Rosetta - Post Separation Ops at Comet 67P C-G, November 14, 2014 - |
Rosetta - Post Separation Ops at Comet 67P C-G, November 14, 2014 - |
Nov 14 2014, 05:17 AM
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#1
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Senior Member Group: Members Posts: 2089 Joined: 13-February 10 From: Ontario Member No.: 5221 |
I think I heard it mentioned during the press conference today, (I can't find it now), about Rosetta itself possibly landing eventually, similar to what NEAR did at the end of the main mission at Eros? Since it's not like there's anywhere else to go with the remaining delta-v left by the end of 2015, and sunlight levels and activity starting to drop after perihelion, and the low gravity makes the difference between orbiting and 'landing' trivial. The whole thing would weigh a kilo or two, right?
Obviously there's a few more pressing concerns right now, but it's something to eventually think about. |
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Nov 14 2014, 10:39 AM
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#2
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Senior Member Group: Members Posts: 2346 Joined: 7-December 12 Member No.: 6780 |
At some point Rosetta will run out of propellant for orbit corrections.
The mass of Rosetta at regular EOM should be 2900 kg - 660 kg - 1060 kg - 100 kg = 1080 kg. (Start weight - propellant - oxidizer - mass of Philae, from here) The surface gravity of the nucleus at 2 km distance from the center of mass should be g = GM/r² = (6.672e-11 Nm²/kg² * 1e13 kg) / (2000 m)² = 1.668e-4 N/kg. The weight of 1080 kg is F = m a = 1080 kg * 1.668e-4 N/kg = 0.18 N. 0.18 N is the force of a mass of 18.37 grams in 9.80665 m/s² gravity. The actual weight of Rosetta would be a little less due to inertial forces by the rotation of the nucleus. To be more precise: The centripetal force for a radius of 2000 m and a rotation period of 12.7 h = 45,720 s is m r 4 pi² / T² = 1080 kg * 2000 m * pi² / (45,720 s)² = 0.0102 N. So we are at 0.18 N - 0.01 N = 0.17 N for Rosetta's EOM weight, corresponding to the weight of a little more than 17 grams on Earth. That's a model estimate, and may differ, depending on the actual landing coordinates. |
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