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

[marcus]

The update just showed the simulated image of Dawn turned so as to be in picture taking mode.
http://neo.jpl.nasa.gov/orbits/fullview2.jpg This happened at around 1:30 PM Pacific time as far as I can tell.
The distance range is given as 149 thousand km.
The table copied back in post #45 estimated that the picture would be taken at 146 thousand km, today.
(EDIT: actually the current status distance changed to 146 thousand km later today. Maybe there is no discrepancy. Unsure about this.)

A discrepancy could suggest that some unexpected factor entered in, but most likely I think it just says what we may have guessed already that the table distances are just estimates. And it's all more or less approximately consistent. I'm very glad we have both the table, and the current status simulated views, and they match up tolerably well.

This "picture taking mode" thing happened the last time Dawn took a picture, I noticed. For some period of time like several hours to half a day the main axis of the spacecraft , along the solar panel arms, is VERTICAL in the simulated view frame. And then when the probe gets back to thrusting it resumes the usual attitude with the solar panel arms nearly horizontal in the view frame.

It could also be that what I'm seeing in the simulated view is not exactly "picture taking" mode. It might be "data transmission"----the attitude has to be turned around so the big antenna is aimed at Earth. But either way it seems to have something to do with getting a navigation shot of Ceres against the background of known stars and transmitting it back to Control at Pasadena.

Damn, some of our family live in Pasadena and show no sign of excitement about this. They are walking on holy ground. If I didn't hate to travel I would go crash on their couch and try to hang around JPL some. Ceres is a PLANET, and it has a lot of WATER ICE. Don't people get this?
Do dolphin's bones lose calcium if their watery habitat is in low gravity? Or are they adapted to more neutral buoyancy and don't need so much gravity? Suppose you had an ice cavern with a lake of chilly water (which serves both to grow algae and to cool the power supply that provides light). Suppose there are fish in that lake. Do the fish suffer from 3% gravity the way today's humans would. The humans would have to do some pretty intense exercising to stay healthy.

 

[Dotini]

Video showing curious bright spot, etc.

This source describes the white spot as "flickering"!
http://www.space.com/28336-mysterious-white-spot-on-ceres.html

http://www.techtimes.com/articles/2...on-ceres-leaves-nasa-scientists-flummoxed.htm

They think it may be a pool of water at the bottom of a crater.

 

[marcus]

BTW if you have been checking out the simulated view current status http://neo.jpl.nasa.gov/orbits/fullview2.jpg from time to time you will have noticed that the brown dot in the center has grown a lot.
Another nice thing, the the cursor that you move around on the frame with your mouse or touchpad is a white dot with a + symbol on it (clicking enlarges, so "+")
The width of the frame is scaled to be 30 degrees.
If you look at the size of that white dot, it is about 1/60 of the width of the frame. IOW that white dot they give you is the size of the full moon in Earth sky
so you can drag it over to the brown image of Ceres in the center and compare.
You will see that Ceres dot is now about 3/5 or 2/3 of the full moon dot. It suggests to me that the guy who does the simulated views for them is thoughtful and consistent in unadvertised ways (this is a form of coolness, the guy has aspects of coolness, check it out.)

 

[OmCheeto]

The update just showed the simulated image of Dawn turned so as to be in picture taking mode.
http://neo.jpl.nasa.gov/orbits/fullview2.jpg This happened at around 1:30 PM Pacific time as far as I can tell.
The distance range is given as 149 thousand km.
The table copied back in post #45 estimated that the picture would be taken at 146 thousand km, today.

That discrepancy could suggest that some unexpected factor entered in, but most likely I think it just says what we may have guessed already that the table distances are just estimates. And it's all more or less approximately consistent. I'm very glad we have both the table, and the current status simulated views, and they match up tolerably well.

This "picture taking mode" thing happened the last time Dawn took a picture, I noticed. For some period of time like several hours to half a day the main axis of the spacecraft , along the solar panel arms, is VERTICAL in the simulated view frame. And then when the probe gets back to thrusting it resumes the usual attitude with the solar panel arms nearly horizontal in the view frame.

It could also be that what I'm seeing in the simulated view is not exactly "picture taking" mode. It might be "data transmission"----the attitude has to be turned around so the big antenna is aimed at Earth. But either way it seems to have something to do with getting a navigation shot of Ceres against the background of known stars and transmitting it back to Control at Pasadena.

I don't think it's "data transmission".
I was just snooping through my Twitter feed, when I saw that https://twitter.com/NASA_Dawn had retweeted a picture of Dr. Rayman from https://twitter.com/NASAJPL, and I realized I wasn't following NASAJPL. There I found a tweet about an interesting website: http://deepspace.jpl.nasa.gov/dsnnow/

If you click on the "Enter now" button, you get a live feed of the Deep Space Network activity. I've been checking in on it, off and on, for about 2 hours now. It was fascinating watching the dishes turn off and on, and seeing the signals switch from each spacecraft to different dishes.

Anyways, I've seen no signals to or from Dawn, over the last two hours.

