Dawn dead in Ceres orbit, ran out of fuel Oct 2018

[marcus]

Published on Jan 19, 2015
The Dawn spacecraft observed Ceres for an hour on Jan. 13, 2015, from a distance of 238,000 miles (383,000 kilometres). A little more than half of its surface was observed at a resolution of 27 pixels. This video shows bright and dark features.
This video was created from an animated gif and has been looped five times.

For the full story see here: http://www.jpl.nasa.gov/news/news.php...

The Dawn mission Twitter has a bunch of images of Ceres that were just put up.
https://twitter.com/NASA_Dawn

The Dawn probe is approaching Ceres (950 km diameter icy miniplanet) from sunward direction, so it sees the fully illuminated face. The range is now about one light second and Ceres looks 36% of the size of a full moon seen from Earth.

The plan is to go into polar orbit starting at altitude 13,500 km above surface, and if all goes well to eventually spiral down to circular polar orbit about 375 km above surface.

The next 2 months are critical. It is not clear to me that they will succeed in achieving stable circular polar orbit. They do not have a large supply of hydrazine propellant for attitude control. The main thrust is solar electric ion drive (about 1/10 of a Newton thrust). In order to have thrust they need the solar panels (a 60 foot spread) oriented to the sun. In order to have ample communication they need the large narrow-beam antenna aimed exactly at the Earth. To maneuver into orbit they need to have the thruster correctly tracking in a constantly changing direction.

The approach trajectory (planned for March) is interesting and is described using two diagrams in the November Dawn Journal (DJ). It involves a highly eccentric loop with some gravity assist to bring the probe up to speed. This is to make up for thrust lost in September when an energetic cosmic ray particle disabled electronics serving the propulsion system.

 

[marcus]

Here is the index of Dawn Journal entries:
http://dawn.jpl.nasa.gov/mission/journal.asp

To see the approach plan, click on November 2014 and scroll down to the diagrams.

For a physical description of Ceres based on what we know so far, see the December 2014 DJ, and scroll down to the physical description in second half.
950 km equatorial diameter spheroid, 2.2 g/cc density, differentiated (crust, mantel, core), rotation period 9 hours, estimated 30% water by mass, escape velocity 500 m/s

For the planned spiral-down to lower altitude orbits, see April 2014 DJ. Roughly the idea is reduce altitude from 13,500 km to around 4400, then 1740, then 375 km. I think some of this is repeated in the August 2014 DJ.

For an account of the Dawn crew's struggle to restore communication, control, and thrust after electronics were stunned by high energy particle(s) in September, see October 2014 DJ.

For current distances and speeds:
http://dawn.jpl.nasa.gov/mission/live_shots.asp
and in particular:
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
This gives the current distance as 303.81 thousand km.
Since the equatorial diameter is 0.95 thousand km, the current angular size is
arctan(.95/303.81)
to get this in degrees, google "arctan(.95/303.81) in degrees".
Double this to get Ceres size as percentage of the moon's, since the angular size of the moon is half a degree.

 

[marcus]

http://neo.jpl.nasa.gov/orbits/fullview2.jpg
Here's how Ceres currently looks from the spacecraft 's standpoint. this image is updated a couple of times a day, possibly more often in future.
Right now (as of 1:42 AM UTC on 20 Jan) the distance to Ceres is given as 300.86 thousand km. So very close to one light second.
You can see Ceres as a brownish dot right in the middle of the frame. The Ceres dot has been growing in size and moving down and to the left against the background of stars. The current speed relative to Ceres is given as 0.14 km/s. Of course both the craft and the planet have the usual orbit speed of some 17 km/s.

The spacecraft image has also been moving down and to the left at the same roughly 45 degree angle at apparently the same rate over the past 10 days or so, against the star background, or in effect the background has been moving up and to the right. The Ceres dot stays centered and the craft stays at the the same position in the frame.

I think roughly speaking we are looking from the sunward direction and Dawn is moving away from us, into the picture frame, directly towards Ceres. That's how I read the picture. See the approach trajectory depicted in Dawn Journal November 2014. The two have been orbiting "neck and neck" with Dawn nearer the sun, separated by a certain distance, and now the spacecraft is closing that distance. It is due to arrive at Ceres around 6 March and be captured at that time in a somewhat elongated orbit.

 

[_Adam]

Very interesting. What will be Dawn's eventual minimum distance to Ceres?

