Where is Dawn Now?

The JPL DAWN mission page has a button for Where is Dawn now?, which is in a drop-down menu if you click on Mission in the main menu on the left of the home page. Dawn of course is the spacecraft that visited the asteroid Vesta, and is now fast approaching Ceres.

If you click on the graphic for Simulated View of Ceres from Dawn you get something like the following screen shot, from Feb 2, 2015
Well, this tells us that it is approaching Ceres, in the middle there, but what stars are those? I couldn’t recognize them, and it was driving me crazy. I thought if I tracked it for a while, I’d see something I recognized, but this got me nowhere.

Note the credit to MYSTIC simulator. This is a tool they hold pretty close to their vest. It is apparently very powerful and used in their actual trajectory planning. Searching for it, I found a link to the NASA/JPL SOLAR SYSTEM SIMULATOR, which is nice, but it provides a much different format, and I was not getting anywhere with it either. Since it was all I could use, I went back a year or so and looked at some wide fields of view, and I was able to track it back to the present, where the background was identifiable as being near the head of Draco, in our Northern sky, but I still had no luck correlating the stars to the MYSTIC image.

The reason for this, I finally realized, is that the SSS tool had not been kept up to date with the latest course corrections ( I presume ) as it gives a different distance from Ceres than MYSTIC, which is official, I guess.

It was finally just dumb luck and persistence that enabled me to spot the correlating stars in my Starry Night software, and I’m sure you’ll agree it’s a good match:


Note that the four stars of the almost-square “box” that Ceres is in ( and remained in for several days, ) is comprised of stars from three different constellations. Two in Serpens Cauda, and one each in Scutum and Ophiucus. A very obscure asterism !

Update: Feb 13

Here’s the latest WIDN image:Feb13

Ceres has “moved” ( in line of sight from DAWN ) “southward” ( in celestial coordinates ) into the constellation of Sagittarius. You might recognize the Sagittarian “teapot” asterism directly behind the bottom half of the representation of the DAWN space craft.

Now here’s the funny thing … DAWN is currently almost directly in line with Ceres from our terrestrial POV, as evidenced by this Starry Night view with Chicago set as the viewing location. Ceres is marked, along with it’s orbit. I adjusted the time to make the view align with the WIDN representation, and this view happens to occur at 9:46PM local time, on Feb 13, 2015. ( I did have to turn off the horizon mask, as this view is looking down through the ground at the given hour. Sagittarius rises just ahead of the sun at this season. )

I think this alignment has to be ascribed to coincidence, except that it does mean that DAWN is “rising” towards Ceres along an orbit which is “sunward” of Ceres’ orbit.


Update Feb 19

Here is an animated gif of a loosely spaced sequence of WIDN images, displaced to show a constant star background. This POV definitely gives the impression ( correct, I believe ) that DAWN is rising up through the ecliptic plane as it closes with Ceres. We can see Ceres sinking to the south below Sagittarius. It’s only going to get more interesting!output_4fYhUU

Feb 20 – A note on escape velocity

The Wikipedia article on Ceres gives a radius of 476 km and an escape velocity of .51 km/sec . This is the speed which gives an object a kinetic energy equal to the negative of the gravitational potential at this distance:

m v2/2 = m GM/R so v2/2 = GM/R

then the escape velocity at some r > R, say, is

vr = vR sqrt( R/r ) … very simple!

We note at Feb 20 03:16:52 we have r = 44580 km and v = .082 km/sec

but .51 km/sec sqrt(476/44580) = .053 km/sec

so DAWN is not gravitationally bound to Ceres at this point.

Note that the exact direction of the velocity is not important. Varying the direction just places the object with escape velocity at different places on various parabolic orbits.

Also note that the engine is not firing at this time. It seemed to be in braking mode the last few days, so has it shed any total energy?

On Feb 9 05:06:33 DAWN had r = 107010 km and v = .096 km/sec

The escape velocity at that distance is .51 km/sec sqrt(476/107010) = .034 km/sec

So the fact that it has shed considerable speed shows that it has gotten much closer to gravitational capture. If it were “drifting” it would have gained kinetic energy equal to the difference of the escape velocities at the two distances, requiring …

(v + delta_v )2 – v2 = .0532 – .0342

or v + delta_v = .104 km/sec

One more thing, braking with the same rocket is more effective at higher speeds, and one usually expects that a braking rocket will fire near close approach. The very low thrust of the ion rocket means that the approach has to be more “gentle”, which accounts for the braking having been in progress already. Let’s watch and see if DAWN does pick up a little more speed before the braking resumes, as it must.