Damn, some of our family live in Pasadena and show no sign of excitement about this. They are walking on holy ground. If I didn't hate to travel I would go crash on their couch and try to hang around JPL some. Ceres is a PLANET, and it has a lot of WATER ICE. Don't people get this?
Do dolphin's bones lose calcium if their watery habitat is in low gravity? Or are they adapted to more neutral buoyancy and don't need so much gravity? Suppose you had an ice cavern with a lake of chilly water (which serves both to grow algae and to cool the power supply that provides light). Suppose there are fish in that lake. Do the fish suffer from 3% gravity the way today's humans would. The humans would have to do some pretty intense exercising to stay healthy.

I cannot answer these questions.

ps. Dr. Rayman says he is not a fan of TMBG. But, I warned him...

Hi Marc,

Sorry to be so assuming about you possibly being a fan of TMBG, but, your latest journal comment

The probe is much closer to Ceres than the moon is to Earth.
And now it is even closer…
And now it is closer still!​

looked just like something from their song “Older”

You're older than you've ever been
and now you're even older
And now you're older still​

You may want to hire a lawyer.

 

[OmCheeto]

The update just showed the simulated image of Dawn turned so as to be in picture taking mode.
http://neo.jpl.nasa.gov/orbits/fullview2.jpg This happened at around 1:30 PM Pacific time as far as I can tell.
The distance range is given as 149 thousand km.
The table copied back in post #45 estimated that the picture would be taken at 146 thousand km, today.
(EDIT: actually the current status distance changed to 146 thousand km later today. Maybe there is no discrepancy. Unsure about this.)

A discrepancy could suggest that some unexpected factor entered in, but most likely I think it just says what we may have guessed already that the table distances are just estimates. And it's all more or less approximately consistent. I'm very glad we have both the table, and the current status simulated views, and they match up tolerably well.
...

Dawn induced panic attack #47.
Just checked "fullview2.jpg", and it's still in photo/data x-mission alignment.
Checked the DSN, and no data was being received.
Somethings wrong!

Just rechecked DSN, and data is streaming in at 125 kb/sec.

Phew!

Anyways, Dr. Rayman, in his last email said;

Hi Om,

... The Where is Dawn Now? feature is not accurate enough to warrant using so many significant figures. See this recent point. And I've explained in many Dawn Journals, sometimes we thrust and sometimes we coast, and the coast periods are neither regular nor of uniform duration. ...

I don't mean to be discouraging. Quite the contrary! I love doing the kind of thing you are having fun with, and I admire your creativity and insight. I would simply point out that it's worth being careful and recognizing the limitations. I'm too busy to provide details, but I hope you find some more in my prior Dawn Journals and, most importantly, that you continue to be so interested in the mission!

Sorry, I'm not a fan of TMBG. I love your PPS! Thank you for being considerate with my time. And now, I will indeed stop responding to your emails.

Marc

As if he could discourage us! And the limitations are only exacerbated by the fact that I lazily didn't bother to note the UTC times, so, in the following excerpt from my spreadsheet, corresponding to the published projected dates and distances, I included both the previous and following days data. Which, up until February 19th, is completely accurate, +/- a day.

Columns B, C, & D are my projections. Column E is Dawn published projections.
And as I mentioned, please ignore all of my data after Feb 19th.

I probably should have changed the "Distance 1000 km" functions after the 19th, as the "x vs y" components of the velocity get really big, inducing planetoid crashing results.
(The "x vs z" number was based on the published distance(hypotenuse) vs 20,000 km, if that makes any sense.)

Though, looking at the projected data and images again, I probably should have used a number closer to 40,000 km as my "x" orbit intercept value.

Oh well, I guess they're lucky I'm not flying the vessel.

And thank god for Emily:

Emily Lakdawalla @elakdawalla · 47m47 minutes ago
(Of course, when I tweet about DSN Now, it is undergoing scheduled maintenance. If you get a 404 on that link, please return in 2 hours)​

She's averted panic attack #48.

 

[marcus]

The pictures are up!
Images taken on 4 February are posted:
http://www.jpl.nasa.gov/news/news.php?feature=4475
That's just some of what's online. It is an ANIMATION that shows a large part of one rotation (not the full 9 hours but quite a bit of it) taken at a distance of 145 thousand km
and the resolution is around 14 km per pixel.
Here are some more links
http://www.jpl.nasa.gov/spaceimages/details.php?id=pia19174
http://www.jpl.nasa.gov/spaceimages/details.php?id=pia19179
I like the watching the animation, but am not sure how to paste it into this post, so here is a still:

Current status http://neo.jpl.nasa.gov/orbits/fullview2.jpg shows the craft has rotated out of picture-taking mode and has resumed normal thrusting, current distance is given as 133.91 thousand km.

That means an angular size of 0.406 degree (getting close to full moon size, which is half a degree)

There seems to be a variety of pictures to choose from, from yesterday's session. The link for this particular one I happened to post is:
http://www.jpl.nasa.gov/spaceimages/images/largesize/PIA19179_hires.jpg
in case anyone wants to forward it or something. If anyone finds other interesting images please share the links! And post them for us if you think appropriate. No naked grain goddesses or fertility symbols please.