 

[marcus]

Very interesting. What will be Dawn's eventual minimum distance to Ceres?

Circular orbit at altitude about 375 km above surface. The Dawn Journal of April 2014 has the plan for gradually spiraling in closer.
Here is the index of Dawn Journal entries so far:
http://dawn.jpl.nasa.gov/mission/journal.asp
Click on April 2014 for the sequence of closer and closer orbits (with some diagrams)
Click on August 2014 for discussion of how Ceres will be studied from the different altitudes.
Click on October 2014 for some cosmic ray trouble they had in September, losing thrust for a few days.
Click on November 2014 for a diagram of the new approach trajectory plotted after they fell behind schedule. (It involves some gravity assist and takes longer to reach the circular polar orbit RC3.)
Click on December 2014 for some history (first half) and some physical description of Ceres (second half).

Journal entries are sometimes long (e.g. December 2014 with all the history) so you may have to scroll halfway down the page to find the part you want. It's OK. It's pretty much all interesting and informative. But be prepared to scroll.

When you click April 2014 you get:
http://dawnblog.jpl.nasa.gov/2014/04/30/dawn-journal-april-30-2/
And the sequence of orbits goes:
"RC3", "survey", "HAMO" and "LAMO" with nominal altitudes above surface (in km) being
13,500, 4400, 1740, 375 km.
LAMO is acronym for "low altitude mapping orbit".
If all goes well and they actually get down to LAMO before the hydrazine (attitude control propellant) runs out, then very small deviations in trajectory will reveal subsurface mass concentrations. The hope is to get a detailed gravity profile of the mini planet, as well as a visual map of the surface. there's an account of the various instruments at
http://dawn.jpl.nasa.gov/technology/science_payload.asp
See also: http://dawn.jpl.nasa.gov/technology/ for other topics like the spacecraft 's solar powered ion propulsion, navigation, general structure.

Adam, I'm glad you asked about the lowest orbit! I had misremembered some altitude numbers, so going back over them gave me a chance to correct some mistakes. Also I want to copy a diagram from DJ November 2014 showing how they plan to approach Ceres and maneuver so as to get into circular polar orbit---the initial target orbit they call "RC3"

 

[marcus]

Adam, in case you are interested there are some physical features of Ceres, current estimates, in another thread. Click here (post#12) and scroll down to post#14 that has the mass and radius.
https://www.physicsforums.com/threads/dawn-distance-to-ceres-distance-to-moon.791038/#post-4977267
Mass is given as 9.43 x 10²⁰ kg. I'd like to see what the circular orbit speed is at altitude 13500, i.e. at radius 13975 km or thereabouts.
I have an idea that the day circles in this approach diagram are about 5000-6000 km apart, along the "RC3" curve. Around 5 or 6 times the diameter of the central dot representing Ceres, which is nearly 1000 km.
To see the approach trajectory diagram, click on "view" here.
View attachment 77558
The sun is far to the left. The probe is coming in from sunward, from the left. Both probe and planet are going into the page at essentially the same solar orbit speed of roughly 17 km/s and in addition the probe is gradually closing the distance between them. I want the "RC3" orbit speed, so:
Putting the thing in brackets into google gives us 67 m/s
[(G*9.43e20 kg/13975 km)^.5]
and multiplying that by the 86400 seconds in a day gives a day's travel of 5800 km along the "RC3" curve
[86400 s*(G*9.43e20 kg/13975 km)^.5]
Here's how google calculator does it:
https://www.google.com/?gws_rd=ssl#q=86400+s*(G*9.43e20+kg/13975+km)^.5

Actually I find it easier to remember the mass as 943 billion billion kg.
Which when using google calculator I can write 943e18 kg
so for example if I want the escape velocity from surface (radius 475 km) I put this in:
(2G*943e18 kg/475 km)^.5
and google tells me 515 m/s, or about half a kilometer a second.

 

[OmCheeto]

One quick note. Up until today, the wiki page on the Dawn spacecraft did not mention the hydrazine thrusters.
And the entry they added today sounds kind of like they are going to run out of gas prematurely:

When the last of its hydrazine fuel is used up, Dawn will become a "perpetual satellite" of Ceres; its orbit is predicted to be very stable.
​

From the November 2014 journal, Marc Rayman had the following to say:

We were so successful in our hydrazine conservation campaign... we anticipate accomplishing all of the original mission objectives regardless of the health of any of the reaction wheels.

