Pluto's Expanding Atmosphere Confounds Researchers, Pluto Atomosphere |
Pluto's Expanding Atmosphere Confounds Researchers, Pluto Atomosphere |
Apr 19 2011, 08:26 PM
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Junior Member Group: Members Posts: 21 Joined: 4-November 10 Member No.: 5509 |
Pluto's Expanding Atmosphere Confounds Researchers
http://news.sciencemag.org/sciencenow/2011...onf.html?ref=hp Could these be evidence of geyers like on Triton? |
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Apr 29 2011, 02:57 AM
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Senior Member Group: Members Posts: 1018 Joined: 29-November 05 From: Seattle, WA, USA Member No.: 590 |
So Andy and I worked through this offline, and we now get the same result through different means, so I'm pretty confident this is correct.
The first step is to get the best numbers possible for the masses of Pluto and Charon and for the distance between them. It turns out there's some recent work using the orbits of Nix and Hydra to do exactly that: http://iopscience.iop.org/1538-3881/132/1/...1_132_1_290.pdf Pluto: 1.28726E+22 kg Charon: 1.69741E+21 kg center-to-center Distance: 1.9571E+07 m My results are: Charon-center to L1: 5.9716E+06 m Charon-center to L2: 7.4611E+06 m Andy's numbers differ from mine by just a few hundred meters now. In this case, his math and Excel were correct from the start; I was the one with the math error. In case anyone's interested, here's how to compute it: let m be the mass of the smaller body divided by the mass of the larger one and let h be the distance from the center of the smaller body to the L1 point as a fraction of the distance from the smaller body to to larger one. Then (1+m)*h^5 - (3+2*m)*h^4 + (3+m)*h^3 - m*h^2 + 2*m*h - m = 0 There are a variety of ways to find h given m. (I used Newton's method, but you can brute force it too.) For L2, the only change is that the quartic and linear terms change sign, like so: (1+m)*h^5 + (3+2*m)*h^4 + (3+m)*h^3 - m*h^2 - 2*m*h - m = 0 I hunted and hunted to find something that laid it out like this, but everything seemed focused on the harder problem of finding L4 and L5 and proving their stability. --Greg |
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May 1 2011, 07:01 PM
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#3
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Member Group: Members Posts: 593 Joined: 20-April 05 Member No.: 279 |
Never too keen on terms to the fifth power, and always keen to ditch "G" my route was:
Masses: Pluto = Ma Charon = Mb Both = Mt Distances: Pluto - Charon = R Pluto - Barycentre = Ra = R*Mb/Mt Charon - Barycentre = Rb = R*Ma/Mt Barycentre to L1 point = X X needs solving in the L1 equation: Mb/(Rb-X)^2 + Mt*X/R^3 - Ma/(Ra+X)^2 = 0 The three terms here are the force of Charon plus the centripetal force at L1, minus the force of Pluto. The easiest way to solve this (for me) was to parametrically define X and iterate towards "= 0" in Excel. Andy |
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