 

[OmCheeto]

The pictures are up! ...

Yay!

...
in case anyone wants to forward it or something. If anyone finds other interesting images please share the links! And post them for us if you think appropriate.

This one is interesting: http://photojournal.jpl.nasa.gov/archive/PIA19174.gif
from the page: http://www.jpl.nasa.gov/spaceimages/details.php?id=pia19174

It seems to confirm my suspicion that Dawn was imaging Ceres for nearly a full rotation. Dawn stopped sending data last night around midnight. 10 hours!

No naked grain goddesses or fertility symbols please.

How about the "Einstein" image, which seems to have answered NASA's 10 year old question?

Was Einstein a Space Alien? (NASA)
March 23, 2005​

Confirmed!

https://pbs.twimg.com/profile_images/474989255773925378/0nhQR66C_bigger.jpeg Cody Healey‏@cody_healey
@NASA @NASA_Dawn Hey, I see a face on that planet. Looks like Einstein!

https://pbs.twimg.com/media/B9GFbgZCcAA0Vsj.jpg ​

I don't follow Cody Healey, and I still don't know how to properly use Twitter, but I think this showed up briefly on Dawn's feed, and I thought it was very funny.

ps. #lookingforwardtononanthropomorphizedpicsofCeres:)

 

[marcus]

Current distance 129.88 thousand km. approach speed 0.100 km/second, so 8640 km per day. When might the next photo shoot be?
The table is basically a list of navigational "photo opportunities" when they plan to stop thrusting and turn the camera at Ceres and get a picture of it against the background of known stars (to use in locating Dawn precisely relative to Ceres, so as to steer the probe better).
The table indicates times when they might share some of these navigational shots with the public.

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
http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/
Click to get Om's dateline+trajectory diagram:
View attachment 78574

 

[marcus]

Let's see if I can get a copy of Om's dateline trajectory diagrams. Both have the sun way off to the left and Dawn coming in from the left. In the upper (respectively lower) Ceres orbit direction is into the frame (respectively straight up in the frame). I'll try to insert these diagrams from Marc Rayman''s Journal which Om made more useful by annotating them, tagging some of the day circles with dates to produce a dateline.

Let's see how that looks. Not bad. In the lower diagram, Dawn enters the diagram at the left edge on 10 Feb, when her distance to Ceres is about 100 thousand km.
For that to work, if distance is now 130 thousand, the probe has to go 30 thousand km between now and 10 Feb. Sort of 4 days, so it has to average sort of 8 thousand km a day. OK, seems do-able.

 

[lpetrich]

There is something interesting about the world Ceres, as I like to call it. It looks like Ceres's spin axis is now nearly perpendicular to the asteroid's current direction to the Sun. That means that it should be possible to map nearly all of it.

 

[marcus]

That's right! There is what appears to be an element of luck. Ceres orbit period is 4.6 years. That means its seasons, like e.g. Winter, last on the order of a year. Dawn might have arrived e.g. in the middle of Winter when the north polar regions are in arctic darkness. The probe's limited supplies only allow it to operate for a few months. So it would be unable to map those regions in darkness. I don't know if the planet's axis inclination was determined earlier by Hubble space telescope and this went into planning the mission, or if it was just luck. Maybe the inclination is a small angle anyway, or we simply happen to be in an "equinoctial" season just now (a Ceres spring or autumn.)

Om's annotated diagram showing the approach timeline from two different perspectives is useful for reference, and since we have turned a page I will bring it forward for convenient reference. Thanks to Marc Rayman for posting the original un-annotated approach trajectories in his April 2014 Dawn Journal.

The lower diagram (looking down on Ceres north pole, sun to the left, solar orbit motion "up" in the figure) shows the probe having deficient solar orbit speed and falling behind until around 24 February when Ceres gravity causes it to start catching up. By 23 April, probe is in a circular polar orbit and beginning to pass under Ceres south pole. The orbit would not look circular in the lower diagram because we are viewing it somewhat edge-on, from the north pole direction.

According to current status page http://neo.jpl.nasa.gov/orbits/fullview2.jpg the distance to Ceres is now (as of 6 February 9AM pacific ) 125.92 thousand km.
2 arcsin(.475 /125.92) = 0.432... deg, about 86% angular size of full moon .

 

[mfb]

Ceres' axial tilt is just 3 degrees, no matter when you arrive you can map nearly everything. The Wikipedia article cites a paper from September 2005, two years before Dawn was started. Hubble could see the bright spot, so it was possible to measure its axis of rotation directly.

 

[lpetrich]

Dawn got complete coverage of Vesta just before it departed for Ceres: Spring arrives to Vesta's north pole, as Dawn departs, plus a request for citizen scientists | The Planetary Society, Dawn Mission: News & Events > Shape of Vesta

For other Solar-System objects, one can get a clue from published maps -- where do they lack detail? Planetary Visions: Texture maps

Mariner 10 flew by Mercury 3 times, but saw only half of that planet's surface. That's from a Mercury solar day being two Mercury years and Mariner 10 being in an orbit with a period of about 2 Mercury years. However, the MESSENGER Mercury orbiter has succeeded in seeing all of that planet's surface (MESSENGER: Global Mosaics of Mercury).

