Talk:Phoebe (moon)
From Wikipedia, the free encyclopedia
 |
This article is within the scope of the Solar System WikiProject, a collaborative effort to improve Wikipedia's coverage of the Solar System. |
| Start | This article has been rated as start-Class on the assessment scale. |
[edit] Density
The density of 0.7 given in the article, also given at Solarviews, seems to be incompatible with being a captured carbonaceous asteroids. I'll take the (estimated) value of 2.3 given at NASA's Solar System Dynamics. -- Looxix 22:41 May 6, 2003 (UTC)
The first sentence "outermost known moon of Saturn" is inconsistent with the list of Saturn's moons.
- Editted first paragraph to reflect new moons. --Patteroast 23:23, 17 Apr 2004 (UTC)
[edit] pronunciation
The moon itself of course is fee'-bee. The adjectival form for Phoebus is Phoebean fee-bee'-un per the OED; the change of gender shouldn't affect the adj. form, so I'll use Phoebean here as well. kwami 2005 June 30 04:45 (UTC)
[edit] Obliquity?
The figure quoted from http://exp.arc.nasa.gov/downloads/celestia/data/solarsys.ssc is extremely doubtful. That source dates back to April 2003, and Cassini got us a close look at Phoebe only in 2004. We need a recent source that actually gives the celestial coordinates of the bloody rotation pole. Urhixidur 00:38, 23 March 2006 (UTC)
One good source is http://www.hnsky.org/iau-iag.htm, but that is also suspect because of its date (2000-2001). The pole is stated to lie at right ascension 355.00, declination 68.70 (epoch J2000).
Ah, here it is: http://www.sciencemag.org/cgi/content/full/sci;307/5713/1237 (Cassini Imaging Science: Initial Results on Phoebe and Iapetus, Porco et al., Science, 25 February 2005: 1237-1242 DOI: 10.1126/science.1107981) --using Google's access, we read "We derived a spin-pole orientation of right ascension = 356.6°, declination = 77.9°" (a good match to the previous source, eh?). Urhixidur 01:00, 23 March 2006 (UTC)
To obtain the obliquity, what we need is Phoebe's orbital pole. http://aanda.u-strasbg.fr:2002/articles/aas/full/1998/04/ds5713/node5.html states that the J2000 Laplace plane at Phoebe's range from Saturn has its pole at R.A. 275.631°, dec. 68.031° (tilted 26.183° to Saturn's equator). The orbit of Phoebe is then given with respect to that plane, as longitude of the ascending node 233.037°, inclination 174.751°. This ascending node longitude is measured from the node of the reference (Laplace) plane on the J2000 Earth equator, not the usual ecliptic reference plane. We now have all we need to figure the orbital pole, and hence Phoebe's obliquity.
The ascending node of the Phoebean Laplace plane is easily located from the latter's pole: it lies at the pole's R.A. + 90° = 5.631°, dec. zero, with inclination 90° minus the pole's dec. = 21.969°.
The spherical triangle we now need to solve has for sides:
- a, the segment of Earth's equator running from the Laplace plane ascending node to Phoebe's orbital ascending node;
- b, the segment of Phoebe's orbit running up from Earth's equator to the Laplace plane, and
- c, the segment of the Laplace plane running from Earth's equator to Phoebe's orbit. The length of c was given above as 233.037°.
- The angle A (opposite a) was also given above = 174.751°.
- The angle B (opposite b) was obtained earlier = 21.969°.
- The angle C (opposite c) will be equal to 180° minus the inclination of Phoebe's orbit.
Adding a to the Laplace plane's ascending node longitude (5.631°) will give the ascending node longitude of Phoebe's orbit. From those two values we can then work out the Phoebe orbital pole coordinates (R.A. = long. - 90°, dec. = 90° - incl.).
So we know A, B, c and need a and C. One identity is: Cos(C) = -Cos(A)Cos(B)+Sin(A)Sin(B)Cos(c) (hence C = 25.456°) and we then need only apply the law of sines to obtain a: Sin(a)/Sin(A) = Sin(c)/Sin(C), hence a = -9.792°.
Thus the (equatorial) longitude of the ascending node of Phoebe's orbit is -4.161°, and its inclination is 154.544°.
Thus Phoebe's orbital pole lies at R.A. 265.839°, dec. -64.544°. All that remains is to find the angle between that point and the rotation pole, at R.A. 356.6°, dec. 77.9°.
This time, our triangle is lacking just a third point, and we choose the Earth's north pole for this purpose (R.A. whatever, dec. 90°). Our new triangle is thus:
- a, the segment running between the Phoebean rotational and orbital poles;
- b, the celestial meridian running from the north pole to the Phoebean orbital pole; and
- c, the celestial meridian running from the north pole to the Phoebean rotation pole.
- The angle A (opposite a) is obviously given by the difference in R.A. between the two Phoebean poles = 90.761°.
- The angles B (opposite b) and C (opposite c) we do not care about.
b is the codeclination of the orbital pole = 154.544°. c is the codeclination of the rotation pole = 12.1°. The relation we use this time is Cos(a) = Cos(b)Cos(c) + Sin(b)Sin(c)Cos(A). Hence Phoebe's obliquity a = 152.14°.
Q.E.D.
Urhixidur 13:05, 23 March 2006 (UTC)