… sure enough, DAWN is at 184 mph at Feb 20 13:22:16 vs. 182 mph at Feb 19 23:50:16
Kind of ironic that they give a more precise speed in mph vs. km/s, which is .08 for both times. 1 mph is ( exactly ) .00044704 km/s

Feb 21 – The Approach Begins

Here’s a diagram I drew “by hand” in MS Paint showing the approach of DAWN to “Ceres Space” I started with the MYSTIC graphics for the times shown. I used the given distances and measured the angle of 5 degrees from the background stars, using my STARRY NIGHT software. This gave me the blue line and the close approach distance of a straight line trajectory.

But what about the gravitational attraction of Ceres? It seems that DAWN is not yet bound to Ceres, but it should be on a discernible hyperbolic trajectory.

After some thrashing around with pencil and paper, I consulted my old copy of Marion’s CLASSICAL DYNAMICS and found the equation laid out for me, and highlighted to boot!

I used my “hand built” yacc based calculator which features a plug-in definition capability. ( It actually pushes the defining string into the input stream when the defined symbol is encountered. )

Here are the defines I used:

define k “g*R^2”
define l “v*rp”
define alpha “l^2/k”
define E “v^2/2-k/r”
define e “(1+2*E*l^2/k^2)^0.5”

“mu”, for the “reduced mass” can be ignored when one body is much less massive than the other, m << M . Then mu ~= m and it cancels out of the expressions for alpha and e.

k is then GM, but I used GM/R2 = g to get k = g R2

Then setting the variables:
g= 0.00028
v= 0.083
rp= 39700
r= 45000
R= 476

( all in kilometers and seconds ) I just typed “alpha” and “e” to get the orbit parameters used in Marion’s equation 10.32, shown above:


Then the distance of close approach and the angular displacement from the current position are given by:

180/pi * acos( (alpha/r – 1)/e )

Perhaps surprisingly, these make sense! It says DAWN would approach to 38400 km at an angular displacement of 36 degrees from the current position, assuming it continued to “coast”.

Here’s the result roughly drawn in purple on the diagram. Of course, this is all in the realm of rough estimation!



… Well, I say we are! According to a definite criterion that I will show you in this most recent animated gif: ( click to see the animation. )

The bright star at top right is Fomalhaut in Piscis Austrinus. It’s actually visible on the southern horizon from mid-northern latitudes, but the POV is shifting to the south. and that’s the constellation Phoenix at the bottom.

But to make my point, I draw your attention to the “Distance to Ceres” caption at the bottom. Note that it “bottoms out” right at 24000 miles, and starts to increase again. DAWN is passing through “periceris” or whatever it might properly be termed. That is, a closest approach to Ceres. Of course, as it continues to “shed energy” it will spiral closer, but right now it is retaining the semblance of an hyperbolic orbit.

As we learned in physics class, the energy of a body at rest infinitely far from a gravitating body is conventionally set to ZERO. If it falls toward a gravitating body it gains kinetic energy but loses potential energy as it sinks into the “gravitational well”. So, any body with net positive energy can escape to infinity with a finite velocity, and is not bound to the gravitating body.

I say all this because the net energy of DAWN ( per unit mass ) is easy to calculate from the MYSTIC captions, and we can gauge its approach to gravitational binding by the steady diminishment that we see in this value, which is equal to 1/2 V_inf2, where V_inf is magnitude of the aforementioned “velocity at infinity”.

The value of V_inf was 115 mph in the last frame of the animation at Feb 24 06:33:54 . As of Feb 25 05:34:47, V_inf was down to 107 mph

March 10 – APOAPSIS

The “arrival” or attainment of zero orbital energy, was well noted on March 6, but there was nothing in the motion or actions of the DAWN spacecraft that would have indicated this event, such as it was. It has been “firing” its thruster steadily before during and after this milestone, as it continues to work towards its Survey Orbit. DAWN is now approaching another event, which has a more obvious connection to  its motion. That is apoapsis, as it is generically termed, or maybe apocerium, if we follow Kepler’s coinage of apohelium, which was later “greekified” to the now conventional aphelion.

