day by a significant amount long before it gets far away, or if it is going to work its way out fast enough that it does not ultimately slow down our day very much.) But certainly, if you slow down the rate of the Earth's rotation by a noticeable amount (let's say you add a minute to our day) ... then "geostationary" needs to be a longer period, so it must be a higher orbit than it was before.
The Earth's rotation rate is currently being slowed with a half life of ~2,600 million years due to tidal friction while the Moon is currently spiraling away from the Earth at ~100,000 km per 2,600 million years.
However, the Moon already has over 91% of the total Earth/Moon angular momentum and can only move out ~70,000 km before it has ALL the Earth/Moon angular momentum and thus become a geostationary satellite.
So I think you are right that tidal dissipation will slow the Earth to the point that the geostationary satellite radius catches up to the Moon in a few billion years time.
MarkBour wrote:
My original question that I was wondering about was whether or not that effect, which must be microscopic in the period of a single year, worked out to anything an artificial satellite would actually notice. As Art answered, drag is more of a problem. I didn't realize that (a) there is much atmosphere at that altitude to create appreciable drag on an orbit. Also, (b) why aren't the air molecules at that altitude also rotating at the same speed as the satellite? Ummm ... really I know very little about the overall motion of the atmosphere, except for a few anecdotal bits I've read.
Actually, particles in the Van Allen belts might well be orbiting on average once a day because they are tied to the magnetic field.
In any event, significant tidal effects on orbital motions take place over tens of million years (for Phobos) or billions of years (for the Moon) and are of no real concern for man made objects.
https://en.wikipedia.org/wiki/Van_Allen_radiation_belt#Implications_for_space_travel wrote:
<<Spacecraft travelling beyond low Earth orbit enter the zone of radiation of the Van Allen belts. A region between the inner and outer Van Allen belts lies at two to four Earth radii and is sometimes referred to as the "safe zone". Solar cells, integrated circuits, and sensors can be damaged by radiation. Geomagnetic storms occasionally damage electronic components on spacecraft. Miniaturization and digitization of electronics and logic circuits have made satellites more vulnerable to radiation, as the total electric charge in these circuits is now small enough so as to be comparable with the charge of incoming ions. Electronics on satellites must be hardened against radiation to operate reliably. The Hubble Space Telescope, among other satellites, often has its sensors turned off when passing through regions of intense radiation. A satellite shielded by 3 mm of aluminium in an elliptic orbit (200 by 20,000 miles (320 by 32,190 km)) passing the radiation belts will receive about 2,500 rem (25 Sv) per year (for comparison, a full-body dose of 5 Sv is deadly). Almost all radiation will be received while passing the inner belt.
The Apollo missions marked the first event where humans traveled through the Van Allen belts, which was one of several radiation hazards known by mission planners. The astronauts had low exposure in the Van Allen belts due to the short period of time spent flying through them. Apollo flight trajectories bypassed the inner belts completely, and only passed through the thinner areas of the outer belts. Astronauts' overall exposure was actually dominated by solar particles once outside Earth's magnetic field. The total radiation received by the astronauts varied from mission to mission but was measured to be between 0.16 and 1.14 rads (1.6 and 11.4 mGy), much less than the standard of 5 rem (50 mSv) per year set by the United States Atomic Energy Commission for people who work with radioactivity.>>
https://en.wikipedia.org/wiki/Tidal_acceleration#Angular_momentum_and_energy wrote:
<<The gravitational torque between the Moon and the tidal bulge of Earth causes the Moon to be constantly promoted to a slightly higher orbit and Earth to be decelerated in its rotation. As in any physical process within an isolated system, total energy and angular momentum are conserved. Effectively, energy and angular momentum are transferred from the rotation of Earth to the orbital motion of the Moon (however, most of the energy lost by Earth (−3.321 TW) is converted to heat by frictional losses in the oceans and their interaction with the solid Earth, and only about 1/30th (+0.121 TW) is transferred to the Moon). The Moon moves farther away from Earth (+38.247±0.004 mm/y), so its potential energy (in Earth's gravity well) increases.
