- 24,753
- 794
I think that's right. It will be in the dark for a part of each orbit and need to hibernate. It's expected to be low on hydrazine for attitude control by that time too.
lpetrich said:Distance: 77.93 k km, 48.43 k mi
Speed: 20 m/s, 61 km/h, 38 mph (17 m/s)
Likely the farthest and slowest, but I'll keep my eyes open.
The planned mission takes about a quarter of the Ceres year by which time Dawn is down into the closest orbit and its orbit PLANE is nearly normal to the sun direction.mfb said:That is very interesting with the orientation of the orbit relative to the sun/Ceres axis. It also means Dawn will run into problems soon after the expected end of the mission.
nope.marcus said:Gorgeous.
I've often read Phil Plait's blog, in the past. It's called (or was when I was reading it) Bad Astronomy. I think the name might be an anatomical pun.
Om, Marc Rayman said something about being able to be in radio contact (using a small antenna) while continuing to thrust. I'm still unclear about that long communication session you reported. Were they verifying a turnaround in radial distance? I'm still somewhat confused. I think turnaround has happened and we are now on the way in!
...
utc thruster mph dist kkm mps
3/18/2015 14:15:25 on 38 78.25 17.0
3/18/2015 20:14:24 on 37 78.32 16.5
3/19/2015 02:13:24 on 36 78.37 16.1
3/19/2015 08:12:23 on 35 78.4 15.6
3/19/2015 14:11:22 on 35 78.43 15.6
om om lpet lpet sim
rec/apr 1.3928 x 2.36 1.3928
hor 3.79 y 4.07 n/a
ver 12.07 z 14.05 n/a
rel vel 12.73 14.82 13.38
raw +/- 1
rec/apr 1.39 1.39
hor 3.79 3.80
ver 12.07 12.86
rel vel 12.73 13.49
But then again, Dr. Rayman probably knows better than I do.marcus said:...
I think turnaround has happened and we are now on the way in!
...
Marc Rayman said:Dawn Blog
March 19, 2015 at 1:21 pm
Hi Andrew,
For others, apoapsis is the most distant point of an elliptical orbit. (The term “apogee” may be more familiar. That is the most distant point in an elliptical orbit around Earth. Apoapsis is more general, applying to orbits around anything.)
As I described in my Dawn Journal above, Dawn reached that highest altitude on March 18, so, yes, its orbit is now taking it closer to the dwarf planet again.
60 mph = 26.8 mpsDr. Rayman said:At today’s throttle level, it would take the distant explorer almost 11 days to accelerate from zero to 60 mph (97 kilometers per hour).
As I commented below, I will provide an update on the mass of Ceres in May. This will give you something else to look forward to!
Marc
Seems fine for now.marcus said:I wonder what the angular size is now. 2arcsin(.475/78.37) in degrees
About 0.7 of a degree. Eventually the angular size should get up to around 60 degrees, if I remember right. That is about 120 times size of moon seen from Earth
What gradually changing headline makes sense to you from a communication standpoint, and perhaps journalistically?
Ah! I see you have already changed the headline. It makes good sense to me. Hope other contributors to the thread (Om, Petrich...) agree as well
marcus said:I wonder what the angular size is now. 2arcsin(.475/78.37) in degrees
About 0.7 of a degree. Eventually the angular size should get up to around 60 degrees, if I remember right. That is about 120 times size of moon seen from Earth
What gradually changing headline makes sense to you from a communication standpoint, and perhaps journalistically?
Ah! I see you have already changed the headline. It makes good sense to me. Hope other contributors to the thread (Om, Petrich...) agree as well
D H said:For example, if the spectrogram shows sharp spikes at 557 GHz and 1113 GHz, that means you are seeing water.
Just checked some 9 hours later and it said 78.25. Here's Om's annotated trajectory to have handy for reference:marcus said:Yesterday current status gave distance 78.43 kkm today 78.37 kkm, so it confirms that turnaround happened even if numbers slightly off from what Rayman posted...
marcus said:. The orbit altitudes are 13500, 4400, 1740, and 375 km. so adding the average radius of around 475 km, we get the orbit radii.
13975 km, 4875 km, 2215 km, and 850 km
...
2arcsin(475/13975) in degrees = 3.9 degrees
2arcsin(475/4875) in degrees = 11.2 degrees
2arcsin(475/2215) in degrees = 24.8 degrees
2arcsin(475/850) in degrees = 67.9 degrees
I've been working on this, off and on, since you asked the question 7 days ago.marcus said:...Anybody know what day in April they will flip it around?