As an aside, this question of what to call the “apo” point of the orbit has been bubbling along for decades, if not centuries. I only knew of aphelion and apogee, to be honest, but now I’m seeing that apoapsis, that is apo-apsis, is supposed to be the standard term, as awkward as it is. Investigating, I find that apogee derives from apogaeum, a nineteenth century analogue to Kepler’s apohelium, based on gaia, for the earth, I presume.

Of course, I became familiar with apogee from the era of “satellites” in the 50’s and 60’s, when I was quite young. I don’t see why the term can’t be applied to any planetary body, just as one speaks of Martian geology.

Well, at any rate, we are approaching the turning point of DAWN from its continuing motion away from Ceres ( despite it’s inevitable capture ) into a motion falling towards it. Following the Where Is Dawn Now? graphic on the DAWN mission page, we see that its motion is at a low point, suggesting a thrown object at its apex.. In the approximately six hours between Mar. 10,2015 00:03:54 and Mar. 10,2015 06:01:34, DAWN has decelerated from 78 mph to 76 mph, and moved 290 miles further from Ceres; from 43060 miles away to 43350 miles away. These are numbers appropriate to highway driving, not space travel!

So what happens when it “makes the turn” and starts falling towards Ceres ? I anticipate that they will turn off the thruster for a while and let it fall, but we’ll see.

March 17 – Making the Turn

DAWN is still approaching apoapsis, but it is very near, as its speed relative to Ceres of a mere 40 mph would indicate. Here is an animated gif comprised of  1 frame per day for the 14 days March 4 thru March 17. I think you can see a slight bow in the arc of the apparent motion of Ceres against the constellation Orion as DAWN adjusts the plane of its orbit. I did the best I could to keep the star background constant, but its still not perfect, as you can see. I hope it’s good enough to allow you to picture the apparent motion of Ceres against the fixed stars from DAWN’s POV. The sun is almost coming into view just above Orion, as you can see by the almost completely dark disk of Ceres.

Note that the distance from Ceres increases from 55020 km to 77740 km during this interval as DAWN climbs to its apex. You can see the disk of Ceres diminish in size as DAWN recedes from it.

output_MbifyqMarch 25, 2015

On March 19 DAWN achieved apoapsis at 48740 miles from Ceres moving at 34 mph, and is now starting to move closer to Ceres. As of today at 10:33 UTC it has “fallen” to 46460 miles away, and is moving at 40 mph! Here we go!

BTW, DAWN is moving towards a point almost directly opposite the sun from Ceres. WHERE IS DAWN NOW? provides views of the sun ( and other objects ) from DAWN’s vantage point, and here is an overlay showing that the sun is just about to come into view in the Ceres frame of reference … along with the earth and mars too!


Note that I had to expand the “sun view” and slightly rotate it to get a matching overlay.


Rocket Science

The explosion of the Antares rocket carrying the ISS supplies on 10/28/2014 was a great disappointment, and a severe setback for the U.S. space program.

Nevertheless, it gets my sleuthing instinct up, which in this case involves some real background study on the Antares rocket.

Also, we have the development in the News of the idea, or “meme” if you will, that the “explosion” was due to a destruct signal.

This idea, which is quite typical of news developments, is distressing to me. It all hinges on rhetoric. In this case the rhetoric of the “kill signal” sent by the range safety officer, and also the identification of “the explosion” resulting thereto.

First reports of the destruct command gave the bare facts, that the range saftey officer had initiated it “before the rocket hit the ground”, but the report I read gave the specific qualification that it was not certain whether this signal accounted for the explosion.

By appearances, the authority of the range safety officer has prevailed in the news reports, so that we are now informed that the rocket “was destroyed in a massive explosion at the launch site after safety officers sent a kill signal.”

Against this idea, we may note that there were at least two explosions. An immediate explosion, apparently of an engine, within one second of the first visible anomaly, and the more famous and spectacular explosion as the rocket fell back to the launch pad. This second explosion is what is being promulgated in the news reports.

Here is an animated gif made from a rough division of the launch video obtained by “double taps” at my terminal. It spans approximately 2 seconds, as timed by the video.

