Talk:Phobos (moon)

Next trick

(moved to Talk:Timeline of natural satellites)

Surface Gravity

I'm going to switch the mm/s² back to m/s². Granted, if this were an independent article mm/s² would be preferable, but it is not; it is part of a large series of articles, and sticking to a single unit (m/s²) is necessary for comparison purposes.

Urhixidur 12:34, 2004 Jul 12 (UTC)

Surface dust impossible?

How can it have surface dust if it's inside its Roche Limit? Wouldn't tidal forces tend to cause the dust to lift from the surface and either dissipate into space, or fall toward Mars? --Doradus 19:17, 27 Aug 2004 (UTC)

I'll take a stab at it: Tidal "forces" aren't really forces, they're more like a difference of force. Let's say I could grab Phobos by two ends and pull it towards me. If I pulled harder on the side nearer to me, Phobos would feel as if it were being pulled apart. This is what Mars's gravity is doing. The force from gravity is weaker on the far side of Phobos because it's farther away from the planet. The apparent force between two sides of Phobos is much larger than parts of Phobos which are near each other. Dust lying on the surface feels a minimal tidal "force" with respect to the surface it is near because they are close. The collective gravity of Phobos, however, will keep the dust on the surface. If Mars had stronger gravity, though, it could pull the dust right off the surface. Peter 00:07, 28 Aug 2004 (UTC)

That doesn't seem right since being within the Roche limit should mean the tidal force at the inner and outer tips are strong enough to overcome Phobos' gravity, by the definition of "Roche limit". I'll take another stab. The tidal force is strongest at the point closest to Mars and the point farthest away from Mars; it falls off to nothing at the "middle" of the moon. So perhaps the dust is only present in a belt around the midsection of the moon, with dust-free "peaks" at the two tips. Bryan 00:35, 28 Aug 2004 (UTC)

I like that explanation. Because Phobos is tidally locked, dust could indeed settle at the midsection. It would be interesting to know whether measurements confirm this. Also, while the tidal force is a repulsive force at the near and far ends of Phobos, I think it is mildly attractive at the midsection, which might help the dust to settle there. (If you imagine the force vectors at the two opposite limbs of Phobos, they are not parallel, but both point toward Mars' center.) --Doradus 03:32, 28 Aug 2004 (UTC)

There does seem to be quite a lot of dust on Phobos (according to the page), so it's probably not impossible. Let's say Phobos is like a stretched spring with one end always pointing towards Mars. The tension on the spring is the tidal "force". The greater the tidal "force", the longer the spring will be stretched. But, let's say we look at only the first half of the spring. How much tidal "force" is there? I think it will be half as strong. If we pursue this exercise down to the scale of dust, the forces will be quite small relative to that applied to the moon as a whole. In the case of Phobos, this force is apparently less than the force of gravity attracting the dust to the moon. Peter 04:50, 28 Aug 2004 (UTC)

No, I still don't think this is right. If this was the case then passing the Roche limit would never result in a body being broken up by tidal force. The cause of this "force" is the difference between the velocity of a particle on the moon's surface and the velocity of a stable orbit at that distance - the stuff on the inner end of the moon is moving slower than it "should" and the stuff on the outer end is moving faster. Remove the influence of the moon forcing them to follow the moon's orbit and those particles would move off into different paths. It has nothing to do with the size of the particle. Bryan 07:47, 28 Aug 2004 (UTC)

Maybe there's another force between the dust particles, like static electricity or some kind of adhesion. Failing that, perhaps Phobos is not inside its roche limit after all. Or maybe we all misunderstand the roche limit. --Doradus 12:39, 28 Aug 2004 (UTC)

I don't believe Phobos is within its Roche limit. It may well be within what the Roche limit would be if it were a liquid moon, but it is less prolate than a liquid moon would be at the Roche limit- using the formulae on the Wolfram page, at the Roche limit the ratio between the semimajor and both semiminor axes would be 1.952- Phobos isn't a perfect prolate spheroid, and the ratio between its largest and smallest axis length is only 1.45. I have a reference from Icarus from 2001 which states that "...assuming a typical friction angle of 30°,it is clearly nowhere near a tidal limit for a MC body, although it may be at the (fluid) Roche limit depending on the exact value for its density." MC appears to be "zero-cohesion Mohr-Coulomb". If Phobos is a 'rubble pile' it would need an 'angle of friction' of at least 7.1° for stability at its current orbit of 2.76 R. (R is the radius of Mars) Assuming a 'typical friction angle of 30°' Phobos should remain tidally stable until its orbit gets as close as 2.12 R.