Venus has 98% coverage from Magellan's radar, all except near its south pole. Likewise, the Moon and Mars have nearly complete coverage.

Jupiter's and Saturn's big moons are well-covered, except for near their poles. However, Uranus's moons only had their southern hemispheres visible. That is because the only spacecraft to visit them, Voyager 2, flew by near Uranus's southern summer solstice.

 

[lpetrich]

Here are some figures on Ceres, from its Wikipedia article.

Stable orbits around an object are all inside its "Hill sphere", and Ceres's Hill sphere has a radius of about 220,000 km.

Surface-satellite orbital velocity and orbital period: 360 m/s and 2.3 hours.

Dawn is currently 122,000 km away from Ceres and traveling around 100 m/s toward it. Escape velocity from Ceres at that distance is about 32 m/s. But it's a month before Dawn has slowed down enough relative to be Ceres to be captured.

We should be getting much better values of Ceres's mass and size before long, however.

 

[mfb]

At 105000km Ceres will look as large as the moon (using its semi-major axis). 2 days to go.

Ceres' mass is known to better than 1%, getting a better measurement know looks complicated - it would need a very good distance measurement to see the small effect of gravity (relative to thrust) with an uncertainty below 1%.

 

[Dotini]

Rayed craters and possible maria make this look like our familiar Moon, IMO.
"Here be dragons -- but not for much longer!"
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html

NASA / JPL / UCLA / MPS / DLR / IDA / Emily Lakdawalla
Dawn's view of Ceres on February 4, 2015 (animation)

This animation consists of 20 individual frames shot as part of Dawn's third optical navigation campaign on approach to Ceres. The images have been enlarged from the original size by a factor of two, and contrast enhanced to bring out more detail.

 

[marcus]

...
...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html

NASA / JPL / UCLA / MPS / DLR / IDA / Emily Lakdawalla

Beautiful job! I like Emily Lakdawalla's comments at her blog.
Dotini, could you please explain how you transferred the animation from her blog (at the link you gave) to your post here? Did you use the "media" button, with the icon that looks like some 35mm film frames. And how exactly does one proceed?

 

[Dotini]

You're going to roll your eyes when I tell you. I have a 20 year old Mac that is so obsolete I don't have cursor control even typing in this reply block, tho I do gain a cursor when I edit.

Anyway, I just highlight, copy and paste.

 

[mheslep]

I see not just scant, but zero mention of Dawn's arrival at Ceres in the mainstream newspapers, which is disappointing. Though, perhaps they're waiting for closest approach. I don't really expect a front page piece every day between now and April.

 

[marcus]

...
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?

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/

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

 

[OmCheeto]

...
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?

Dr. Rayman, on Friday stated:

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

My guess is that the washing out of the polar regions may be due to the lack of available light.

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/

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.

 

[marcus]

My guess is that the washing out of the polar regions may be due to the lack of available light.

Good guess! I'm glad you got a response from Dr. Rayman about the axial tilt. Just a few degrees (Wikipedium on Ceres say about 3 deg) and it is currently Southern summer--around the solstice in fact.

Om your extra interest and energetic research makes all the difference! Keeps me feeling optimistic and excited by what we are learning in this thread.

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition. That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

 

[marcus]

About capture, I measured on your annotated diagram and it looks to me like in either diagram the distance from Ceres at time of capture is 40% more than the distance on 1 March which is 49000 km. So around 6 March, capture, distance must be R = 68000 or 69000 something like that.
So we can calculate the upper limit on speed (2GM/68000 km)^.5
remembering that the mass of our planet is M = 943 billion billion kg.
Just paste this into google: (2G*943e18 kg/68000 km)^.5
When I do that, google comes back with 43 m/s
Actually it says "((2 * G * (943e18 kg)) / (68 000 km))^.5 = 43.0233559 m / s" :D

Current status says the current speed is about 100 m/s. So if there is going to be capture on 6 March then by this rough calculation the speed relative to Ceres has to come down to below 43 m/s.

I didn't measure very accurately just held a plastic ruler up to the diagrams on the screen. It was about 40% farther on 6 March than on 1 March.
Current angular size: 2 arcsin(.475 / 108.83) = 0.500148519 deg (as of 3PM pacific, on 8 Feb)

BINGO, half a degree. Full moon size!

 

[marcus]

It's always fun to see better and better pictures, but I would say that the next big landmark day for me will be FEBRUARY 24 WHEN DAWN IS 28 kkm directly North and

28 kkm directly behind Ceres in their race around the sun.

so that the hypoteneuse distance works out to 40 kkm (28²+28²=40²).

The idea is that 28 thousand km lag should be the FARTHEST BEHIND CERES THAT DAWN EVER TRAILS.

It has to cling fiercely to Ceres' coat-tails (if Grain&Fertility godesses can be said to have coat tails).
It has to stop drifting behind at that point and from that day forward begin catching up.

The combined x,y,z distance will generally be greater than that but from then on the behind distance along the Ceres orbit track has to be gradually reduced until they are neck and neck and
Dawn is safely in a circular polar orbit around the planet.

 

[OmCheeto]

Good guess! I'm glad you got a response from Dr. Rayman about the axial tilt. Just a few degrees (Wikipedium on Ceres say about 3 deg) and it is currently Southern summer--around the solstice in fact.

Om your extra interest and energetic research makes all the difference! Keeps me feeling optimistic and excited by what we are learning in this thread.

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition. That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

Well, I figured the speed had to be somewhere between the orbital and escape velocities. But one of the only relevant equations I found was for a Kepler orbit, which required a "semi-major axis", which, given my complete geometrical incompetence, is completely useless. How on Earth do you extract a "semi-major axis" from a curve segment?

I think I'll just sit back now, and wait for the pretty pictures.

 

[mfb]

Every speed below the escape velocity will give an orbit. We don't know the planned 3D motion so it is hard to estimate the speed of Dawn. I guess the speed relative to Ceres will drop below the escape velocity at that point. It will make Dawn the first object to have been in orbit around two different celestial objects (not counting Earth).

 

[marcus]

Om, However it happened we got fairly close agreement for the speed V around "capture" on 6 March. I was just roughly estimating and got 43 m/s and you got 45 m/s I think.
Whatever the speed actually is, if capture occurs that day then the speed must be just slightly LESS than escape speed (43 or 45, something around there).

 

[lpetrich]

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition.

Yes, that's a reasonably good approximation, though it's a spherical-cow sort of formula. Or, in this case, a spherically-symmetric asteroid all alone.

I'll estimate the sizes of the two main sources of departure from that ideal state: Ceres's equatorial bulge and the Sun. Ceres should have an equatorial bulge from its rotation, and one can estimate its size from the hypotheses of hydrostatic equilibrium and constant density.

Ceres's equatorial bulge:
$$ \text{Oblateness gravitational coefficient } J_2 \sim \text{flattening } f \sim \left( \frac{T_{SS}}{T_{rotation}} \right)^2 $$
SS = surface satellite

For Ceres, it should be about 0.06 or 1/17. To do a better job, one should use the formulas for Maclaurin spheroids, as they are called.
$$ \text{EB's relative effect } \sim J_2 \left( \frac{r_{equatorial}}{r} \right)^2 $$

The Sun:
$$ \text{Sun's relative effect} \sim \frac{M_{Sun}}{M} \left( \frac{r}{r_{to\ Sun}} \right)^3 \sim \left( \frac{T_{SS}}{T_{Sun\ orbit}} \right)^2 \left( \frac{r}{r_{equatorial}} \right)^3 $$

At 100,000 km (100 megameters):
EB: 1.4*10^(-6)
Sun: 0.026

That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

A good approximation ought to be constant acceleration, since Dawn's ion engines have *very* low thrust, and since they are typically run almost continuously.

 

[mfb]

The uncertainty on the mass of Ceres is much larger than the effect of the equatorial bulge.

Dawn has a constant acceleration as measured by Dawn, but not in the system of Ceres.
103400km now (90m/s relative velocity), which means Ceres appears larger to Dawn than the moon to us (using the semimajor axis).

Oh, and Ceres is a very good approximation to a spherical cow in frictionless vacuum. It does not give milk, however.

 

[OmCheeto]

Every speed below the escape velocity will give an orbit. We don't know the planned 3D motion

I kind of do.

The x-axis would be a line from the sun to Ceres.
The y-axis would be through the north pole.
The z-axis is along the line of motion of Ceres.

I derived this from plotting out the points from the published images.
Much massaging has been done, and I do not recommend piloting crafts in this manner.
The original y and z values jumped around so badly, that it looked like pinball.
Then I found a couple of buttons on my spreadsheet: Trendlines & Polynomial

so it is hard to estimate the speed of Dawn.

It will be fun to see how close I got, just by eyeballing the graphs.

I guess the speed relative to Ceres will drop below the escape velocity at that point.

I'm afraid I still don't understand how this works.
But as wiki states; "The consequences of the rules of orbital mechanics are sometimes counter-intuitive."
I have found this to be quite true.
When I first saw stated that PE tends towards zero at infinite distances, I scoffed.
Everyone here one Earth knows the equation is PE = mgh, and therefore, as h approaches infinity, so should PE.
Things are different, out in space.

It will make Dawn the first object to have been in orbit around two different celestial objects (not counting Earth).

Yay!

 

[marcus]

Since we just turned a page I will bring forward the essentials. Here's Emily Lakdawalla's version of the latest shots.
...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html

NASA / JPL / UCLA / MPS / DLR / IDA / Emily Lakdawalla

Here's Om's date/distance labeled version of Marc Rayman's approach trajectory diagrams. In upper, the trajectory path is projected onto the plane normal to Ceres orbit motion. So Ceres' motion is into the page. The upper diagram does not show how the probe falls behind Ceres (almost 30 thousand km) until 25 Feb, when it begins to catch up. Because that falling behind would be out of the page. The lower diagram (which Dawn entered on the left edge today, 10 Feb) shows the path projected down on Ceres orbit PLANE, with the sun off to the left and the planet's orbit motion directly up on the page. This is the same path projected on a different plane. It shows how the probe falls behind at first by nearly 30 thousand km--I estimate 28,000 km but that remains to be seen.

Here's the link to current status:
http://neo.jpl.nasa.gov/orbits/fullview2.jpg

 

[marcus]

25 Feb is an important upcoming date because it is when the spacecraft will hopefully stop falling behind Ceres and (with Ceres' gravity helping) begin to catch up. At that time, I estimate that Dawn will be 28 kkm behind Ceres (measured on the orbit plane projection) and 28 kkm up measured on the upper diagram---the projection on the plane orthogonal to planet motion. Combining those two distances at right angles to each other gives 40 kkm, which agrees with Marc Rayman's figure for 25 Feb.

http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/

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, which listed planned navigational photo shoots. (I added three rows, without photo data, to help connect with the trajectory diagrams.)
Distance from Dawn to Ceres in (kilometers)
Ceres diameter in pixels
Resolution in (kilometers) per pixel
Resolution compared to Hubble
Illuminated portion of disk

 

[marcus]

Current status is for 11 Feb at 4AM UTC, which is 10 Feb at 8 PM pacific time. (8 hours earlier than UTC)
The distance given is 93.02 kkm (thousand kilometers).
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
2 arcsin(.475 / 93.02) = 0.585... deg
117% the size of a full moon.

Those estimates I gave for 25 Feb are just rough approximations. I said 28 kkm up (off the orbit plane) and trailing 28 kkm behind (for a combined distance of 40 kkm.
But it could, for instance be a bit less up and a bit more back.
It could be say 27.1 kkm up and 29.4 kkm back

The main thing is that around 25 Feb it stops falling behind and begins to catch up to Ceres in their race around the sun, and the farthest it falls behind is something short of 30 thousand kilometers.

If you have been periodically checking the current status page, with the simulated view of Ceres seen from Dawn, you may be enjoying seeing that brown dot in the middle of the frame grow large.
The white dot you get as a cursor, on that page, is full moon size (it's a 30 degree field of view and the dot is 1/60 of the frame width). So it gives something to compare with.

 

[marcus]

Current status ( http://neo.jpl.nasa.gov/orbits/fullview2.jpg ) now shows the Dawn spacecraft in picture-taking mode--having temporarily stopped thrusting.
And having rotated so as to change orientation--point the camera I guess.
So we can expect a bunch of new photographs (perhaps in a day or two, after processing) unless they are just calibrating instruments this time.

This particular photo-shoot is designated RC1. There should be a noticeable improvement in resolution over last time (assuming we get to see the results.)

Current status says distance as of 7PM pacific on 11 Feb (3h 12Feb UTC) is 86.43 kkm
2 arcsin(.475 / 86.43) = 0.6298 deg
126% of full moon
It gives the approach speed as 90 m/s. That's around 8 thousand km a day. So it is reasonable to suppose that some of the pictures, maybe all, will be shot at the nominal range of 83 kkm .
==========
Looking ahead, on 19 Feb, and 25 Feb, range will be 46 kkm and 40 kkm so angular size will be
2arcsin(.475/46) and 2arcsin(.475/40) respectively
237% and 272% of full moon size, respectively.

 

[marcus]

Om, I recall you checked out the Deep Space Network (DSN) to see if it was receiving images from Dawn, at one point, during a past photo shoot.
I got curious and tried to do this. I saw three locations---Goldstone, Madrid, Canberra. Each location had several antennas.
http://eyes.nasa.gov/dsn/dsn.html
I see! you click on the number under each antenna and it tells you which spacecraft the antenna is receiving data from. At the moment none of the antennas were taking Dawn data.

EDIT: I checked more recently (9PM pacific time) and saw that Dawn was talking to Canberra. Also an antenna at Madrid appeared to be standing ready to take over.

Checked current status as of 5PM pacific 12 Feb, distance 80.03 thousand km
2 arcsin(.475 / 80.03) = 0.680136325 deg
136% of moon-size :D

 

[OmCheeto]

Om, I recall you checked out the Deep Space Network (DSN) to see if it was receiving images from Dawn, at one point, during a past photo shoot.
I got curious and tried to do this. I saw three locations---Goldstone, Madrid, Canberra. Each location had several antennas.
http://eyes.nasa.gov/dsn/dsn.html
I see! you click on the number under each antenna and it tells you which spacecraft the antenna is receiving data from. At the moment none of the antennas were taking Dawn data.

EDIT: I checked more recently (9PM pacific time) and saw that Dawn was talking to Canberra. Also an antenna at Madrid appeared to be standing ready to take over.

Checked current status as of 5PM pacific 12 Feb, distance 80.03 thousand km
2 arcsin(.475 / 80.03) = 0.680136325 deg
136% of moon-size :D

I checked at around 5:30 pm PST yesterday, and communication had started.
This morning, around 2 am PST, communication had by that time, switched to Madrid.
Currently, there is no communication, and I see that Dawn's ion drive is back on. (Duh!)

I should mention, that my mathematical skills have become quite rusty over the last 30 years, from dis-use. But the plotting, and mathematical magic, of Dawn's trajectory have really re-sparked my interest. Not to mention, that I've discovered 3D rendering software on my laptop.

I discovered that it can even create an animation of the above. But at 20 megabytes, it's a bit above my storage limit.

 

[mfb]

Two images from different viewing angles, switching back and forth, would be sufficient to get a nice 3D impression.

 

[OmCheeto]

Two images from different viewing angles, switching back and forth, would be sufficient to get a nice 3D impression.

Never mind. I just checked my account. I'm not even close to my limit. :)

But my latest attempt came out to 34 megabytes! It took me 8 minutes to upload.

3D Animation

 

[lpetrich]

I've measured the positions off of marcus's most recent diagram. They are as a tab-delimited spreadsheet text file that I've attached to this message.

Columns:
Date
Distance in kkm, where given
X1 raw -- first picture horizontal
Y1 raw -- first picture vertical
X2 raw -- second picture horizontal
Z2 raw -- second picture vertical

The second row is my measurement of Ceres's position in the picture

Scaling my measured positions to Dawn's distances from Ceres has proved more tricky than I expected, so I'll use Mathematica for that.

BTW, OmCheeto, what software did you use for:
- Measuring Dawn's position off of marcus's picture
- Plotting Dawn's position in 3D
?

I wrote an image measurer for myself since I couldn't find a good one that enters a position with each click on the picture being measured. It's OSX-native, so to port it to Windows or Linux, you'll need GNUstep.

 

[marcus]

Om,
Impressive bit of animation, thanks for sharing it. The figure in your post #87 rotates! I'm curious: what type of account puts a storage limit on this? Is it a PF limit, or one connected with your ISP (internet service provider)? Or has it to do with some "cloud" thing that you upload to, analogous to YouTube for videos or SoundCloud for music.

 

[marcus]

LPetrich,
I printed off your numbers, and I am trying to judge the difference in scale. For Ceres the upper X1 and lower X2 are 238 and 321
that is, in the lower image the planet is farther to the right. So if I take Ceres as my origin, or zero, then the farthest right the probe goes is around 20 March
Say 582 - 238 = 344 on the upper
and 560 - 321 = 239 on the lower
As I interpret it, this determines the relative scale of the two diagrams. that is "239" on the lower is the same distance as "344" on the upper diagram.
Does this make sense to you?

You are doing the work, so maybe I should not make suggestions. But my inclination would be to subtract off the Ceres coordinates and make Ceres the origin. And then multiply the lower diagram distances by 344/239 to scale them up to be the same size as in the upper diagram. I wonder if this seems reasonable to you?

BTW the originals of the two diagrams are in the November Dawn Journal, as far as I know.
http://dawnblog.jpl.nasa.gov/2014/11/28/dawn-journal-november-28/
The date labels are figured out from taking the capture date to be 6 March, which is what Marc Rayman estimated it would be. this is marked on both trajectories

 

[lpetrich]

Those are raw pixel coordinates, right off the images. I did it that way so that it's easy to compare with others' measurements. Yes, it's a raw-data release. :)

My measurement of Ceres is in the second row. So you can subtract that out, at least if you think that it's not grossly in error.

I found that the first diagram has a true distance per pixel about 3/2 the second diagram, without a large error. So my measurements of Ceres's position in the pictures is not far off. I also found that 1 kkm = 3.22 second-diagram pixels, also without great error. I guess I should upload my adjusted version.

 

[OmCheeto]

Om,
Impressive bit of animation, thanks for sharing it. The figure in your post #87 rotates! I'm curious: what type of account puts a storage limit on this? Is it a PF limit, or one connected with your ISP (internet service provider)? Or has it to do with some "cloud" thing that you upload to, analogous to YouTube for videos or SoundCloud for music.

The limit is from my ISP. I've had my own webspace since around 1996.
I broke my 50 megabyte limit a while back from all the images I was posting to PF.
I didn't realize until today, that my $5/month upgrade bumped me up to a 750 mb limit.
I guess I'm still kind of stuck in my 4 kilobyte ram and 300 baud mindset, from when I got my first PC.

That little "movie" just bumped my usage from 90 mb to 120 mb.

 

[marcus]

Om, Petrich, Mfb,
as of right now Ceres is 50% larger than the full moon. This is the according to the current status report as of 10:30 PM pacific on 13 Feb
which is the same as 6:30 AM UT on 14 Feb. this is as I am posting this
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
says distance is 72.73 kkm which corresponds to 0.75 degree or 150% of moon's angular size

I'm impatient to see the photos that were taken yesterday.

 

[OmCheeto]

Om, Petrich, Mfb,
as of right now Ceres is 50% larger than the full moon. This is the according to the current status report as of 10:30 PM pacific on 13 Feb
which is the same as 6:30 AM UT on 14 Feb. this is as I am posting this
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
says distance is 72.73 kkm which corresponds to 0.75 degree or 150% of moon's angular size

I'm impatient to see the photos that were taken yesterday.

My guess is, that they are going to wait until 1 minute after midnight, such that it will be a Valentines day present.
Set your alarm, and take a nap! :D
That's what I'm doing.

 

[lpetrich]

Standard gravitational parameter - Wikipedia

In celestial mechanics, the standard gravitational parameter μ of a celestial body is the product of the gravitational constant G and the mass M of the body.

μ = GM

For several objects in the Solar System, the value of μ is known to greater accuracy than either G or M. The SI units of the standard gravitational parameter are m³s⁻².

For Ceres, it is 63.1(3) km³s⁻², or an error of 1/20.

I've included my calculations in my latest attachment. It's another TSV text file, and I've zipped it to save space.

In addition to my raw positions, I've calculated:
Positions in pixels with Ceres's position subtracted out
Scaled and combined 3D positions - kkm
Distances for them -- kkm
Smoothed 3D positions (5-point with quadratic fit) -- kkm
Distances for them -- kkm
3D velocities (2-point differences) -- m/s
Velocity magnitudes -- m/s
Orbital velocities -- m/s
Escape velocities -- m/s

I get capture at March 5 - 6, and the last velocities are close to the circular-orbit velocities for their distances.

 

[marcus]

Thanks Petrich! I just printed it out and will have a look. Almost midnight here, so may not get back to you until tomorrow.
It looks like you ran a successful check because capture worked out right (around 6 March) and the final orbit velocities are right.
I don't understand what you did very well at present, but I think it will become clearer after a little time.

 

[mfb]

I used your numbers to calculate accelerations from gravity and from Dawn. And while the second derivative of those numbers is not very reliable, the data points are good enough for an interesting result: gravitational acceleration from Ceres exceeds the average acceleration from Dawn's ion drives during the fly-by end of this month, probably 2-3 days before Feb 20 where you have the first velocity data.

Expressed in (Mm)^2*m/(s*day), GM=5450. That leads to 3m/(s*day) at 42.5km distance, while I get ~2m/(s*day) average for the ion drives, increasing to 4m/(s*day) close to RC3 (but there I would need a better integration scheme to get more reliable values).

Quick cross-check: Using the values of 3000 I_sp and 1 kW estimated power I found, continuous thrust would allow ~5800kg*m/s momentum change per day. Dawn is lighter than 1000 kg now, so those velocity changes look reasonable.
Also, most of the time the calculated ion drive acceleration is against the direction of motion, something you would expect from an enery/fuel-saving approach.

 

[lpetrich]

I used for smoothing a quadratic function fitted to 5 points with equal weight. That resulted in this filter kernel: (1/35)*{-3,12,17,12,-3}

For derivatives, I used this filter kernel: (1/2)*{-1,0,1}

I calculated Dawn's orbital energy, and it declines until about March 17, then it levels off until about April 7, then it starts declining again. Since Dawn's ion engines are typically run for long stretches of time, I conclude that there will be a gap in its running between March 17 and April 7, a gap that will let Dawn fall toward Ceres.

Orbital energy:
$$ E = \frac12 v^2 - \frac{\mu}{r} $$
Strictly speaking, energy per unit spacecraft mass. It changes at this rate:
$$ \frac{dE}{dt} = {\mathbf v} \cdot {\mathbf a} $$
for acceleration a. Doing the case of going from one circular orbit to another in an orbit that is as circular as possible, I find:

(Final orbital velocity) - (initial orbital velocity) = +- (propulsion delta-V) (+ for inward, - for outward)

Also, Dawn will get to about 40 kkm at about February 23, giving Ceres an angular size of 1.5 degrees or 3 times the Moon's from the Earth. It will then move away to about 80 kkm at about March 18, with Ceres being 0.75d across or 1.5 times the Moon's.. Late in April, it should be less than 20 kkm away, giving an angular size greater than 3d or 6 times the Moon's.

Dawn should reach its lowest Ceres-relative velocity at about March 21, a little more than 10 m/s (36 km/h, 22 mph). It is currently traveling at 90 m/s (320 km/h, 200 mph) relative to Ceres. Late in April, it should get up to about 60 m/s (220 km/h, 130 mph).

It will spend some time in a high-altitude orbit before going into a low-altitude orbit. The lowest possible one is for a surface satellite, and that sort of satellite will have orbital velocity 360 m/s (1300 km/h, 800 mph) and period 2.3 hours. So after getting into a high orbit, Dawn will expend about 300 m/s delta-V for getting into a low orbit. But it's not expected to do any more traveling, since escaping Ceres will require a similar amount of delta-V and going to some other sizable object will require even more delta-V.

I've uploaded the Mathematica notebook that I've been working on for these calculations.

 

[OmCheeto]

Just for the record, I fully approve of your numbers, lpetrich.

Ceres and Dawn are moving "into" the image.

Dawn and Ceres are moving "up" in the image​

Except for the apex, our numbers are very close.
And I should mention that your apex numbers are better than mine.
I think I massaged mine a bit too much, as my raw data at that point matched yours much more closely.

I used for smoothing a quadratic function fitted to...

You apparently know your maths...
I, unfortunately, have lost too many brain cells to figure out how to interpret some of this "magic".