jonesdave116
Philosopher
- Joined
- Feb 26, 2015
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To close, why not take up Tom Bridgman's challenge? In Electric Comets III: Mass vs. Charge.
Funny you should mention that!
So, from Tom Bridgman's blog (http://dealingwithcreationisminastr...14/12/electric-comets-iii-mass-vs-charge.html), we have Qq = 4*Pi*epsilon0*G*m*V*(density - density1), where density is the density of the comet assumed by EU, given a conservative value here of 2000 kg m^3, and density1 is the measured density, 470 kg m^3.
Q = charge on Rosetta, q = charge on 67P, epsilon0 = vacuum permittivity = 8.854 187 817... × 10^−12 F/m, G = gravitational constant = 6.67384 × 10-11 m3 kg^-1 s^-2, m = mass of Rosetta = ~1500 kg & V = volume of the comet = ~2.14 x 10^10 m^3.
Plugging in the figures gives Qq = 3.72 x 10^-4 Coulombs.
Now we have to make a couple of estimates for the capacitance of Rosetta and 67P. This is based on calculations done by Prof. Harvey Rutt, FREng, FInstP, FOSA, FIEE, SMIEEE, CEng, CPhys, and posted in the Rosetta blog. For 67p this comes out at 0.3 microfarads, based on a 2.5 km spherical capacitor. As Harvey said, this is an overestimate. In fact a 2.0 km radius would have been an overestimate, but we'll stick with this figure.
Rosetta is somewhat more difficult, but was based by Harvey on a 3m diameter sphere, and two 14m x 2m rectangles. This gives 750 picofarads, rounded up to 1 nanofarad. I've checked these figures myself, and they look reasonable.
Now, in a post on the Rosetta blog we were told that the voltage on Rosetta was between -5V & -10V. We'll work with -10V.
As Q (charge) = C (capacitance) x V (voltage), we can now work from there to see what charges and voltages are implied on the comet.
From Q = CV, the charge on Rosetta will be -10V x 1 nanofarad = -10^-8 Coulombs. We know that Qq = 3.72 x 10^-4 Coulombs, therefore q (the charge on 67P) = 3.72 x 10^-4 Coulombs/-10^-8C = -10^4 Coulombs.
This also enables us, from Q = CV, to work out the voltage that would be carried by 67P; rearranging gives V = Q/C, = -10^4C/3.0 X 10^-7F = ~33 gigavolts. As Harvey said, I think we might have noticed that! He also went on to state that a weak 1 amp discharge would discharge this insane voltage in about 3 hours, and the electrostatic force would disappear.
So, I would say that the EU 'theorists' may well have done these calculations themselves, realised their implications, and kept stum about it to their acolytes, on the basis that their main raison d'etre is to sell books and DVDs, and this most certainly wouldn't help. Best just to muddy the waters with talk of Langmuir sheaths and double layers and non-linearities to cover the fact
that they meant 'electrostatic repulsion', which is easily shown to be BS.
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