[edited & bolded for clarity, brevity, and yippie factor]​

And from the December 2014 journal, regarding its post mission orbit:

You may be confident that Ceres will be Dawn’s permanent home. I wrote about the long-term prospects for the interplanetary traveler here. It is indeed important to ensure it does not impact Ceres, as you mentioned, but we do that by choosing a final orbit that will remain stable.​

 

[marcus]

Here is the index of Dawn Journal entries so far:
http://dawn.jpl.nasa.gov/mission/journal.asp

When you click April 2014 you get:
http://dawnblog.jpl.nasa.gov/2014/04/30/dawn-journal-april-30-2/
And the sequence of orbits goes:
"RC3", "survey", "HAMO" and "LAMO" with nominal altitudes above surface (in km) being
13,500, 4400, 1740, 375 km.
LAMO is acronym for "low altitude mapping orbit".

we can find the circular orbit speeds for each of these planned orbits, to get an idea of how much thruster work it will take to spiral down to the lower altitude ones. First, since the average radius is around 475 km, I want to convert these altitudes to orbit radii.
13975 km, 4875 km, 2215 km, and 850 km

...I find it easier to remember the mass as 943 billion billion kg.
Which when using google calculator I can write 943e18 kg
so for example if I want the escape velocity from surface (radius 475 km) I put this in:
(2G*943e18 kg/475 km)^.5
and google tells me 515 m/s, or about half a kilometer a second.

(G*943e18 kg/13975 km)^.5 = 67.1 m/s

(G*943e18 kg/4875 km)^.5 = 113.6 m/s

(G*943e18 kg/2215 km)^.5 = 168.6 m/s

(G*943e18 kg/850 km)^.5 = 272.1 m/s

 

[marcus]

When we started the thread the angular size of Ceres was 36% that of the moon seen from Earth.
Now that the current distance is only 247.66 thousand km, it's 44%.
arctan(.95/247.66) in degrees = 0.21978 ≈ 0.22 And the moon's size is half a degree.
I want to estimate what the angular size will be, in degrees, from the planned orbit heights. 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

[EDITED to correct angles at OmCheeto's polite suggestion]

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

 

[marcus]

The four planned orbits are called RC3, survey orbit, HAMO, and LAMO (i.e. "low altitude mapping orbit")
April 2014 Dawn Journal ( http://dawnblog.jpl.nasa.gov/2014/04/30/dawn-journal-april-30-2/ ) says that it will take about a month to descend from RC3 to Survey.
And 6 weeks to spiral down from Survey to HAMO
And about 2 months to spiral down from HAMO to LAMO. they don't plan to go lower.
==quote DJ April 2014==
[Picture caption: Dawn will make five spiral loops during the month it will take to fly from its RC3 orbit ( 13,500 kilometers) to survey orbit (4,400 kilometers). ]

The first coils around Ceres will be long and slow. After completing its investigations in RC3, the probe will spiral down to “survey orbit,” about 4,400 kilometers above the surface. During that month-long descent, it will make only about five revolutions. After three weeks surveying Ceres from that new vantage point, Dawn will follow a tighter spiral down to the (misleadingly named) high altitude mapping orbit (HAMO) at 1,470 kilometers.

In the six-week trip to HAMO, the craft will wind around almost 30 times. It will devote two months to performing extensive observations in HAMO. And finally as 2015 draws to a close, it will fly an even more tightly wound course to reach its low altitude mapping orbit (LAMO) at 375 kilometers, where it will collect data until the end of the mission.

The ship will loop around 160 times during the two months to go from HAMO to LAMO.
==endquote==

 

[marcus]

Dawn took some more pictures of Ceres on 26 Jan. There was a "tweet" from someone involved with Dawn image processing named Vishnu Reddy

Vishnu Reddy @moonyguy · 3h3 hours ago
We are getting the good stuff from OpNav2 from @NASA_Dawn of #dawn_ceres, first images look very impressive

The simulated picture of the spacecraft and planet shows it in a peculiar orientation, with the thruster apparently turned off. This would seem to represent how the spacecraft looked while it was taking pictures today.

The angular size of the planet is now arctan(.95/229.42) in deg = 0.237 degrees
47% of a full moon.

