Meanwhile, I'll address some of the claims floating around about the measured mass and density of Comet 67P (Wikipedia). By measuring the motion of the Rosetta spacecraft around the comet, they have been able to estimate the mass of the nucleus at about 1e13 kg or 10 trillion kilograms. Current measurements suggest this value is accurate to within about 10%, so future estimates should fall in the 9-11 trillion kilogram range, with some smaller probability that they fall outside that range.

Rosetta Blog: Determining the Mass of Comet 67P/C-G

This will not be the final word on the comet mass as at various times during the mission, Rosetta will be moved to locations where more precise measurements are possible. Note that the standard model for comets will also expect some rate of mass loss.

Using that mass estimate and the measured volume of the comet constructed from imagery, we get a mean density of 0.4 gm/cm^3, significantly less than water and around what you expect for the 'dirty snowball' model.

When I saw this estimate, I considered the possibility that the Electric Universe crowd might claim that this low mass was an artifact of assuming that the attraction was purely gravitational and did not include contributions due to electrostatic repulsion which might occur in one of the many 'electric comet' models. Shortly after I started work on this analysis, I was notified that this was indeed being claimed by some Electric Comet supporters.

While Electric Universe supporters continue to make excuses, people with real training in physics can use such claims to determine physical parameters. So we ask the question:

**How much charge would that take to make a dense object appear to be a given lower mass object?**

First, we setup the force equation so the gravitational acceleration between the mass of the spacecraft, m, and the perceived mass of the comet, M', vs the real mass, M.

This equation balances the forces so 'real' masses M & m appear to have apparent masses, M' and m (assuming the spacecraft mass, m, is unchanged), if they are carrying charges Q and q, respectively. The fact that both force laws are inverse-square allows distance between the masses, r, to cancel out. This leaves:

Then we can re-arrange the terms to solve for the product of the charges Q & q:

Using the fact that mass is density times volume, we can recast this equation into a form more useful for exploration of our question:

Now we have an expression that can tell us about the charge on the comet nucleus and the spacecraft, based on other numbers which we can measure as well as some hypothetical 'real' densities. We don't obtain the actual value of the charge of the comet nucleus, but we now have a constraint that can be combined with other data to tell us more.

From the form of the equation itself, we see that for the apparent density to be less than the real density, Qq will always be positive, meaning that the charge on the spacecraft and comet must be of the same sign, both positive or both negative. This makes intuitive sense, as the electrostatic force will be repulsive in both cases. But if the charges are of opposite signs, the apparent density will be higher than the real density since now the two objects must be attracting each other through the electrostatic force.

Let's plug in some numbers. We have the 'apparent density' of the comet, based on the allegedly flawed assumption that gravity is the only important quantity for steering the Rosetta spacecraft.

We have a mass for the spacecraft, m=1230 kg (assuming the spacecraft is the dry mass defined in Wikipedia)

Apparent comet nucleus density = 0.4 gm/cm^3 = 400 kg/m^3

Combined with the mass, we can estimate the volume of the nucleus.

V = (1e13 kg / 400 kg/m^3) = 2.5e10 m^3

Let's assume the real density is closer to that of rocks, say 2 gm/cm^3 or 2000 kg/m^3. Then, using the SI units where epsilon0 = 8.854187817e-12 farads/m, we find

Qq = 4*3.14159* (8.854187817e-12 F/m ) (6.673e-11 N(m/kg)^2)*(1230 kg)* (2.5e10 m^3) (2000 - 400) kg/m^3

= 3.65e-4 F*N*m

We can also plot the charge constraints, examining the requirements for several different densities. In the graph below, we plot the spacecraft charge on the x-axis vs. the charge on the comet nucleus (y-axis) for several different values of 'real' density. Plotted logarithmically, different values of Qq form straight lines for the different density values.

While we don't know the charge on either the comet nucleus or the Rosetta spacecraft, we now know the product of these quantities. This provides us with options to examine.

One thing we should note is the surprisingly small value of this quantity. If the comet carried a charge of 1 coulomb, then the spacecraft would be charged to 3.65e-4 coulombs. You can charge a small capacitor to one coulomb. It isn't that much charge. If the comet were charged to 100 coulombs, then the spacecraft would be charged to 3.6e-6 coulombs, or 3 millionths of a coulomb. The higher the charge on the comet nucleus, the smaller the charge on the spacecraft must be to explain the density discrepancy Electric Comet supporters advocate. Similarly, we could place more charge on the spacecraft than the comet.

