67P/Churyumov-Gerasimenko. A Single Body That’s Been Stretched- Part 15.



On 5th December 2014, the ESA Rosetta Blog published ‘Cometwatch 2nd December’. The photo, taken on the 2nd, showed the candidate landing site known as Site A and had the following commentary:

“The cliff walls that drop down onto this plateau seem to show slightly brighter sections, perhaps reflecting compositional differences, or fresher material that has yet to be degraded by exposure to the space environment.”

Rock C accounts for perhaps 20% of those brighter sections and certainly could be considered as fresher material. Though the manner by which it came to be exposed may come as a surprise.


In Part 14, two giant monoliths dubbed rock A and rock B were shown to have been broken free at the ‘shear line’ when the head lobe tore away from the body and rose to its present position. These two rocks then drifted 170 metres across the crater floor of Site A before landing and tipping on their sides. The crucial aspect of this analysis was that their 170-metre displacement was parallel to the rotation plane of the comet. This meant that, given the present positions of the two rocks, one could trace a line along the rotation plane in the forward direction of rotation and discover where they had come from.

Now we’re going to apply the same logic to rock C, the third of at least four rocks that drifted across Site A. It’s referred to as a rock, for brevity, but sometimes as a monolith which is far more apt for a rock of this size. And of course, we know its composition is not like that of a terrestrial rock. As with rocks A and B, the process is as follows.

1) Identify the monolith.

2) Trace a line from its current position, along the rotation plane in the forward direction- actually, this will be the corridor, the width of which is roughly the length of rock C.

3) Look for a match between the shape of the monolith and the shape of ‘rock’ formations along the corridor.

4) Check to see if it flipped over when it landed and thus identify the base i.e. the correct orientation for seating it back in position.

But it has to be said, rock C isn’t as neat as rock A as we’ll soon see. That doesn’t mean we won’t get a definitive fit. It’ll just take some explaining. For a start, most photos show it as being spear head shaped but one aerial view shows the spear point as being squared-off. However this aerial photo is much grainier and subject to white-out. The spear shaped examples are much clearer and closer up and there are at least four different ones, taken from different angles with different lighting. So we’ll assume it’s a spear shape. In addition to this, the dotting tool is playing up, occasionally smudging the dots. They also have to be small dots for this post because they are delineating very fine features and larger dots obscure the very thing you’re looking out for. Everyone will need to zoom right in to see these finer points, on almost every photo.

So we’ve already identified the monolith in the header photo. The 50 million kg mass estimate is based on a judgement of its dimensions based on roughly known dimensions of site A. Rock C’s dimensions are about 200m x 30m x 25m. Multiplying by the 470kg per cu metre estimate for density we get about 70 million kg. Seeing as it tapers considerably, I settled on 50 million kg as a conservative estimate. The photo below shows rock C’s position on the comet and gives an idea of how far it floated.


‘A’ denotes Philae candidate Landing Site A. It’s the flat crater floor. Missing Slab A comprises the entire crater, out to the orange-dotted rim and across to the terracotta shear line. If you haven’t yet read about Missing Slab A (Part 9), its importance for rock C will soon become apparent.

Yellow triads- rock C (lower); its seating position (upper).

Orange- Missing Slab A perimeter.

Terracotta- the shear line where the rim of the head lobe used to sit. It actually kisses the two yellow dots of the rock C seating position but is left short so as not to confuse.

Here is rock C’s rotation plane corridor traced back along its path which is the same as saying it’s a corridor traced forward along the comet’s rotation plane. This is from the aerial photo:


r denotes rock C; s denotes its seating position.

Yellow- same dot positions as yellow dots in header photo.

Big dark blue dots- rotation plane corridor for rock C.

Bright green- direction of rotation of comet along the blue lines.

Terracotta- shear line where rim of head lobe sheared from the body. (All other annotations relate to rocks A and B from Part 14).

Now we can look along the corridor for a match between our spear shaped monolith and any spear shaped formation in the crater rim. And there’s…nothing.