 

[OmCheeto]

And the new image came out about 2 hours ago:

On a geeky sidenote:

(G*943e18 kg/13975 km)^.5 = 67.1 m/s

initial orbital speed ---^

I had another panic attack yesterday when I looked at your initial orbital speed, and noticed that Dawn was still traveling at 120 m/s, and saw that the ion propulsion system generated a force of a single sheet of paper. So I looked for the acceleration rate, and found that at full throttle, Dawn's three ion thrusters can produce an acceleration rate of 7 meters/sec/day, so it will only take about 7 days to reach orbital velocity. But then I went back and calculated the acceleration from the 10th through yesterday, and found that Dawn was decelerating at about 2.5 meters/sec/day, which implies that they've been using only 1 thruster.

So I plugged 2.5m/s/day into a spreadsheet and found that at that rate, Dawn will slow down to orbital velocity by around February 16th. At which point it can coast into orbit [edit #2 ignore the following per marcus's next post] al rendezvous around the 23rd of February.[/edit #2]

[edit]: Original image would not render.

 

[marcus]

Thanks Om! Good work. Glad you worked through the numbers. I saw the new (somewhat higher resolution) clip of Ceres rotating here:
https://twitter.com/NASA_Dawn/status/560077265334071297
Here is the plan of approach viewed from above the north pole of Ceres, with orbital motion direction from bottom to top of frame. The sun is way far to the left of the diagram. Dawn approaches from sunward direction. Dawn enters at left edge of diagram on or about 10 February. small circles mark the days. Capture is on 6 March, if things go as planned.

That is when probe will be near enough and slow enough to finally be in orbit.
You may have read about this: they lost a few days of thrust back in September 2014 due to cosmic ray knocking out some electronics so they are not going quite as fast in solar orbit as originally planned. So as Dawn approaches it will FALL BEHIND slightly, and not be as near to Ceres as desired. This falling behind is shown by Dawn trajectory sloping DOWN in the diagram. (Up is the 17 km/second solar orbit direction)

So the approach will be tricky and involve a kind of "gravity assist". On 10 February in about 2 weeks, when Dawn enters at the left edge of the frame, it will be about 100 thousand km from Ceres. You can see that it begins to fall behind (goes lower in the frame). But eventually by about 23 February, Ceres gravity begins to bring it up to speed (it begins to go higher in the frame, catching up to Ceres.)

When it finally achieves circular orbit closer in, this looks odd in the diagram because it is a polar orbit and we see it looking down on Ceres north pole. So we see the circular orbit slantwise, from an angle.

 

[OmCheeto]

Thanks Om! Good work. Glad you worked through the numbers. I saw the new (somewhat higher resolution) clip of Ceres rotating here:
https://twitter.com/NASA_Dawn/status/560077265334071297

That was the original image I tried to embed: http://photojournal.jpl.nasa.gov/archive/PIA19171.gif
Unfortunately, it rendered in preview, but not upon final post. Hence, my rapid edit.

Also, on a hyper-geeky side note, would you double check your observed angle calculation for the lowest orbit:

2arctan(475/850) in degrees = 58.4 degrees​

I come up with 2arcsin(475/850) in degrees = 67.9°

 

[marcus]

You are right! they should all be 2arcsin not 2arctan :)
I was wondering if I should bother to change them. Will go back and edit post #9.

 

[OmCheeto]

You are right! they should all be 2arcsin not 2arctan :)
I was wondering if I should bother to change them. Will go back and edit post #9.

Bucket list: After 7 years, I'd really like to contribute...

 

[marcus]

Poor Om!
You had me laughing aloud again. But actually you are contributing "hand over fist" (abundantly) to our coverage of the approach to Ceres, and you surely know it. Thanks for both the most recent photo and correcting that trig error.

BTW did you notice that in the photo the axis of rotation is tilted about 30 degrees to the left? Or even a little more than 30, I can't tell.

Swollen things rotate they don't move just to the right, they move to the right and UP.

I wonder if that is just a random orientation of the spacecraft , causing the orientation of its camera, or if it has some meaning. It is about the same angle by which the Ceres orbit plane is tilted in this diagram http://neo.jpl.nasa.gov/orbits/fullview2.jpg

In that current status diagram Ceres and Dawn move down and to the left, against the stellar background. Sort of 35 degrees down from horizontal. Could there be a connection? I have no guess.

BTW the current status diagram says the current distance is 216 thousand km.
So the angular size (now I will use arcsin :w) is
2arcsin(.475/216) in deg = 0.252 degrees. Smack on half the size of a full moon.

