APOD: Phobos: Moon over Mars (2017 Jul 21)

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APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by APOD Robot » Fri Jul 21, 2017 4:05 am

Image Phobos: Moon over Mars

Explanation: A tiny moon with a scary name, Phobos emerges from behind the Red Planet in this timelapse sequence from the Earth-orbiting Hubble Space Telescope. Over 22 minutes the 13 separate exposures were captured near the 2016 closest approach of Mars to planet Earth. Martians have to look to the west to watch Phobos rise, though. The small moon is closer to its parent planet than any other moon in the Solar System, about 3,700 miles (6,000 kilometers) above the Martian surface. It completes one orbit in just 7 hours and 39 minutes. That's faster than a Mars rotation, which corresponds to about 24 hours and 40 minutes. So on Mars, Phobos can be seen to rise above the western horizon 3 times a day. Still, Phobos is doomed.

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by bystander » Fri Jul 21, 2017 4:34 am

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by RocketRon » Fri Jul 21, 2017 5:29 am

So how is it that Earth's moon was once close in, and is slowing moving out.
And Phobos was further out, and is slowly moving in.
??

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by gwrede » Fri Jul 21, 2017 6:53 am

What is the light area on the right side of Mars? Is it snow, heavy overcast, or what?
- Where's Latin?
- Alas, it's out of scope.

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by neufer » Fri Jul 21, 2017 9:31 am


RocketRon wrote:
So how is it that Earth's moon was once close in, and is slowing moving out.
And Phobos was further out, and is slowly moving in. ??
Because Phobos completes one orbit faster than a Mars rotation such
that Mars's reactive tidal distortions lag behind Phobos pulling Phobos back.

The Moon completes one orbit slower than an Earth rotation such
that Earth's reactive tidal distortions precede the Moon pulling it forward.
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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by Asterhole » Fri Jul 21, 2017 11:57 am

RocketRon wrote:So how is it that Earth's moon was once close in, and is slowing moving out.
And Phobos was further out, and is slowly moving in.
??
Earth's Moon came from the Earth itself when a titanic meteoric impact billions of years ago caused a large chunk to be cast off. In contrast, Phobos - and Deimos - probably were asteroids that became captured by Mars' gravitational field.

We know that not all of the satellite moons of the Solar System were formed in a like manner. Some, such as the larger Jovian and Saturnian moons most likely coalesced in orbit around their planets. Others like Saturn's Phoebe and Neptune's' Triton were captured as Mars' were. And still others - we just don't know.
They're all wasted!

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by Cousin Ricky » Fri Jul 21, 2017 12:33 pm

APOD Robot wrote:That's faster than a Mars rotation, which corresponds to about 24 hours and 40 minutes. So on Mars, Phobos can be seen to rise above the western horizon 3 times a day. Still, Phobos is doomed.
“Still”? Shouldn’t that be “therefore,” Phobos is doomed?

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by rstevenson » Fri Jul 21, 2017 2:00 pm

Cousin Ricky wrote:
APOD Robot wrote:That's faster than a Mars rotation, which corresponds to about 24 hours and 40 minutes. So on Mars, Phobos can be seen to rise above the western horizon 3 times a day. Still, Phobos is doomed.
“Still”? Shouldn’t that be “therefore,” Phobos is doomed?
adverb: still
2.
nevertheless; all the same.
"I'm afraid he's crazy. Still, he's harmless"
synonyms: nevertheless, nonetheless, regardless, all the same, just the same, anyway, anyhow, even so, yet, but, however, notwithstanding, despite that, in spite of that, for all that, be that as it may, in any event, at any rate.

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by neufer » Fri Jul 21, 2017 2:36 pm

APOD Robot wrote:
[Phobos] completes one orbit in just 7 hours and 39 minutes.
That's faster than a Mars rotation, which corresponds to about 24 hours and 40 minutes.

So on Mars, Phobos can be seen to rise above the western horizon 3 times a day.
To be clear:

1) Deimos rises above the eastern horizon once every 5.328 days/sols on average.
2) Phobos rises above the western horizon 2.217 times a day/sol on average.