I may make some significant edits at this point, I think the Phobos page as it stands is in error. Does anyone have any citations (preferably in another peer-reviewed academic journal) which states that Phobos is within its Roche limit? --Noren 21:28, 1 Sep 2004 (UTC)

(William M. Connolley 09:08, 3 Sep 2004 (UTC)) Tidal forces don't fall to zero at the "middle" - they are attractive there. See Image:Tidal-forces-calculated.png from tidal force.

I'm not sure if this was added after or before you guys had this disucssion, but it says in the artilce "it has been calculated that Phobos is stable with respect to tidal forces, but it is estimated that Phobos will pass the Roche Limit for a rubble pile of its description when its orbital radius drops to about 8400 km, and will likely break up soon afterwards." So, basically, it is not inside the Roche Limit yet.

The page as it existed during most of the above discussion claimed that Phobos was within its Roche limit. I added the phrase you quote to the main page shortly after posting the above to this discussion, the 8400 km figure used on the page was derived from the 2.12 R (mars) given in the above reference. I initially wanted to be sure I wasn't missing any evidence to the contrary by asking for it here... shortly after that I decided to go ahead and be bold and edit it myself. --Noren 17:47, 4 Oct 2004 (UTC)

Millisecond precision in orbit period

Yes, we do know it to about 1 ms. See http://exp.arc.nasa.gov/downloads/celestia/data/solarsys.ssc. Remember that you can observe over a great number of revolutions and the divide by that number, so precision grows with time.

Urhixidur 17:12, 2004 Sep 2 (UTC)

I believe we can time a single revolution to that accuracy. Will subsequent orbits vary by less than 1ms? Surely Mars is not that uniform, perturbation by Jupiter must be on that order of magnitude. Unfortunately I don't know how to calculate either of those effects, so I guess I'll take your word for it. --Doradus 19:11, 2 Sep 2004 (UTC)

I agree with Urhixidur. The reverse or the above is true- I expect we can't time a single revolution to that accuracy(I doubt the position is known that well), but we can time many of them and take the average, just as Urhixidur describes. The orbital period should be quite uniform, I believe. Mars' gravity should be the same during each orbit- no change in period resulting from there- and the gravity gradient from Jupiter should be extremely small. The constant component of any pull from Jupiter or another body will effect both Mars and Phobos equally, and thus cannot influence the period. Diemos might conceivably be relevant, however, though I think the orbit period is by definition an average anyhow. --Noren 20:34, 2 Sep 2004 (UTC)

I think you're right about Jupiter's pull. I disagree on the other points, but not strongly enough to kick up a fuss.  :-) --Doradus 04:25, 3 Sep 2004 (UTC)

Formatting

I am having trouble getting the formatting to work on 800x600 resolution. That image in the middle kind of mucks things up. I kept tweeking with it and got it to work for this resolution, but don't know the impact at other resolutions.

Mass in Earths

is given by 1.07 × 10¹⁶ kg (1.8 µEarths).

There seems to be a misscalculation here. I'm getting 50.58 nano Earths.

Could someone please check this and either correct the article or my calculation? Thanks 84.160.196.73 09:38, 13 Mar 2005 (UTC)

Well, I also did the same calculation and came up with yet another result, namely 1.8 nanoEarths. I went ahead and put this in the article; if I'm mistaken then it should be reedited, of course.

${\frac {M_{Phobos}}{M_{Earth}}}\cong {\frac {1.07\times 10^{16}}{5.9736\times 10^{24}}}\cong 1.79\times 10^{-9}$ -Noren 06:02, 15 Mar 2005 (UTC)

I confirm Noren's result.

Urhixidur 13:07, 2005 Mar 15 (UTC)

In a related change, the volume in Earths was also mistakenly expressed in microearths rather than nanoearths. (As something of a sanity check, I expected comparative volume and mass to be of the same order of magnitude, and they previously were not.) -Noren 00:36, 19 Mar 2005 (UTC)

pronunciation

I had added the alternate pronunciation fob'-us, but it's pretty rare, so I'm taking it out. It's given as an alternate pronunciation in Gayley's Classic Myths in English Literature and in Art. In the OED, not even the much more common cognate "phobic" has a variant with that vowel. Put it back if you like... --kwami 07:52, 7 Apr 2005 (UTC)

The adjectival form Phobian is fairly common, and consistant with the Greek. kwami 2005 June 30 06:19 (UTC)

As seen from...

"As seen from Phobos, Mars would be 6400 times larger and 2500 times brighter than the full Moon as seen from Earth" it isn't supposed to be as seen from Mars, Phobos...?

No. Phobos looks tiny as seen from Mars. Mars looks huge as seen from Phobos. So it's correct. The Singing Badger 01:04, 19 Apr 2005 (UTC)