 

[phyzguy]

For a physical description of Ceres based on what we know so far, see the December 2014 DJ, and scroll down to the physical description in second half.
950 km equatorial diameter spheroid, 2.2 g/cc density, differentiated (crust, mantel, core), rotation period 9 hours, estimated 30% water by mass, escape velocity 500 m/s

marcus: Do you know where the estimate of 30% water by mass comes from? How do we know this? Do we think there is a subsurface ice layer, or is the water chemically tied up?

 

[marcus]

I think they expect a mantel (at least 100 km thick) which is a mix of ice and rocky material---not pure ice, but also not chemically tied up.

the mass has been measured by near encounters with small asteroids
the equatorial and polar radii have been measured (oblateness)
and the rotation rate has been measured.

comparing oblateness and rotation rate allows one to estimate the extent of DIFFERENTIATION. You know that silicate rock is about 3.5 density and ice is about 1, and Ceres average is 2 g/cc. So it is either uniform average 2, or it is differentiated. with thin crust encasing a less dense mantel. (crust, mantel, core)
An undifferentiated body will tend to be more oblate because it has more mass farther out, at the same rotation rate.
therefore ceres is probably differentiated, given its small oblateness at the observed rotation rate, because that way it would have less mass farther out.

 

[marcus]

Phyzguy, for reference there are some recent estimates of planet features in post #14 of an earlier thread:
https://www.physicsforums.com/threads/dawn-distance-to-ceres-distance-to-moon.791038/#post-4977267
==quote==
Mass was determined by observing NEAR ENCOUNTERS with much smaller objects in the asteroid belt.
Here is a sample paper of that type:
http://www.doiserbia.nb.rs/img/doi/1450-698X/2005/1450-698X0571037K.pdf
The roundish oblate spheroid figure---the equatorial diameter is larger than the polar diameter---was measured rather precisely by Hubble Space Telescope
http://arxiv.org/abs/0711.1152
Equatorial radius 479.7 km and polar radius 444.4
The rotation period (9 hours) was determined using HST by observing surface features.
The same rate of rotation will produce MORE oblateness in an homogeneous body than if the body differentiated into layers, because the differentiated body has more of its mass in the central core and a smaller moment-of-inertia. The homogeneous body has mass distributed radially farther out, so will tend to become more flattened by rotation
...
For extra detail I will quote from the November 2007 paper:
==quote http://arxiv.org/abs/0711.1152 ==
Adopting a mass for Ceres of M = 9.43 ± 0.07 × 1020 kg (average of most recent measurements [Viateau & Rapaport, 1998; Michalak, 2000; Kovacevic & Kuzmanoski, 2007]), we find a mean density ρ = 2 206 ± 43 kg.m−3. This value is relatively high for a hydrated G-type asteroid like Ceres, but can be explained by a low porosity [see Britt et al., 2002], and is similar to the density of the icy outer Jovian satellites Ganymede and Callisto.
==endquote==
==endquote==
I'll try to remember polar equatorial and mean radii in round numbers. Equatorial 480 km, polar 445 km, mean 475 km
And diameters in round numbers. Equatorial 960 km, polar 890 km, mean 950 km.

 

[phyzguy]

Marcus, thanks for the details. I wonder if that white spot we see is a crater that has exposed fresh ice. This is really exciting - every day or so we see more detail. I can't wait for the kind of images we got at Vesta.

 

[marcus]

I know what you mean! The Vesta images were superb quality and fascinating. I hope, first of all, that Dawn makes it into the first circular polar orbit (after having lost thrust for several days in September and now needing gravity assist to catch up). And secondly that it has sufficient resources for attitude control to spiral down from that first 13500 km altitude polar orbit to the so-called LAMO orbit at altitude 375 km.

Correct me if I am wrong, you may have watched the stay at Vesta more closely. I think in spiraling down, which will take several months of thrusting (according to Dawn Journal April 2014) you want to keep your orbit plane roughly normal to the sun. Facing the sun so your panels are always getting full sunlight. And you have to keep your thruster pointing in the direction of orbit. as if to slow you down (but the actually effect will be to speed you up as you spiral in.)

the orbit speed, and the amount of energy required to get you down to the next level, keep increasing. I calculated some orbit speeds at the various planned levels, to get an idea of how much thruster impulse is required, and why the last step, down to altitude 375 km, takes so long to accomplish.
I think the last step also takes 160 revolutions, i.e. loopings around the planet.

