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Post by jonjel on Nov 14, 2014 13:03:31 GMT 1
A fantastic bit of science and engineering to land the probe on the surface of the comet, even if it did fall over!
But I have been puzzling a bit. We are told that the 'weight' of the probe now it is on the surface is only about a gram. So three parts of bugger all gravity. Yet the space craft is in orbit around the comet, passing within I think about 10 miles.
I understand the principles of orbit, a bit like a rifle bullet falling to ground but in effect continually falling round the corner. But with so little gravity does that mean the orbit is remarkably slow, or what?
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Post by principled on Nov 14, 2014 20:14:44 GMT 1
Good point Jonjel. I thought "Let's do some research", and came across what I thought was an ideal research paper (calculating the orbit around Eros)....until I read the abstract. I am now completely confused! The problem seems to be not only the lack of gravity but also the irregular shape and motion of the body.
P
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Post by alancalverd on Nov 17, 2014 18:20:01 GMT 1
Orbital speed = (GM/r)0.5 where M is the mass of the comet (assuming it is much larger than the mass of the orbiter). In this case M is about 1013kilogram and G = 6.7 × 10-11 m3 kg-1 s-2 so at a radius of 16 km the probe is travelling at about 0.2 m/s. That should make landing dead easy, even compared with landing a helicopter in a slight breeze!
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Post by jonjel on Nov 19, 2014 12:37:06 GMT 1
Alan. The maths is beyond my simple brain, but I am sure it is correct.
You said 'the probe' is travelling at 0.2M/sec.
Did you mean the probe, or the 'mother ship'? That would make each orbit very slow indeed, 1Km would take around 1.3 hours.
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