But there is what appears to be a very good fit to the spear shape that’s displaced one rock C length towards the shear line. It’s marked ‘s’, with the same three yellow dots as in the header photo. Its pointed end is just touching the rotation plane corridor so that’s how we know it’s displaced by one rock C length from its corridor. That would be a 200-250 metre displacement. This seating position doesn’t look very promising in the aerial photo, being grainy and whited out along the left hand edge as you look down but it does show some undulations which will be worth revisiting. Anyway, we know from the header photo that there’s quite a good shape match. Also, the length of the rock is 8% longer than the seating position as they are viewed in the header photo. It’s less foreshortened due to sitting up on the slope so this is a very good fit, lengthwise.

In fact, this potential seating position is kissing both the shear line and rock B’s seating position (see part 14 for the seating position of the rock A/B formation). That’s intriguing in itself because it’s at the shear line where you’d expect most disruption and if rocks A and B were right next door, it makes this site all the more likely to have suffered some upheaval as well.

But we can’t just say rock C must have propelled itself 250 metres at right angles to the rotation plane before deciding to do a 90° turn and fly along it. We have to have a legitimate scenario that allows for this apparent behaviour. The following is a technical explanation of how this might have happened. It’s fairly long and has no pretty, dotted pictures. If you’d like to see the matches first, along with with a multitude of pictures, by all means scroll to “THE ROCK C MATCHES”. However, it would be as well to come back and read this explanation for the rotation plane drift, as this will serve as a technical grounding for Part 17, which will deal with rock D. Other future parts to do with the translational displacement of the comet’s z axis of rotation during the head lobe rise will also rely on an understanding of all the elements laid out in the next few paragraphs. This explanation also strengthens the case for the missing slabs (Part 9).

So, how can we explain rock C’s rotation plane drift? This is where we have to look at every scrap of evidence available, including the fact that all these fragments are hypothesised to have been dropping off the underside of missing slab A when it broke from the shear line and hinged up along the back of Site A (Part 9).

If rock C really was seated at the shear line in the suggested position it means that it was displaced about three times further down the comet’s rotation line than rock A, about 500 metres or more as opposed to rock A’s estimated 170 metres. Since it’s already been established that the comet rotated underneath the rock fragments at a constant rate as they “tried in vain to keep up” (Part 14), it means that if one rock was displaced three times further, it had to have been ‘airborne’ for three times longer. In order for it to stay up longer it would have to drop from a higher altitude. In other words, the translational, horizontal speed along the rotation plane was constant for all rocks but if some dropped from higher up they got faster and faster as they accelerated for longer in the vertical axis. This is the same phenomenon as when a stone is thrown horizontally off a cliff. Ignoring air resistance, the stone maintains the constant horizontal speed throughout the drop while the vertical speed increases at a rate of about ten metres per second for every second of the drop. For these fragments floating across site A, the speed increase is in the order of 0.1 millimeters per second during each second of the drop so they had long drop times.

In the case of fragments dropping off the underside of slab A as it lifted up at the shear line and hinged at the back of Site A, each fragment dropped randomly from a different height, stayed up for a unique length of time and attained a unique final vertical speed that corresponded to that time window. But all fragments drifted back horizontally along the rotation plane at the same speed.

If rock C stayed up for three times longer than rock A, it had to have dropped from a height that was nine times greater than rock A’s drop height. This is because the distance travelled by a rock under a given acceleration is proportional to the square of the time period that it is in free fall. This is according to the law S= 1/2at^2 where S is the distance dropped (height of release), a is the comet’s acceleration due to gravity and t is the time in free fall. 1/2a is the same for all rocks so there is a neat relationship between drop height S and t squared. Therefore, if rock C was in free fall or ‘airborne’ for three times longer than rock A, it had to have been released at a height that was 3^2 or “three squared” higher than rock A. That’s why it would have been released from about nine times higher.

If it was released from nine times higher, it means that the release occurred when the slab had hinged much further and higher than the point at which it had released rock A. This in turn means that it would have translated rock C across the comet’s rotation plane and at right angles to it as it hinged away from the shear line. This is because the shear line and the hinge happen to be roughly parallel to the rotation plane. The hinging process must by definition have transported any attached fragments across the rotation plane as the slab rose.