If you want and haven't already, have a look at the April 2014 journal that describes that:
http://dawn.jpl.nasa.gov/mission/journal.asp

Here are the orbit velocities at the 4 different levels:
(G*943e18 kg/13975 km)^.5 = 67.1 m/s

(G*943e18 kg/4875 km)^.5 = 113.6 m/s

(G*943e18 kg/2215 km)^.5 = 168.6 m/s

(G*943e18 kg/850 km)^.5 = 272.1 m/s

Those are for the initial circular orbit, the survey orbit, high altitude mapping orbit (HAMO), and low altitude one (LAMO)

 

[OmCheeto]

Poor Om!
You had me laughing aloud again. But actually you are contributing "hand over fist" (abundantly) to our coverage of the approach to Ceres, and you surely know it. Thanks for both the most recent photo and correcting that trig error.

BTW did you notice that in the photo the axis of rotation is tilted about 30 degrees to the left? Or even a little more than 30, I can't tell.

...

I'm afraid that after playing with the Rosetta trajectory simulator for so many hours, I lost all interest in trying to keep myself oriented in space.​

It did though answer one question I always had about the earth-sol ecliptic in relation to the sol-galactic ecliptic. We're traveling almost sideways!

Another weird thing I heard the other day, which kind of ties into my question as to why Hubble took such a crumby picture of Ceres, was that the sky we see with our eyes isn't actually what it would look like if you had really sensitive eyes. I thought it was a joke the other day when I saw that someone posted that, viewed from Earth, Andromeda is 6 times larger than the moon. But it was Phil Plait who said it, so instead of calling him a prankster, I checked his maths. It's true!Object______degrees____size relative to Ceres when viewed from Earth
Andromeda______3.16______15,400
Moon___________0.52_______2,520
Ceres_______0.00021___________1
Crab Nebula___0.097_________472

ps. Sorry to go off on a tangent.
pps. Did you know that Jan 14th was the 10th anniversary of Huygens landing on Titan? I don't think I even knew we'd landed on Titan. I am so far behind in this stuff.

 

[mfb]

The visible angles for Ceres will be larger than calculated, as those calculations approximate it as a disk. A good approximation now, but a bad one for the final low orbit.

Based on the recent picture, it doesn't look like Ceres has many large-scale structures on the side they saw.Edit: For future use:
2arcsin((radius of ceres)/14500km) * (moon semimajor axis)/(diameter of moon)

 

[marcus]

Hi mfb, glad you are reading. Fixed the angular sizes for the various orbits yesterday:

When we started the thread the angular size of Ceres was 36% that of the moon seen from Earth.
Now that the current distance is only 247.66 thousand km, it's 44%.
arctan(.95/247.66) in degrees = 0.21978 ≈ 0.22 And the moon's size is half a degree.
I want to estimate what the angular size will be, in degrees, from the planned orbit heights. 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

[EDITED to correct angles at OmCheeto's polite suggestion]

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

While we are at it I think the current angular size must be around 53% of the full moon. .. Let me check.
I just checked http://neo.jpl.nasa.gov/orbits/fullview2.jpg and it is 203.61 thousand km.
so we can start getting used to the more correct formula and say
2arcsin(.475/203.61) in deg = 0.2673 degree,
so 53% of full moon size

 

[mfb]

Ah right, the arcsin is correct, sorry.

Dawn is http://neo.jpl.nasa.gov/orbits/fullview2.jpg , that gives 52% the size of moon (using its current distance). Enough to be clearly visible as a disk for a human eye.

 

[marcus]

Not a problem! Glad you are interested in the mission and following this too! I just checked and the distance to planet is 201.17 thousand km
2arcsin(.475/201.17) in deg = 0.27057... deg
I notice you are using Wolfram alpha, and that has the current distance to the moon (which varies) and gives the angular size currently as 0.514 degree.
2arcsin((radius of moon)/(distance to moon))*180/pi = 0.514
I have just been using a convenient approximation of half a degree for the moon seen from earth, in all cases. So we would get slightly different answers.

 

[mfb]

Doesn't make a large difference. 53.7% with the updated distance to Ceres (below 200,000km now) and the semimajor axis of moon. 54.5% with the current distance to moon.

