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ESA indicates that the first bounce lasted two hours and reached a height of 1 km. With the extremely weak surface gravity and low escape velocity of the body (< 1 m/s), and other publicly available information regarding the comet's mass properties and the landing site location, can we estimate how close the lander came to drifting away into space?

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  • $\begingroup$ You'll need an answer to this question first. $\endgroup$ – Jerard Puckett Nov 13 '14 at 15:10
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    $\begingroup$ Also, from what was said at the 20:00 conference, it wasn't so much a bounce... Philae landed and anchored the drills correctly, but the harpoons didn't fire. As it began spinning down its flywheel, the whole lander was jerked off its moorings and sent flying, spinning in the opposite direction. If it was just a bounce, we knew it was to land at 1m/s and followed the procedure closely, so the bounce-back would be very unlikely to be more than 1m/s. But in this situation the "flywheel-propelled launch" speed is indeterminate. $\endgroup$ – SF. Nov 13 '14 at 15:34
  • $\begingroup$ @SF. Interesting. It ought to be possible to get some idea of the "launch" speed from the time aloft, but the nonspherical gravity field is a complication. $\endgroup$ – pericynthion Nov 13 '14 at 15:45
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    $\begingroup$ @pericynthion: Knowing the new landing site and time to reach it would give us a rather good estimate (the horizontal component of the speed), if the distance was reached in a single bounce. But it was in two, so the best we'll get is the average speed between the two. $\endgroup$ – SF. Nov 13 '14 at 15:54
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    $\begingroup$ @TildalWave if it was equipped with an accelerometer capable of measuring in the 1E-6 m/s^2 range.. do you know whether that's the case? $\endgroup$ – pericynthion Nov 13 '14 at 19:52
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We can do a very rough back-of-the-envelope approximation, at least. Say 67P is a sphere 1.7 km in radius with mass of 1x1013 kg; Wolfram Alpha says surface gravity is ~2.3x10-4 m/s2 and escape velocity = 0.886 m/s. Gravity 1km above that (at periapsis) would be 9x10-5 m/s2.

I'm too lazy to do the calculus of a ballistic trajectory through that gravitational gradient so let's just use the average, 1.6x10-4 m/s2. If the first bounce is a 2-hour parabola, then periapsis is at 1 hour (3600s), vertical velocity is 0, and by v = v0 + at, initial velocity was therefore 0.576 m/s; periapsis works out pretty close to 1 km with that average acceleration as well, so that seems sane.

Running the same equations for just the small lobe of the comet produces similar results; the lower mass is somewhat cancelled by the lower starting altitude.

So this was a fairly close call - Philae may have taken off at about 2/3 of escape velocity!

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It was never close. Rosetta did not have escape velocity when it ejected Philae. So as long as Philae was ejected in the retrograde direction, Philae would not have escape velocity either. After the bounce, Philae would have even less energy due to landing gear attenuation and would therefore be even farther from escaping than it was pre-bounce.

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  • $\begingroup$ The dynamics aren't necessarily that simple, given the separation impulse (do we know what it was, as a vector quantity?) and the apparent reaction force from the flywheel spindown. $\endgroup$ – pericynthion Nov 15 '14 at 2:16
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    $\begingroup$ They are that simple. As long as the separation impulse was small -- which it must have been since Rosetta is still captured -- then nothing else matters. The only other thing that I can think of that might cause Philae to escape would be geyser impingement. $\endgroup$ – Erik Nov 15 '14 at 2:18
  • $\begingroup$ What about the flywheel? $\endgroup$ – Russell Borogove Nov 15 '14 at 3:39
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    $\begingroup$ The flywheel is onboard Philae -- how does that change Philae's energy relative to the comet? $\endgroup$ – Erik Nov 15 '14 at 12:23
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The new report says Philae was moving at 38 cm/s.

Using Russell's figure of 0.886 m/s as escape velocity, we have 43% of escape velocity.

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  • $\begingroup$ Either way, much too close for comfort! $\endgroup$ – a CVn Nov 14 '14 at 20:32

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