- #71
OmCheeto
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Dr. Rayman, on Friday stated:marcus said:...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html
Emily says Dawn is approaching Ceres "from the south". That means we should be able to see more small-circle rotation detail at the bottom of the Ceres image. I think I do. This agrees with the upper Om-figure of post #62. Going by that figure, Dawn crosses the equatorial plane of Ceres around 15 Feb.
Or let's say the orbit plane determined by direction to sun (left) and orbit direction (into the page).
But the axial tilt is only about 3 deg, so the planes are nearly parallel. Until mid-February Dawn is looking slightly from the south and sees more of the south polar region. Increasingly after mid-Feb it will be looking more from the north.
Please correct if I'm missing something. In the animation, lighting also seems stronger near the north pole, with northern regions blanked out in glare. We are told the planet's axial tilt is only 3 degrees, could the current tilt (small as it is) be towards the sun?
My guess is that the washing out of the polar regions may be due to the lack of available light.Marc Rayman says:
February 6, 2015 at 10:49 am
Tom,
You’re right that the direction of Ceres’ axis is not precisely known (yet!), but it is tipped only a few degrees from the plane of its orbit. (This might help others understand a bit more about the axial tilt.) It seems to me then that in one practical (albeit unconventional) sense, Ceres is always near equinox. That is, the sun is always near the equator, so, yes indeed, most of the surface is lit during a Cerean day. That would also mean the poles are never more than “just barely illuminated.” Astronomers have several estimates of the pole location but using one popular one, a colleague calculated for me that the sun right now is about 4 degrees south latitude and near the southern summer solstice.
I should add that Dawn is not over Ceres’ equator, so we are not seeing the northern and southern hemispheres equally. It is approaching over the southern hemisphere, and it was above (roughly) 22 degrees south latitude for OpNav 3. Dawn will cross the equator between RC1 and RC2.
Marc
Current angular size: 2 arcsin(.475 / 110.67) = 0.491832996 deg (as of 9AM pacific, on 8 Feb)
Om has some more figures tabulated in post #55 but here is the table of optical navigation photo shoots (with a few extra rows added) from 29 January D.J.
http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/
Code:Jan 25 (237,000) 43 (22) 1.3 96% Feb 1 (168,000) Feb 3 (146,000) 70 (14) 2.2 97% Feb 10 (100,000) Feb 12 (83,000) 121 (7.8) 3.8 98% Feb 19 (46,000) 221 (4.3) 7.0 87% Feb 25 (40,000) 253 (3.7) 8.0 44% Mar 1 (49,000) 207 (4.6) 6.5 22% Apr 10 (33,000) 304 (3.1) 9.6 18% Apr 15 (22,000) 455 (2.1) 14 50% Apr 23 (14,000)
Relevant column headings as given in the original table:
Distance from Dawn to Ceres in (kilometers)
Ceres diameter in pixels
Resolution in (kilometers) per pixel
Resolution compared to Hubble
Illuminated portion of disk
In an effort to figure out what "Capture by Ceres' gravity" means, I spent the last few days doing some "orbital mechanics" studying.
The only thing I've concluded, is that Newton was a FREAK!
Even with spreadsheets, the internet, and one of the worlds most powerful laptops, I have not a clue what I'm doing.
But in my feeble attempt to solve the capture problem, I discovered lots of weird and wonderful things, which is always a good thing.
About the only useful thing I can add, is my z-y plot of Dawn, as it approaches Ceres, as viewed from the Sun. (z becoming the x-axis in the 2D plot below)
Ceres is at the origin.
Position labels are date and distance in kkm from my digitization.
3-6 & 3-20 are extractions from one of Dr. Rayman's journals.