 

[marcus]

Still, no reason I always have to compare with the moon, the moon in the sky is just something around half a degree. That gives a rough sense of what a degree is.
We can make "degree" our scale and get a visual sense that way.
Today when I checked and the distance was 193.92 thousand km.
2arcsin(.475/193.92) in deg = 0.281 degrees. So that's how big the thing Dawn sees is. :)

 

[marcus]

The January edition of Dawn Journal came out today!
http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/
Marc Rayman, the mission director, makes terrible puns but he also has really interesting things to tell us.

It's a once-a-month thing. I've found April 2014 Journal interesting, also October, November, December 2014. Here is the entire listing of links to past DJ.
http://dawn.jpl.nasa.gov/mission/journal.asp

 

[mheslep]

This is to make up for thrust lost in September when an energetic cosmic ray particle disabled electronics serving the propulsion system.

I'm curious how one could possibly determine that a cosmic ray was in fact the cause of the failure on Dawn and not some other reason, such as a long term collective effect of less energetic radiation, or simple mechanical fatigue brought about by, say, temperature swings.

 

[Delta31415]

this is very interesting.

 

[OmCheeto]

I'm curious how one could possibly determine that a cosmic ray was in fact the cause of the failure on Dawn and not some other reason, such as a long term collective effect of less energetic radiation, or simple mechanical fatigue brought about by, say, temperature swings.

I've submitted your question to Dr. Rayman. Though, it is still in moderation mode.

OmCheeto says:
Your comment is awaiting moderation.
January 30, 2015 at 9:23 am
Hi Marc!
We’re following the mission at a physics forum I belong to, and the following question came up today:

“I’m curious how one could possibly determine that a cosmic ray was in fact the cause of the failure on Dawn and not some other reason, such as a long term collective effect of less energetic radiation, or simple mechanical fatigue brought about by, say, temperature swings.”

I figured you would be the only one who could answer this.

Thanks!

Om

Not sure how to interpret his response times. Sometimes they are very fast, and sometimes they take 24 hours, and sometimes:

17. Matt Gibbons says:
January 11, 2015 at 12:31 am
Hello Dr. Rayman,

Seems I read, ...

Matt Gibbons
Bellingham, WA

Reply

• 
  Marc Rayman says:
  January 11, 2015 at 10:38 am
  Thank you for your message and your interest, Matt.
  
  I too have occasionally seen this claim...

he responds before the question was asked.

ps. I think all times are listed as PST, since he works at JPL.

 

[marcus]

I'm curious how one could possibly determine that a cosmic ray was in fact the cause of the failure on Dawn and not some other reason, such as a long term collective effect of less energetic radiation, or simple mechanical fatigue brought about by, say, temperature swings.

You are in luck, Mheslep! You can satisfy your curiosity to a large extent by reading the October 2014 Dawn Journal. I can't give a better explanation than Marc Rayman.

Spacecraft people are used to these fairly rare events and seem able to recognized them. A foreign particle hits a transistor and changes a bit, a bug appears in a program, the program has to be reloaded and the system rebooted. Or some such narrative.
So the electronics in spacecraft are SHIELDED in a certain way, and programs are made RESISTANT in certain ways, which are special to high altitude equipment that is outside the Earth atmosphere protection from cosmic rays.
Whatever they do, it seems to reduce the prevalence of this kind of thing.

Other spacecraft have experienced the same thing. and Dawn itself had a similar event occur in 2011. Marc Rayman describes that. they were lucky that time and caught it early and were able to reboot almost immediately. Same sort of thing happened as in 2014 but they lost less thrust time.

and you are quite right to note that it might have been SEVERAL particles hitting the same transistor at once. But the simplest assumption is it was just one high energy proton. the most common type of cosmic ray event.

I can't say anything based on my own direct personal knowledge because I'm way non-expert--I'm just relaying my impression, what I gather from following the "news". You might read October 2014 DJ, and look at wikipedia on cosmic rays, especially if they have a section on cosmic rays and spacecraft experience and shielding. Let us know if that leaves your curiosity still unsatisfied. Maybe someone who knows more will respond here, or I can do some digging.

 

[marcus]

I like Om's way of responding to Mheslep's question! I did not see it until I finished writing and posting my own response.
For sure, why not write to Rayman's blog?! but I'll leave my answer standing anyway since it suggests some partial answers to think about for the time being until Rayman or one of his group responds on the blog.