Ergo: "Phobos can be seen to rise above the western horizon 3 times a day" on only ~145 days a year.
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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by ems57fcva » Fri Jul 21, 2017 3:57 pm

neufer wrote:
APOD Robot wrote:
[Phobos] completes one orbit in just 7 hours and 39 minutes.
That's faster than a Mars rotation, which corresponds to about 24 hours and 40 minutes.

So on Mars, Phobos can be seen to rise above the western horizon 3 times a day.
To be clear:

1) Deimos rises above the eastern horizon once every 5.328 days/sols on average.
2) Phobos rises above the western horizon 2.217 times a day/sol on average.

Ergo: "Phobos can be seen to rise above the western horizon 3 times a day" on only ~145 days a year.
Neufer - You beat me to this one. The statement that "Phobos can be see to rise ... 3 times a day" is incorrect and caused by not taking into account the rotation of Mars. Overall, I would amend your statement to being that Phobos rises once every 2.217 days as seen from a location on Mars. Phobos has low eccentricity, making it so that when it rises probably differs by no more than a few seconds from the average.

The important thing is that Phobos can be counted on to rise only twice a day, with a 3rd time coming once every 4 or 5 days/sols.

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by Donald Pelletier » Fri Jul 21, 2017 4:57 pm

The time interval beetween two rises of Phobos is 11h 6min, so on Mars you see 2 rises of that moon per day, rarely 3 (https://en.wikipedia.org/wiki/Phobos_(m ... cteristics)

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by neufer » Fri Jul 21, 2017 6:07 pm

Donald Pelletier wrote:
The time interval beetween two rises of Phobos is 11h 6min, so on Mars you see 2 rises
of that moon per day, rarely 3 (https://en.wikipedia.org/wiki/Phobos_(m ... cteristics)
Well, if one is pole-ward of 70.4º one can never see Phobos at all.

Technically, (although somewhat misleading) the APOD statement stands:

"Phobos can be seen to rise above the western horizon 3 times a day."
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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by MarkBour » Fri Jul 21, 2017 6:41 pm

So our moon is slowly being urged out toward the altitude of a geostationary orbit. The description I read of this, which made some sense, is that the Earth's tidal bulge is what causes this effect on the Moon. And because of the Moon's effect on Earth's rotation, the geostationary distance is slowly increasing. We have quite a few tiny artificial satellites in geostationary orbits. Is the slowing of Earth's rotation enough that it causes any noticeable need for adjustment in these satellites? Or would it auto-correct their orbits, if the change was noticeable?
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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by neufer » Fri Jul 21, 2017 8:52 pm

MarkBour wrote:
So our moon is slowly being urged out toward the altitude of a geostationary orbit.
Our moon is slowly being urged out away from the lower quasi-stable altitude of a geostationary orbit.

(Deimos, likewise, is slowly being urged out away from the lower quasi-stable altitude of a geostationary orbit.)
MarkBour wrote:
The description I read of this, which made some sense, is that the Earth's tidal bulge is what causes this effect on the Moon. And because of the Moon's effect on Earth's rotation, the geostationary distance is slowly increasing. We have quite a few tiny artificial satellites in geostationary orbits. Is the slowing of Earth's rotation enough that it causes any noticeable need for adjustment in these satellites? Or would it auto-correct their orbits, if the change was noticeable?
Tiny artificial satellites (including the ISS) must deal with drag first & foremost.
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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by MarkBour » Fri Jul 21, 2017 11:19 pm

neufer wrote:
MarkBour wrote: So our moon is slowly being urged out toward the altitude of a geostationary orbit.
Our moon is slowly being urged out away from the lower quasi-stable altitude of a geostationary orbit.
...
  • Doh! I see where I got this major misconception, thanks for the correction.
  • The Moon is already way, way farther from Earth than geostationary orbits.
  • So, I want to try again ...
  • At the same time, fair warning. The following is nearly the opposite of a helpful article on this issue. It is rather the kind of thing one might scribble on some scratch paper while trying to work it out.
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.) 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.