Only after its release would rock C have then been free to undergo the apparent backwards travel along the rotation line due to the comet rotating underneath it. If the release was nine times higher than rock A it probably would have been transported a significant distance across the rotation plane on the underside of slab A. Then it would have been released and traveled backwards along the displaced corridor, the one that’s plotted in blue dots above. Viewed from above, rock C would be floating along the corridor at a constant speed. Viewed from the side, it would be dropping like the stone thrown from the cliff: starting high and fairly horizontal then curving increasingly towards the vertical as the influence of the comet’s gravity sped it up more and more.

For rock C to have been released in such a way as to float up range along its displaced rotation plane corridor and land where it is now located, slab A would have been tilted quite high as it made the release. Provisional calculations suggest 45°-50°, based on the displacement of rock C’s corridor and the width of slab A but these are just informed guesses because their is no accurate distance scale for these features. If slab A’s width was 750 metres then it would have released rock C at a height of around 550-600 metres if angled at 45°-50°. And if, as we established above, rock A was released from one ninth of this height, this would imply it was released at a height of around 60 metres. That figures because, doing the trig, slab A would have hinged only 4.5° for that 60 metres and translated rock A a mere 2 or 3 metres across the rotation plane. Rock A did indeed stay within its rotation plane corridor after release.

So we have arrived at a scenario to explain rock C’s displacement at right angles across the rotation plane, its release at ~550 metres altitude and subsequent drift along the displaced corridor. However, this isn’t just a ‘what if’ juggling around of rocks. The 9-fold increase in release height of rock C was needed to explain the fact that it had ended up three times further displaced down the rotation plane than rock A. And that height of 550 metres involved slab A needing to be angled at around 50 degrees. This is the angle that just so happens to transport rock C across the rotation plane by the required 250 metres. Lastly, it constrains rock A (and B) to be released low down and remain in its original rotation plane corridor. The complex behaviour of three different rocks perfectly complement each other in a single episode while obeying the laws of gravity and trigonometry. The drop times would be about 55 minutes for rock C and 18 minutes for rock A if the standard acceleration due to gravity for the comet is used. This is in keeping with the Philae lander bounce distances and times. However, the fast rotation on release of the head lobe would reduce the effective acceleration by a half or more even though Site A was near the spin axis. If it’s reduced by a half then the drop times are increased by root 2 to 78 and 26 minutes.


Now lets look at the orientation of rock C to establish whether it flipped over. Rocks A and B flipped over 90° because they were notionally cuboid in shape (see Part 14 for mechanism of the flip). Since rock C is spear shaped it has two narrower, blade-like edges. If it landed on one of these it would probably topple either way, possibly explaining any 180° flip. However there are numerous other explanations for a 180° flip.

The only surface we can analyse as a possible seating surface is the side that’s facing upwards. If the monolith did indeed come from the suggested seating position it’s a 50:50 chance it was the upper spear-shaped side that we see and a 50:50 chance it was the lower spear-shaped side that’s obscured. So we have no choice but to see at least whether this upper side fits.

Here are two photos with some initial match points which look promising. Each coloured dot on rock C corresponds to the same coloured dot in its suggested seating position.


In the following close-ups you can see what appears to be the triangular mark left on the rock by the triangular ridge that the pointed end was sitting on, or rather, overhanging slightly. You can see transverse lines between the triangle marks on the rock. There may also be one or two transverse lines on the seating position in the corresponding place but it’s hard to tell (so they’re not annotated). The bright green dots show the rock’s pointed tip curving down and its seating position curving up. That’s the clincher for this rock matching this spot, like the sliced-through mini crater for rock A. The curve on the point of the rock is even more prominent in another photo further down.


Key (full zoom required):

Red dots- triangular formation.

Orange- transverse lines (smudged to yellow in places)

Green- curved-down end of rock and its curved-up seating position.

Rock C has well-defined undulations, which can be better discerned in this next view. It’s more of a side-on perspective. A yellow dot is placed under the rock. This is the photo that shows the pointed tip drooping down markedly and to the same degree as its seating position curves up (not annotated). The curve shape is similar too.