The rotational angular momentum of Earth decreases and consequently the length of the day increases. The net tide raised on Earth by the Moon is dragged ahead of the Moon by Earth's much faster rotation. Tidal friction is required to drag and maintain the bulge ahead of the Moon, and it dissipates the excess energy of the exchange of rotational and orbital energy between Earth and the Moon as heat. If the friction and heat dissipation were not present, the Moon's gravitational force on the tidal bulge would rapidly (within two days) bring the tide back into synchronization with the Moon, and the Moon would no longer recede. Most of the dissipation occurs in a turbulent bottom boundary layer in shallow seas such as the European Shelf around the British Isles, the Patagonian Shelf off Argentina, and the Bering Sea.
The dissipation of energy by tidal friction averages about 3.75 terawatts, of which 2.5 terawatts are from the principal M2 lunar component and the remainder from other components, both lunar and solar.
An equilibrium tidal bulge does not really exist on Earth because the continents do not allow this mathematical solution to take place. Oceanic tides actually rotate around the ocean basins as vast gyres around several amphidromic points where no tide exists. The Moon pulls on each individual undulation as Earth rotates—some undulations are ahead of the Moon, others are behind it, whereas still others are on either side. The "bulges" that actually do exist for the Moon to pull on (and which pull on the Moon) are the net result of integrating the actual undulations over all the world's oceans. Earth's net (or equivalent) equilibrium tide has an amplitude of only 3.23 cm, which is totally swamped by oceanic tides that can exceed one metre.
This mechanism has been working for 4.5 billion years, since oceans first formed on Earth. There is geological and paleontological evidence that Earth rotated faster and that the Moon was closer to Earth in the remote past. Tidal rhythmites are alternating layers of sand and silt laid down offshore from estuaries having great tidal flows. Daily, monthly and seasonal cycles can be found in the deposits. This geological record is consistent with these conditions 620 million years ago: the day was 21.9±0.4 hours, and there were 13.1±0.1 synodic months/year and 400±7 solar days/year. The average recession rate of the Moon between then and now has been 2.17±0.31 cm/year, which is about half the present rate.
From the observed change in the Moon's orbit, the corresponding change in the length of the day can be computed: +2.3 ms/century
However, from historical records over the past 2700 years the following average value is found: +1.70 ± 0.05 ms/century
Opposing the tidal deceleration of Earth is a mechanism that is in fact accelerating the rotation. Earth is not a sphere, but rather an ellipsoid that is flattened at the poles. SLR has shown that this flattening is decreasing. The explanation is that during the ice age large masses of ice collected at the poles, and depressed the underlying rocks. The ice mass started disappearing over 10000 years ago, but Earth's crust is still not in hydrostatic equilibrium and is still rebounding (the relaxation time is estimated to be about 4000 years). As a consequence, the polar diameter of Earth increases, and the equatorial diameter decreases (Earth's volume must remain the same). This means that mass moves closer to the rotation axis of Earth, and that Earth's moment of inertia decreases. This process alone leads to an increase of the rotation rate (phenomenon of a spinning figure skater who spins ever faster as she retracts her arms). From the observed change in the moment of inertia the acceleration of rotation can be computed: the average value over the historical period must have been about −0.6 ms/century. This largely explains the historical observations.>>
Phobos: Moon over Mars
) The counterintuitive part is that as this happens, the angular velocity of the Moon around the Earth is actually going to decrease. It's counterintuitive to me because we have this force urging it forward, but the net effect of that force is that it will get worse, and always be behind the tidal bulge. It is not getting a higher angular velocity and approaching equilibrium. The only thing it's orbital angular velocity in our sky is approaching in this process is 0, and it will be getting a little more like the fixed stars. It is always going to proceed across the night sky, even after eons of this effect. Indeed, it will cross our night sky more quickly, as do the objects that are not orbiting us (Sun, planets, stars).