Next, what does that actually do to the Moon? It's one of those great counterintuitive things about orbital mechanics that even a neophyte such as myself has learned about. If you add forward velocity to the Moon, it moves to a higher orbit. So the Moon is slowly moving away from us. (I couldn't mess up that part, since I'd heard it throughout the various articles and comments. :-) ) 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).

I suppose some of the thoughts that come next from this are interesting --- 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). So, although I may again have already taken a couple of wrong turns, I'm assuming this effect dampens 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. 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.
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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by Boomer12k » Fri Jul 21, 2017 11:45 pm

I wonder if it will break up and become a ring....

That is an incredible image of Mars...

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by neufer » Sat Jul 22, 2017 4:55 pm

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)
:arrow: 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/r6.

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/r6 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.>>
Last edited by neufer on Sun Jul 23, 2017 4:40 pm, edited 5 times in total.
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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by Guest » Sat Jul 22, 2017 11:17 pm

gwrede wrote:What is the light area on the right side of Mars? Is it snow, heavy overcast, or what?
I had the same question. I replied via email and was told to ask the question on Asterisk. presume that means right here. So how can it be out of scope?

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Re: APOD: Phobos: Moon over Mars (2017 Jul 21)

Post by bystander » Sat Jul 22, 2017 11:42 pm

gwrede wrote:What is the light area on the right side of Mars? Is it snow, heavy overcast, or what?
Guest wrote:I had the same question. I replied via email and was told to ask the question on Asterisk. presume that means right here. So how can it be out of scope?

Clouds above Syrtis Major.

http://heritage.stsci.edu/2016/15/
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Tidal deceleration

Post by neufer » Sun Jul 23, 2017 12:56 pm

https://en.wikipedia.org/wiki/Tidal_acceleration#Tidal_deceleration wrote:
Tidal deceleration comes in two varieties:
........................................................
1) Fast satellites: Some inner moons of the giant planets and Phobos orbit within the synchronous orbit radius so that their orbital period is shorter than their planet's rotation. In other words, they orbit their planet faster than the planet rotates. In this case the tidal bulges raised by the moon on their planet lag behind the moon, and act to decelerate it in its orbit. The net effect is a decay of that moon's orbit as it gradually spirals towards the planet. The planet's rotation also speeds up slightly in the process. In the distant future these moons will strike the planet or cross within their Roche limit and be tidally disrupted into fragments. However, all such moons in the Solar System are very small bodies and the tidal bulges raised by them on the planet are also small, so the effect is usually weak and the orbit decays slowly. The moons affected are:
  • Around Mars: Phobos
    Around Jupiter: Metis and Adrastea
    Around Saturn: none, except for the ring particles (like Jupiter, Saturn is a very rapid rotator but has no satellites close enough)
    Around Uranus: Cordelia, Ophelia, Bianca, Cressida, Desdemona, Juliet, Portia, Rosalind, Cupid, Belinda, and Perdita
    Around Neptune: Naiad, Thalassa, Despina, Galatea and Larissa
Some hypothesize that after the Sun becomes a red giant, its surface rotation
will be much slower and it will cause tidal deceleration of any remaining planets.
........................................................
2) Retrograde satellites: All retrograde satellites experience tidal deceleration to some degree because their orbital motion and their planet's rotation are in opposite directions, causing restoring forces from their tidal bulges. A difference to the previous "fast satellite" case here is that the planet's rotation is also slowed down rather than sped up (angular momentum is still conserved because in such a case the values for the planet's rotation and the moon's revolution have opposite signs). The only satellite in the Solar System for which this effect is non-negligible is Neptune's moon Triton. All the other retrograde satellites are on distant orbits and tidal forces between them and the planet are negligible.
........................................................
  • The planet Venus is believed to have no satellites chiefly because any hypothetical satellite would
    have suffered deceleration long ago, from either cause; Venus has a very slow and retrograde rotation.
Art Neuendorffer