You can see corresponding, gentle troughs and ridges on the seating position and they follow the correct transverse angles (the angles reverse though, because of flipping the rock 180° to seat it). However, this part of the seating position has some white-out exposure so it’s not easy to get a definitive match to these undulations. This photo shows the ridges in the seating position and the corresponding troughs on the rock, all in one frame (zooming required).

The aerial view with the rotation plane corridor also shows these transverse ridges on the seating position. They run across the top of the ‘s’ and through the bottom of it. They also continue to the right of it and correspond quite well in angle and length to the dotted lines in the larger depiction, above. Here’s the aerial view, reproduced. The ridges are the lighter sections with slight shadows above them. They are left unannotated due to dots detracting from the ability to see these faint features:

Finally, on the header photo at the near end of the seating position there is a dip in the form of a right angle that looks as if it’s part of a square. Here’s the header photo, reproduced:

Three red dots- these form the right angle.

Orange dot- completes the square.

It does seem to form a square. It looks as though rock C would have sat on that obvious right angle and might harbour some square mark on its underside (now its top side) along that edge. The trouble is, that part of the rock isn’t visible in any of the photos shown here so far. So in pursuit of an extra match, I trawled all the photos for a view of this side of rock C. I found just one candidate. It doesn’t just show a square mark at the base but a corresponding square chunk carved out from top to bottom as if some giant square post had been driven through that side of rock C and on down into the comet surface. So perhaps that should be the clincher instead:


Red- the right angled chunk taken out of the side of rock C.

Yellow- rock C is the upper triad. (The lower two triads, from part 14, are rock A plus its seating position).

Terracotta- the shear line.


As is the case for rocks A and B, it is difficult to ascribe the displacement of rock C to any process other than being lifted a substantial distance from the comet and then being dropped. This is because it ended up 500 metres or more from its seating position and is now found to be along the rotation plane from that position, albeit displaced sideways by 250 metres. As with rocks A and B, there is no evidence of a pit where sublimated gases had built up under rock C causing a pressure build-up. Moreover such a pressure build-up is hard to explain, given the 70-80% porosity of the comet (noted as a problem in Thomas et al, January 2015). If it did occur, it would have needed to catapult a 50 million kg monolith to 600 metres altitude, leaving no collateral damage at all at the seating position. And as before, there is no evidence of a meteor impact at the seating position.

In conclusion, the only reasonable explanation for rock C’s behaviour is a massive shearing event that provided the heavy lifting.

In the next post, Part 16, rather than looking at rock D we shall return to the reality of simple, obvious matches between head and body. This is because a wealth of recent images show new matches while confirming old ones. This evidence has been sidelined while we’ve been dealing with more technical issues for the last 10 Parts and so it needs to get an airing. Rock D will be the subject of Part 17.

Copyright ESA/Rosetta/NAVCAM – CC BY-SA IGO 3.0



2 thoughts on “67P/Churyumov-Gerasimenko. A Single Body That’s Been Stretched- Part 15.

  1. After you gave such obvious clues, that monolith was really the only candidate. I think in many ways it is a more obvious match, as rock A and rock B are so close together in many images they appear to be one monolith. Every previous time I had seen rock C, it appeared to be part of the site A fracture. As soon as you look at a couple of close ups to work out its 3D profile, it’s separate monolithic attributes become obvious. It’s flip over is helpful as matches can be seen in plan mode, often in the same image, and the mirror attributes are a proof of principle for the head-body matches in your first few posts. I don’t know how many times you have to find damning evidence that proves stretch theory in different ways, but this is yet another proof independent of the multitude of previous proofs.


    • Marco
      I think the OSIRIS photos will show up many more smaller fragments at the back of Site A. The only one you can see for sure is rock D but there are others that look to be separate monoliths. The NAVCAM resolution doesn’t quite confirm this though.

      You say, “I don’t know how many times you have to find damning evidence that proves stretch theory in different ways, but this is yet another proof independent of the multitude of previous proofs.”

      Part 16 will be yet another independent proof, both in terms of corroborating the 22 independent matches from Parts 1 & 2 and new matches as more of Anuket becomes illuminated.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s