...
I did that about a month ago and attached the excel file in my post back then. Thrust and motion are not well aligned right now - they still have to change the orbital plane a bit (and the detour allowed to do so very well as Dawn is so slow). Starting April 4th according to the numbers, thrust will get more and more against the flight direction, one week later its acceleration is well aligned with the flight course (in opposite direction of course).OmCheeto said:lpetrich's numbers are still tracking exceptionally well, and I was going to create a daily vector arrow thrust map based on the difference between his numbers, and the gravitational pull of Ceres.
mfb said:I did that about a month ago and attached the excel file in my post back then. Thrust and motion are not well aligned right now - they still have to change the orbital plane a bit (and the detour allowed to do so very well as Dawn is so slow). Starting April 4th according to the numbers, thrust will get more and more against the flight direction, one week later its acceleration is well aligned with the flight course (in opposite direction of course).
Distance 71010km, speed 52 mph (84km/h, 23m/s) in three hours.
mfb said:I used your numbers to calculate accelerations from gravity and from Dawn.
Jimster41 said:I had no idea Ceres was so interesting... what's the theory as to why it's so warm (-38 F)?
Ceres Surface Properties by High-Resolution Imaging from Earth
The measured temperature of the warmest area is 235 ± 4 K
Is it unexpectedly warm for 2.8 AU? I had just assumed that its surface would be at the equilibrium temperature.Jimster41 said:I had no idea Ceres was so interesting... what's the theory as to why it's so warm (-38 F)?
The equilibrium temperature for a black body as far out as Ceres has an equilibrium temperature that is significantly lower than the temperature of the sun (no matter where you measure it).OmCheeto said:And I've heard that interstellar space has a temperature equivalent to that of the sun.
I didn't believe it when I saw it. (It was here at PF)
I googled it.
It's true.
marcus said:At our distance from sun (AU) without atmosphere greenhouse etc. I think the equilibrium temp is about 255, I could be wrong. So roughly, 60% of that is 153 K. around 168 K. Sort of what one would expect. Maybe the fact that it is a bit warmer has to do with the difference in albedo. Ceres absorbs over 90% of the incident sunlight.
mfb said:I did that about a month ago and attached the excel file in my post back then. Thrust and motion are not well aligned right now - they still have to change the orbital plane a bit (and the detour allowed to do so very well as Dawn is so slow). Starting April 4th according to the numbers, thrust will get more and more against the flight direction, one week later its acceleration is well aligned with the flight course (in opposite direction of course).
Distance 71010km, speed 52 mph (84km/h, 23m/s) in three hours.
mfb said:The velocity values are from leptrich.
What is special about the x direction?
Marc Rayman says:Tim G says:
March 26, 2015 at 6:36 am
Marc, the MYSTIC simulator indicates that Dawn’s
specific orbital energy, v^2/2 -GM/r, has been creeping up.
I noticed that it dipped below zero on March 6th when the mission
planners celebrated Dawn’s capture.
Is this related to saving time because saving xenon propellant isn’t the only issue?
March 26, 2015 at 11:46 am
Hi Tim,
...Rest assured, however, that Dawn is securely in orbit and is not going to leave. (So its orbital energy will remain negative.)
You are quite right that the energy is increasing... To summarize, Dawn would naturally fall toward Ceres after reaching the high point of its elliptical orbit ..., but we use the ion engine to accelerate toward Ceres even faster. We have plenty of xenon and there is no need to conserve it this late in the mission. So we reduce the time it takes to travel to lower altitude (which is part of what we have to do to reshape the orbit into RC3) by propelling the spacecraft toward it. You can also see this in the animation above of the approach trajectory. Note how the thrust does not reverse quite when Dawn reaches that maximum orbital altitude. Rather, the ion beam is still directed away from Ceres, meaning the ship is still pushing itself toward Ceres. Only after it has accelerated its descent does it thrust in the reverse direction to brake smoothly into its low, circular orbit.
lpetrich said:As to OmCheeto's noisy graph, that's a problem with calculating numerical derivatives. Doing so magnifies errors. That's why I used a filter function that uses 4 neighbors instead of 2 neighbors. Another way of reducing noise is to fit the numbers to some curve, like a cubic spline, and then take derivatives of that curve.
I found that surprising so I googled it too. Weird as it sounds, according to https://en.wikipedia.org/wiki/Interstellar_medium it's true - within the Milky Way most of the interstellar medium is at >6000K, and about half is coronal gas at >106K. Clearly not at equilibrium with the CMB...OmCheeto said:And I've heard that interstellar space has a temperature equivalent to that of the sun.
didn't believe it when I saw it. (It was here at PF)
I googled it.