Of course, this claim has loads of other implications, none of which we've seen explored by the Electric Comet advocates or their 'theorists'. Let's examine some of these questions. We'll see if we get any real answers from Electric Comet supporters. Some of these I'll explore in a future post.

The comet nucleus is traveling through the solar wind, which is a plasma, free electrons, free protons, and a fraction of heavier ions.

- How much of this plasma does the comet nucleus intercept? How long would it take for the charge on the nucleus to neutralize? You need to compute or at least estimate the projected geometric area of the comet nucleus.
- Step it up a notch and consider not just how much of the solar wind will be intercepted by the comet's geometrical profile, but how much additional charged material is attracted to the nucleus from beyond this region due to the electrostatic attraction of the comet. Now how long will it take for the nucleus to neutralize charge? What happens then, and why?
- Is the amount of additional charge attracted (most likely solar protons) sufficient to explain the amount of OH and water production measured per the Thornhill model for the comet OH emission? (see Electric Comets II. Of Water & Ice).
- The constraint above means a very small amount of charge could be on comet as well as spacecraft. In the solar wind plasma, one might expect that non-uniformities in the solar wind might fluctuate substantially, perhaps even occasionally charging up objects with the OPPOSITE charge. If it manages to do that on only one of the objects in the comet/spacecraft system, this repulsion can suddenly become attraction. Therefore, in the electric comet model it is possible that the comet might suddenly appear MORE massive. With such small values of charge on the comet or spacecraft sufficient to explain the density discrepancy, it may certainly be possible for one of the objects to accumulate charge of the opposite sign and make the comet appear not just more massive, but significantly more massive. What navigational issues does this present for Rosetta?

- How would/could a net charge on the comet nucleus be maintained in interplanetary plasma?
- Where's the battery or generator that maintains the potential between the comet nucleus and the solar wind?

**Exercise for Readers**

It might be tempting for Electric Universe supporters to use the analysis above as evidence that planetary bodies might carry significant electric charge which alters our estimates of planetary masses. While the analysis above works easily for two bodies, I'll suggest it as an exercise for the reader to determine what happens for three or more bodies which will disrupt simplistic attempts to re-interpret solar system dynamics.

**Other Notes:**

- For another challenge for the geniuses of the Electric Universe, see my response to a comment from Lodaya.
- Rosetta Blog
- Plasma instrument on Rosetta
- A long-running and recently restarted Electric Comet discussion is ongoing over at the International Skeptics Forum.

**1) The irregular shape of the nucleus makes it impossible to determine precise values for the area/spacecraft mass/some other variable.**

- We're just trying to get estimates here. Real scientists and engineers do these types of estimations all the time. You can at least estimate a maximum and minimum for the area and examine the implications of that.

**2) The problem is non-linear and therefore unsolvable.**

- Real scientists and engineers solve non-linear systems all the time. Multi-target spacecraft trajectories, weather forecasting, etc. 'Non-Linear' is not a total roadblock for people who know what they're doing.

**3) This model does not include effect 'X', therefore it does not apply to Electric Comets and they will not address it.**

- Then include effect 'X'.

**4) These problems come from someone who is an enemy of Electric Universe theory/fails to treat Electric Universe theories with appropriate respect so therefore their objections can be ignored.**

- Bad news dudes, if other scientists who actually do spacecraft missions saw your claims and actually looked at them, they would insist on answers to these questions and more before they trust electric comet advocates to multi-million dollar/euro equipment.

**5) Electric Universe 'theorists' have no way to verify that these claims would actually be made by other scientists so they will not be addressed by Electric Universe supporters.**

## 6 comments:

Interesting points you make about the excuse set these people use. Having had a read over at the International Skeptics Forum, I became interested in excuse No. 1, mainly due to an EU acolyte who keeps infesting a general science news website I visit, as well as the 'discussion' going on at ISF.

His throwaway remark about the solar wind reacting with silicates to create water molecules was just too stupid not to investigate. So I did some back of a cigarette packet calcs to find the discrepancy in the numbers required. I allowed a very generous cometary surface area of 12km^2, based on the widest and longest points of 67P. I gave a solar wind speed of 400km/s, and a not unreasonable 1 proton per cm^3 for the solar wind density at 2.7 AU. Let's call the comet surface rectangular, for the sake of envisioning this. So, every second we have a 'block' of space containing 1 proton per cm^3 that is 400km long, 4km wide and 3km deep hitting the surface of 67P. Convert that into cm^3, and we end up with 4.8 x 10^18 protons hitting the comet per second. That's a lot. Until you decide to find out how many molecules are in a litre of water (the average rate given off by 67P in July-August). There are ~3.3 x 10^25! That is 7 orders of magnitude more. Not even allowing for the fact that you need 2 protons to make a water molecule, or the fact that the peak emission rate of the comet on its last visit was ~500 l/s. Given my generosity in the figures, we are looking at a discrepancy of 9-10 orders of magnitude.

So, pretty close by their standards!

To Ian Whittaker,

Yes. Myself and others have done a similar estimate.

The next level might be to estimate the area required to accumulate sufficient hydrogen ions to explain the water production. From that, you might estimate the amount of charge needed on 67P to attract that much material.

Then there's the question of how long would it take to neutralize that charge with the incoming material and how does the nucleus maintain electrical potential.

It is interesting how many questions in science can be addressed in the form of 'how much input do you need to generate a specific quantity of output' and knowing the requisite conserved quantities. It is really the same basic question of inventory management which competent business people must solve on a regular basis.

Well, Tom, it looks like we have another figure to plug into the equation. The GIADA dust analysis results were reported a little while back, and in that report they mentioned a voltage on the spacecraft of around -5 - -10V.

A poster on the Rosetta blog, Harvey, who I have good reason to believe is Prof. Harvey Rutt, from the Univ. of Southampton, U.K., did some calculations based on that figure, and made a couple of assumptions for the capacitance of the comet and spacecraft, and came up with the figures of a charge of 10^-8C on Rosetta, and 10^4C on the comet. Using these figures he calculated the voltage on the comet to be ~35GV. That was using the density as measured. If you want it to be rock at ~2000kg/m^3, then it gets even more astronomical.

Do his equations and results look valid to you?

http://blogs.esa.int/rosetta/2015/04/09/giada-investigates-comets-fluffy-dust-grains/

Towards the end of the thread.

To Ian,

As also noted in that thread, these voltages are consistent with ionization by solar wind and solar photons. This liberates electrons which form a thin 'atmosphere' above the surface, which can't travel too far due to the remaining net positive charge on the surface.

Harvey's updated computation is here.

The computations are only as correct as the model of charge configuration matches reality. A capacitor needs a structure to hold the charged plates apart since they have opposite charges. If they make contact, they ground out the entire system. What serves that role for 67P? In a plasma environment, charges are moving around pretty freely unless they have energy to maintain the separation (such as energy from solar wind and solar photons).

And what is generating the 35 gigavolts potential? Where's the battery or generator? The EU theology basically assumes batteries and generators magically exist everywhere. Note that this energy takes some pretty hefty particle accelerators to achieve on Earth (Wikipedia: Tevatron). SOMETHING has to be generating the electric field over this region! EU 'theorists' never specify WHAT is doing that. In addition, such a large-scale electric field will have many other effects on the heliospheric plasma (and spacecraft traveling through it) which always seem to be undetectable...

Saying the voltage on the comet is meaningless as voltage is the DIFFERENCE in potential between two points. Where is the zero point? Perhaps it's the Sun? Or is this between the comet nucleus and Rosetta?

If so, it means a singly charged particle starting at the Sun accumulates 35 GeV of energy by the time it reaches the comet. These energies make the particles extremely relativistic, speeds very close to the speed of light. This means EVERY single electron, EVERY single ion, is subjected to this potential difference, and they will be driven in opposite directions, one sunward, the other towards the comet.

At 35 GeV, this is NASTY hard radiation, well above the pair-production threshold (see Transforming Energy into mass: Particle Creation) for electrons and protons. If that were the case, we should see large amounts of antimatter where the electrons and ions strike. This is even worse than the EU radiation environment I examine in my "Death by Electric Universe" series.

So 'Harvey' does not seem to know what happens in a plasma when huge voltages are applied across it.

To be fair Tom, regarding your last sentence, I think Harvey knows exactly what these charges/ voltages would imply. Having followed his posts for a number of months, I know he is no fan of EU nonsense, and showed these calculations purely to point out how ludicrous the explanation of electrostatic repulsion actually was. The same arguments you use are the same he has used in numerous posts on that blog.

I think it was more of a "well, if this magically could happen, this is the charge/ voltage that would be required."

As he pointed out in a later post in that blog, despite the insane voltage, it would discharge within seconds.

To Ian,

My apologies if I mis-read Harvey's position in the discussion.

I read sections of the thread but not the entire thing since it is rather long. Many of his posts I read were just reporting the results of various iterations of the computation, attaching little implications with the results, perhaps leaving those to the reader.

Tom

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