Well, it simply defies credulity that the apparent explosion at the base of the rocket, in the vicinity of the engines ( there are two of them ) was initiated by a manual signal after the observation of the anomalous “flare” of the rocket plume. In fact, the “108 %” announcement is made right at this time, and verbal acknowledgement of the disaster is rather delayed.

So all I’m saying is that the 2 second interval labeled 4:14 thru 4:15 in this video is a true record of the “accident” itself. I don’t think this is particulary controversial, outside of the impression given by the news reports.

So then the question is, what happened? The flare indicates to me that there was a breach in the LOX circulation through the bell. Aside from the flare, this would have caused a drop in pressure, obviously, in the LOX feed to the engine, and to the turbine driving the pump. So things go south real fast with the flame front going “upstream” into the engine.

My conviction is that there are more than a few individuals on the development Orbital Sciences staff that knew immediately what had happened. The company has stated that the facts will be known in “days not weeks”, so any day now …

Pascal’s Diamond

The relevance of Pascal’s Triangle to a “best of seven” tournament, such as the World Series, is obvious. But there is that wrinkle of not actually playing all seven games. In the past I had contented myself with various adjustments, but I never actually drew up “Pascal’s Diamond”, which is based very simply on the same generating rule as the Triangle, but terminated appropriately, as shown here:
Note that it’s equivalent to Pascal’s Triangle up to game 4 ( as indicated by the black numbers ) but terminates along the diagonals indicating 4 wins by either team. We immediately see that the nominal probability ( assuming “chance” outcomes ) of a 4,5,6, or 7 game series is 1/8, 1/4, 5/16, or 5/16, respectively.

Note that the equal probability of a 6 or 7 game series is a reflection of the “chance” outcome of game 6, which terminates the series at 4-2, or forces a game 7 with equal probability.

I decided to compare these probabilities with the last 91 world series results, in terms of games played, or equivalently, “games won by the series loser”. I found these to be 18, 18, 20, and 35. This compares to a nominal expectation of 11.37, 22.74, 28.43, and 28.43, for 91 “chance” games. So, this looks a little out of whack! But is it really? …


Well, we don’t have to do a canned analysis, even if we find ourselves being driven that way. We can make up any sort of test and apply it experimentally to sets of 91 “chance” series, and determine the “probability of rejecting the null hypothesis”, in this case the hypothesis that the real outcomes obey the same statistics as “chance” would have it.

Let’s just try looking at a generated sequence of 91 chance-ruled series. Here’s 10 of them, with the average over all 10 at the bottom. Each row represents the number of 4,5,6, and 7 game series in a trial of 91.

$ series 10
9 20 31 31
9 18 38 26
12 19 32 28
14 14 37 26
12 20 31 28
12 19 33 27
12 24 32 23
11 23 27 30
11 25 27 28
8 26 26 31

11.000 20.800 31.400 27.800

Well, notice the 14, 14, 37, 26 … qualitatively very similar to the actual historical “trial” of 91 world series, so impressionistically, we don’t have grounds to believe the games are ruled by anything but chance.

I’ll just add that I did go “back to the books”, or Wikipedia as we do these days, and tutored myself on some of the sigma-based statistical tests, and I’m ready to report that it’s all very interesting!

The thing that wouldn’t stop

The “thing” is namely the “June 27th flow” ( that’s when it started ) of Kilauea on the big island of Hawaii. This flow seems unusual because it is a very thin tendril that has been “guided” by a long crack in the ground, forming an extended lava tube of about 10 miles length, as shown in this Sept 15 map. ( Click to enlarge. )
It looked for a while like it was headed for the Kaohe Homesteads, but then it spilled out of the crack and bypassed them, following the fall line of the land. Fall lines are shown on the map as blue lines, like streams. Now it looks like it is headed right for the town of Pahoa, which resembles a “shore town” in that it is pretty much built along a single thoroughfare, route 130. It’s projected to reach that road in about 10 days, and following the fall line it’s on, I see ground zero at this small block of homes, shown in Google Maps Street View.
The official projection is something around there, but I couldn’t quite decipher it. It seems the only hope would be for it to stop, but it has been going for a month or so now and it’s hard to see that happening.

Here’s an animated gif view of the approaching “thing” in the form of a plume of smoke wending through the forested landscape. It’s a little hard to follow because the POV shifts backward as it advances. It starts as it advances along the “crack” and ends as it’s bypassing the Kahoe Homesteads.

Here is the official projection of the flow into Pahoa.
It is very close to what I thought, which is after all just a matter of following the downward slope. Of course it remains to be seen if the flow itself follows this simple projection.

In the following Google Maps view, the street view I showed above is looking southwest along La’Au Place. The projected path in the satellite view is just south of that, on the other side of the grove of trees. Note the the dark rectangular roof visible in both views.
UPDATE Oct 24, 2014
After a seeming stasis, with little advance through most of October, the flow front has reactivated in the last few days: ( click to enlarge )
It has come within a few hundred yards of the “transfer station” ( a recycling drop-off ) on Apa’a St. on the outskirts of Pahoa, so it seems like some sort of incursion is imminent. ( click to enlarge )
Update Oct 26, 2014
It didn’t take long for the flow to cross Apa’a St. (click to enlarge )
It’s headed for the cemetery, of all things, and going down the cemetery road. Crossing this street has made the news, and brought out the comments: e.g. “What do you expect if you live on the side of a volcano.” Let us not focus unduly on the ignorance of the commenter. This is the way the news operates. Everything is in the abstract. Details are presented without context, so this sort of interpretation is to be expected. OTOH we may note that the presence of a cemetery on the line of flow suggests that this event is without precedent in living memory. A brief perusal of the maps of this flow backs this up. It is very unusual in its nature and its direction of flow.

Well, Kilauea is there, and I think the residents are duly philosophical about living next to it. It’s a very narrow flow, and they may have reason to hope that it will stay narrow as it cuts a path to the sea, and remain in the nature of a temporary disruption.

Update 10/29/2014

Are you superstitious ? I hope not … because I may be!


Asteroid 2014 R6

Last March I posted NOTES ON CLOSE ENCOUNTERS, describing some thumbnail math relating to the hyperbolic shape of an asteroid trajectory passing near the earth, as viewed in the earth frame of reference.

I noted that the encounter with asteroid 2014 DX110 had a very “flat” ( high eccentricity ) trajectory, based on the speed and near approach distance of 1.3125 and 54, expressed in units of earth escape velocity and earth radius. ( Not earth orbit radius! ) The rule-of-thumb eccentricity came to 185 in that case.

The predicted encounter with 2014 RC set for this Sunday has a much less “flat” flyby trajectory, based on the available graphic shown here:


From the graphic we have, in terms of my previous post, r0 = 25,000 mi = 6.3 ; v0 = 40000 km/hr = 1 ( i.e. very nearly earth escape velocity. ) So this is a lot closer and a little slower than 2014 DX110. Turning to the formula

s2 = r02 / ( 1 – 1/ ( r0 v02 ) )

… we get an “impact parameter” ( projected asymptotic approach distance ) of 1.09 r0, so 2014 R6 will actually be “drawn” about 9% closer by earth’s gravity, compared to a straight line fly-by, and the hyperbolic trajectory has an eccentricity of 11, versus 185 and a deviation of just 1/2 % in the case of 2014 DX110.

Euclid & The Duchess

In Chapter IX of ALICE IN WONDERLAND Alice has a conversation with “the Duchess”, wherein the Duchess relates to her several “morals,” including the following:

“Be what you would seem to be”— or if you’d like it put more simply —“Never imagine yourself not to be otherwise than what it might appear to others that what you were or might have been was not otherwise than what you had been would have appeared to them to be otherwise.”’

It has occurred to me, and I feel firm in this, that Lewis Carroll meant this as a mock paraphrase of Proposition 6 of Book II of Euclid’s Elements:

If a straight line is bisected and a straight line is added to it in a straight line, then the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half equals the square on the straight line made up of the half and the added straight line.

I forget the exact occasion, but it was as I was perusing Euclid that I was struck by the similarity of construction. The phrase, “straight line” , corresponds roughly to “otherwise” and “others” in the Duchess’ moral, and both expressions suggest a lilting meter. ( … and I hope I don’t have to remind anyone of Charles Dodgson’s mathematical vocation ! )

Furthermore, the syntax is similarly baffling, Euclid can actually be parsed, but not so the Duchess. Her moral contains three long phrases, each coherent on its own but with overlapping connections that make the whole thing incoherent:

Never imagine yourself not to be otherwise than what it might appear
to others …

it might appear to others that what you were or might have been was not otherwise than what you had been …

what you had been would have appeared to them to be otherwise.

However, note that Prop. 6 does contain a “garden path” , as described by Stephen Pinker in THE LANGUAGE INSTINCT.

This is the phrase, “the rectangle contained by the whole with the added straight line …” which one is inclined to take as a complete description of a rectangle. However, the following phrase, “… and the added straight line together with the square on the half … “ is rendered nonsensical if one goes down this path, because a line cannot be taken together with a square. Of course, (ahem) the phrase, “with the added straight line” refers to the base of the rectangle while, “and the added straight line” specifies the height.

To me, it all adds up, especially considering the preamble. “Be what you would seem to be” would stand for the algebraic expression:

(a+b)2 = a2 + 2 a b + b2 … OR … a ( a + 2 b ) + b2
… as per the theorem.  ( Note the “bee” and “two bee” in the moral. )

Also, “If you’d like to put it more simply”, alludes to the historically primitive character of geometry vis a vis algebra, which is somewhat more abstract. So Euclid’s book II may be construed as simpler than modern algebraic notation, being put in more primitive, or “simpler” terms.

Well, Heath was 40 years after ALICE, but I think that the english version of basic Euclid was probably already a matter of tradition by this point, although I really couldn’t say, of course.


All the above I’ve been carrying around for some years, but in anticipation of writing this up, I was spurred to check the original greek of Proposition 6:

… so feast your eyes! heh heh. Well, I spent some time on it but I don’t want to belabor the issue. You can note “tetragono” ( square ) and “gramme” ( line ) and “orthogonion” (!) orthogonal.

The thing I want to say about it is that the “garden path” does not exist in the greek, and the syntax seems to be a little more technical. “apo” means the base of a square, so it announces the specification of such, which is terminated by “tetragono” so that’s it, with no ambiguity.

The “garden path” clause specifies the base and height of a rectangle delineated by TO [ base spec. ] KAI [ height spec. ]

The base spec. is “upo tes holos syn te proskeimene” ( base of the whole with the extension ) and the height spec is “proskeimenes periexomenon orthogonion” ( encompassing orthogonal extension ? ) then comes “meta” = “together with” … ( the square on the half ) so there seems to be a more elaborate system of “terms of art” in play.

Consider a Spherical Square …

… of course I allude to the punchline, “Consider a spherical horse”, although there is such a thing as a spherical square. Namely a quadrilateral drawn on the surface of a sphere with equal sides and equal angles. Many years ago I was drawn into contemplation of this by a statement in Peterson’s Field Guide to the Stars and Planets, by Donald Menzel, that “The sky contains approximately 40,000 square degrees …” as excerpted here:
A more accurate value is given by 4Pi/(Pi/180) = 41252.96125, but this is still approximate, as it does not allow for the curvature of the “square degree” itself. So I endeavored to derive an exact value for the area of a spherical square degree, and did so, as recorded in this note:

I was quite pleased with this result at the time, as it was simple and even had a certain elegance, and in fact I had difficulty reproducing the calculation before I rediscovered my note. I did remember the result verbally: “four arcsine of sine squared alpha over two”, where alpha is the measure of the square across the center.

Applying this formula gives the value of 41254.00842 for “square degrees in the sky”, differing by about 1 in the units place from the linear approximation.

I recall that I knew at the time that the formula was “four arcsine of tan squared beta over two”, where beta is a side of the square. This gives 41250.86683, undershooting the linear approx. by about twice the amount that the alpha formula overshoots. I’m sure there’s a story there, but let’s move on.

I was recently drawn into contemplation of the spherical square by a NYT Quote-acrostic based on a quote by Calvin Trillin, which alluded to the infamous Indiana Pi Bill ( misattributing it to Texas ). The cited Wiki article offered in explanation of the bill’s content, a crank mathematical screed, that “it is clear that the assertion is simply that the area of a circle is the same as that of a square with the same perimeter”

I started wondering if this might not be true for some spherical square and circle. I finally realized that it is true of the “degenerate” spherical square whose sides are quadrants of an equator, but too late! … I was back on the case.

In my calculation, I had retreated to the comfort zone of spherical coordinates, and resorted to integral techniques which, while well within the bounds of Introductory Calculus, were less than straightforward.

I knew that the problem could be solved by methods of Spherical Trigonometry, but I didn’t see that these were any simpler.

Well, this time around I came up with a simple construction that makes it a “one step problem” , applying the rule of “spherical excess” for the area of a spherical triangle:

Well OK, a “two step problem”, we still have to solve for theta. This is an easy application of Napier’s rules for right spherical triangles applied to either the light blue or lavender auxillary diagram in the figure, and giving

or …

Additionally, we may solve for the distances x and y, which are the half-lengths of the sides of the spherical rectangle. Applying Napiers rules, as above, we get:

Note that when psi = chi and equivalently x = y, we have the identity

… as alluded to above.

Notes on Close Encounters

The passage of 2014 DX110 is just the latest in a series of  near earth encounters that have been observed, and it raises the question, what is the likelihood of one of these actually hitting the earth, supposing a continuing sequence of them? Of course this has been going on for a long time, and they DO occasionally hit the earth, but how might we model these encounters statistically?

Let’s take the lunar orbit standard and suppose that a medium sized asteroid passes within “the lunar hoop” perpendicular to its relative direction of motion, say once a year. Then if each of these objects has a constant probability per unit area of passing through a particular point in the hoop, we need only compare the cross section of the earth to the size of the hoop to estimate the fraction of these encounters that result in a collison with the earth, and this fraction is just the square ratio of the earth’s radius to the moon’s orbit, or about 1/602 = 1/3600 .

Gravitational Focusing

But then we might ask, what about the gravitational attraction of the earth? Wouldn’t it tend to “focus” objects passing nearby and increase the effective collision radius?

I think the answer is “somewhat”, but there is a simple treatment of the problem that is interesting more for the understanding it affords than any modification of such an estimate.

Let’s directly apply conservation of energy and angular momentum to get a simple and exact answer to the idealized “two body problem” of a small object passing by the earth in an inertial frame of reference. This is actually a reasonable approximation, I think.

Conservation of energy is expressed by

1/2 v2 – MG/r = constant

and conservation of momentum is expressed by

v rperp = constant

In particular, if an object approaches the earth from a great distance with speed v1 and s is its projected straight line distance of closest approach ( measured to the center of the earth, ) then its angular momentum wrt the center of the earth is v1 s , and if r0 is the actual hyperbolic distance of closest approach, where it has velocity v0, we must have

v1 s = v0 r0

A hyperbolic Interlude

The situation is illustrated in the following diagram, which shows the hyperbolic path of an object passing the earth ( in blue ) and the asymptotes of the hyperbola in gray:

The diagram has been drawn to scale for a hyperbola with parameters a=7, b=24, which gives the focal distance of sqrt( 72+242 ) = 25, and hence the rational eccentricity of 25/7 . It’s part of a sequence of pythagorean triples with the lowest element being the sequence of odd integers beginning with three, and the two larger elements differing by 1 ..

3 4 5
5 12 13
7 24 25
9 40 41

This is of purely heuristic interest, but it gives us a sequence of “rationalized” hyperbolas with linearly increasing eccentricity.

Note that our ratio of interest, s/r0 , is also rational. 7 24 25 is the 3rd entry in the list, and the generalization holds that this ratio is (n+1)/n for the nth entry. We’ll come back to this …

Dynamics continued

We can simplify our notation by an appropriate choice of units. We note that the formula for conservation of energy, with constant set to zero :

1/2 ve2 – MG/re = 0


1/2 ve2 = MG/re

… is the defining equation for the escape velocity from the surface of the earth, and we can express the potential energy in terms of the escape velocity:

MG/r0 = 1/2 ( re / r0 ) ve2

Then if we measure r in units of re, and v in units of ve, Conservation of energy between the far motion at velocity v1 and zero potential energy, and the close approach v0 can be simply expressed as:

v12 = v02 – 1/r0

Note we have “multiplied through by 2” to get rid of the factor of 1/2 in each term.

Now we can use the equation for conservation of angular momentum to express v0 in terms of v1, or vice versa, and with some elementary rearrangement of terms we get these two equations for s in terms of r0 and v1, or r0 and v0 :

s2 = r02 ( 1 + 1/ ( r0 v12 ) ) = r02 / ( 1 – 1/ ( r0 v02 ) )

Notice that for large v1, i.e. much greater than earth escape velocity ( implying large v0 as well,) we have r0 = s, so the object just zooms right by. Otherwise, we have an easy formula for determining s in terms of r0 and v1 or v0 .

Impact criterion

If we set r0 = 1, i.e. the radius of the earth, we have a very simple formula for the apparent or effective size of the earth in terms of v1, expressed in units of earth escape velocity.

s = sqrt( 1 + v1-2 )

For example, for the earth to “appear” twice its diameter, the incoming speed would have to be 1/sqrt(3) = .58 , or 58% of earth escape velocity.

The case of 2014 DX110

Well what about 2014 DX110 ? It’s stated that its flyby speed was 33000 mph or 14.7 km/sec, compared to 11.2 km/sec, so v0 = 1.3125

It came within 54 earth radii at close approach meaning r0 = 54, and we can use the v0 version of our formula to get

s = 54 / sqrt( 1 – 1/ ( 54 x 1.31252 ) ) = 54 x 1.0054

s/r0 = 1.0054

… and finally

To find a rationalized hyperbola which “approaches” this case, we note ( as per above ) that (n+1)/n = 1 + 1/n, so we want n = 1/0.0054 ~= 185, so we take the 185th in the sequence of Pythagorean triples:

371 68820 68821

and eccentricty e = 68821/371 ~= 185.5

Sol 550 : A MAHLI Panorama

Here is a hotlinked sol 549 Navcam image of a small ridge that captured the interest of Curiosity. I think it’s about 2 feet long:

On sol 550 it took a series of MAHLI closeups, and also this Navcam image of the robotic arm ( with the MAHLI on it ) during the process of doing so:

BTW, here is an image which shows the arm in the stowed position, prelaunch:

There are three joints on the arm itself, with all the axes parallel, so it moves in one plane with motion of these three joints. There’s a vertical “post” that carries the arm assembly and allows it to swing out and away from the base. You can see it projecting downward near the left front wheel.

You can also see the name plate on top of the upper beam of the stowed arm, and the forward projecting arm joint at the right side of the rover. These are visible in many of the downward looking Navcam and Mastcam images, including the first image of this post.

If I had understood all this correctly, I never would have made that blooper in my SPOT OF BOTHER post!

Anyway, the MAHLI took a series of images of this little ridge on sol 550, and they appeared to be suitable for a panorama, so I tried it. It worked pretty well, but it is not a true panorama. The images on the left were part of a “scan” where the MAHLI changed position but kept pointing in the same direction, more or less. Then the images at the right were made by swinging the POV outward. It stitches down the centerline pretty well, but you can see some doubling up of particular features above and below. Well, a nice result and it shows incredible detail. As always, CLICK TO ENLARGE!

Mandelbrot II : Julia Islands

I mentioned the image gallery section of the Wiki Mandelbrot Set article as the starting point for my renewed interest in this topic. It is based on a zoom sequence terminating on a “Julia Island”, which is a component of the Mandelbrot set resembling a “Julia set”, which is based on the actual sequence of an iteration rather than the in/out criterion of the Mandelbrot set. So there’s a little math magic in this resemblance.

Here is my recreation of the zoom, based on the linked tool I mentioned, which by no means outstrips the images in the article ( which can be expanded, ) but it does go a little deeper, and it keeps the Julia Island location at center. This was easy to do by working backwards from the deepest image and keeping the same coordinates by simply changing the scale in the URL field.

Here’s a “hot link” of the lowest level of the Wiki zoom :Wikipedia commons( To see its full glory visit the site. ) It roughly corresponds to the antepenultimate frame of the animated gif. The commentary in the main article notes that there is a “satellite” ( i.e. “mini-Mandelbrot set” ) at the center of the island structure that is “too small to see at this magnification”. Just visible there is the large square structure in the center of the last frame of the gif, where you can almost see the satellite at the center of it, as a tiny dark speck.

This conclusion is supported, if not confirmed, by perusal of a nearby and entirely analogous structure, which is a little bit larger. Here’s a slower zoom, taking 1/2 per frame instead of 1/10, showing a descent through the analogous square structure to reveal the Mandelbrot satellite.

Note the emergence of a new “sea” of similar spirals and structures as part of the INFINITE descent. It’s fascinating and incredible and, like the song says, “a little bit frightening.”