It's true.
wabbit said:I found that surprising so I googled it too. Weird as it sounds, according to https://en.wikipedia.org/wiki/Interstellar_medium it's true - within the Milky Way most of the interstellar medium is at >6000K, and about half is coronal gas at >106K. Clearly not at equilibrium with the CMB...
Now within the solar system in our neighborhood I suppose this translates to :
- an extremely dilute coronal gas at >106K
- a gas of photons flowing out of the sun, at ~250K
And the effective temperature from averaging both would be ~250K
Is that right?
Still sounds weird that in the ISM the coronal gas would win over the photons. I suspect that the ISM temperature reported is just its matter temperature, and its average effective temperature is actually close to 3K.
I think your image shows yet another kind of temperature, related to vibrations within a molecule. ISM temp is probably just ordinary gas temperature, related to the kinetic energy of the gas molecules (or atoms, or ions).OmCheeto said:It's the average jigglinesss.
wabbit said:I think your image shows yet another kind of temperature, related to vibrations within a molecule. ISM temp is probably just ordinary gas temperature, related to the kinetic energy of the gas molecules (or atoms, or ions).
As to Ceres, is its temperature different from the one quoted by Marcus from equilibrium with solar light?
marcus said:...To recap what was said earlier in the thread, space has about 400 million CMB photons per cubic meter. And those photons have a near-perfect thermal distribution, with a temperature of 2.728 kelvin.
It's the first thing I check for, every morning.marcus said:
OmCheeto said:... want to pick the best one.
What does everyone think?
...
3. Orbital plane.
When is Dawn predicted to enter it?...
Along with capturing distance, speed, thrust status, and communication status, two to three times a day, I've also been capturing the images.marcus said:I'd be most interested in knowing more about the time-table and manner of Dawn getting into the right orbit plane. the plane can't be be exactly in line with the sun direction (else Dawn would fall in shadow during orbit). It has to be turned "clockwise" (looking down on Ceres north pole) by some 5 to 7 degrees. It looks to me like this will involve a complicated maneuver---a slightly "S-shaped" approach.
As it stands, if the probe comes straight in, without that kind of weaving or zig-zag maneuver, it would get into an orbit plane that was actually turned slightly COUNTER-clockwise relative to sun direction, seen looking down along Ceres rotation axis.
As of this moment, 16:11 UTC (19:11 PDT), Canberra #34 and Goldstone #15 indicate carrier signal. When I checked about an hour ago, 15:03 UTC, Canberra #34 was in standby.BTW I see that current status (fullview2) shows the probe in communication mode this morning with thruster turned off. But when I checked DSN I did not see any antenna talking with Dawn at that time.
I'd like to see an article on "radio-navigation". Perhaps I'll google that later.You mentioned doppler measurement of the carrier wave. I think that is what Rayman means by "radio-navigation". It's remarkable that they can derive useful info about where the probe is and where it's going by keeping track of radial velocity relative to Earth stations, by doppler.
Unfortunately, my timeframe for editing has expired. I wonder, if I'm very polite, and offer them cookies, that one of the mentors would edit my post:Loathe to seem nitpicky about spelling but we shouldn't offend adherents of Greek&Roman pagan religion by misspelling the goddess Demeter's name. Better edit and change "dometer" to "demeter" lest you incur the wrath of the goddess.
(Yes, we've all googled "complex impact crater morphology", ...)
Nice link. The Wippykidia referenced a book
http://www.lpi.usra.edu/publications/books/CB-954/CB-954.intro.html
and I found interesting diagrams and discussion of the central uplift formation in large craters---chapter 3 page 26
http://www.lpi.usra.edu/publications/books/CB-954/chapter3.pdf
Kepler Crater - Central Peak (NASA, Earth's moon)
...
Complex craters occur above a certain diameter crater, the cutoff diameter is dependent on gravity, so it varies from planet to planet (or moon to moon).
...
Despite the label "central peak," a central peak is not always exactly in the center of a crater, nor is it always symmetrically shaped; Kepler crater is an example. Instead of having a nice central peak, Kepler crater has an irregular off-center peak. This form is most likely due to the crater being close to the boundary diameter between a simple and complex crater. Larger craters, such as King crater, can also display oddly shaped central peaks that are likely the result of an oblique impact.
Cookies!OmCheeto said:I wonder, if I'm very polite, and offer them cookies, that one of the mentors would edit my post: