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I Amount of excess charge to overcome Earth's gravity

  1. Mar 14, 2017 #1
    Years ago I read an article illustrating the vastly different strengths of the electromagnetic and gravitational forces. The article gave a figure for the number of excess electrons dispersed throughout the earth that it would require to overcome the gravitational binding of earth's matter and disperse that matter into space. I don't remember what that exact number of electrons was, but I do remember that it was surprisingly small. Does anyone know what the number of electrons would be?
  2. jcsd
  3. Mar 14, 2017 #2
    My estimate is 5.146 x 1014 C, or 3.21 x 1033 electrons. That works out at 5.38 x 108 electrons/kg, or 8.61 x 10-11 C/kg.
    How I got it: Estimate gravitational energy by imagining a sphere of radius r and density ρ, allow matter to fall from infinity onto surface to make a shell of thickness dr, integrate energy from r = 0 to R. Do the same for charge. Equate the expressions, get Q2 = M2G*4πε0
  4. Mar 14, 2017 #3
    Edit: I misread the question. I answered a different question of how much excess charge such that the gravitational attraction is cancelled out. Of course, there's still the Coulomb force pushing everything apart which overcomes gravity quite easily.

    I believe you should use the Reissner–Nordström metric here. https://en.wikipedia.org/wiki/Reissner–Nordström_metric
    $$ds^2 = \left( 1 - \frac{r_\mathrm{S}}{r} + \frac{r_Q^2}{r^2} \right) c^2\, dt^2 -\left( 1 - \frac{r_\mathrm{S}}{r} + \frac{r_Q^2}{r^2} \right)^{-1} dr^2 - r^2\, d\Omega^2_{(2)}$$

    We want to find Q such that ##\frac{r_S}{r}## exactly balances out ##\frac{r^2_Q}{r^2}##.
    ##r^2_Q=\frac{Q^2 G}{4\pi\epsilon_0c^4}##

    Let's choose ##r=r_E=6371\text{km}## and ##M_E = 5.972\times 10^{24}\text{kg}##.
    Solving for Q
    ##\frac{Q^2 G}{4\pi\epsilon_0c^4} = \frac{2GM_E}{c^2} r_E##
    ##Q^2 = 8\pi\epsilon_0 c^2 M_E r_E##
    plugging and chugging
    Q=2.76*10^19 coulombs
    which is 1.72*10^38 electron charges. Much much larger than mjc123's result.
    Better check my math though. I don't like SI electromagnetic units.
  5. Mar 14, 2017 #4
    Thank you both, mjc123 and Khashishi.
  6. Mar 14, 2017 #5


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    bolding mine

    Well, half of our answers look similar:
    3.21E+42 (my answer)
    3.21E+33 (your answer)

    And Kashishi's answer appears to be the exponential average of our two answers. (not sure if there is such a thing)
    (42+33)/2 = 37.5
    Somehow, I don't think we've collectively found the answer.
  7. Mar 14, 2017 #6
    The gravitational binding energy is ##\frac{3GM^2}{5R}##
    The electric binding energy is ##\frac{3Q^2}{20\pi \epsilon_0 R}##
    Equating the two gives:
    ##Q^2 = GM^2 4 \pi \epsilon_0##
    ##Q = \sqrt{4 \pi \epsilon_0 G M^2}##
    plugging gives 5.15*10^14 C = 3.2*10^33 electrons. That's the same answer as mjc123, which is correct.

    My previous response was just looking at how much charge it would take for the charge to mess up the gravitational field while ignoring the direct electrostatic force.
  8. Mar 15, 2017 #7


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    I concur. I transcribed Coulomb's constant(ke) incorrectly as 8.99E-9 originally, when the value should be 8.99E9.

    The method I used was to equate the two force equations for gravity and charge:
    F = G*m1m2/r^2 = keq1q2/r^2
    and then do a bunch of maths.

    Soln: 3.21E33 electrons
  9. Mar 15, 2017 #8
    OK, I'm not a physicist (which is why I asked this question in the first place), but would this approach work:

    The electromagnetic force is 10^36 stronger than the gravitational force.
    The mass of the earth is 6 x 10^24 kg.
    The mass of the electron is about 9.1 x 10 ^-31 kg

    So, the mass of electrons required such that the repulsive force of their electromagnetic charge equals the attractive gravitational force of the mass of the earth is 10^-36 less than the mass of the earth. This is 6 x 10^-12 kg of electrons. This is about 7 x 10^18 electrons.

    This is far fewer electrons than the other answers, so I suspect that I'm wrong here. What do you think?
  10. Mar 15, 2017 #9
    The electromagnetic force for two electrons is 4*10^42 times stronger than gravitational force between them, not 10^36.
    Nevertheless, that still leaves many orders of magnitude. The problem with your calculation is that the electromagnetic force is only between the excess electrons. There is no force between the electrons and the rest of the Earth. You have assumed that each excess electron repels the entire Earth, not just the excess electrons.

    Edit: corrected value above
    Last edited: Mar 15, 2017
  11. Mar 15, 2017 #10
    That's a meaningless statement without context. Force between WHAT? Depends on their mass and charge. I've just checked; you appear to be thinking of the ratio of the electromagnetic and gravitational forces between two PROTONS. It'll be different for electrons because of their different mass. I worked it out as ca. 4 x 1042.

    See equations above; the self attraction or repulsion depends on the square of the mass or charge. So you need to take the square root of the force ratio, or ca. 2 x 1021. that gives an electron mass of ca. 3 x 103 kg, or approx. 3 x 1033 electrons.
  12. Mar 15, 2017 #11


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    In defense of Cato, when I googled: The electromagnetic force is 10^36 stronger than the gravitational force
    I came up with 3 reliable looking sources:

    A "dot edu" site:
    "Is the electromagnetic force really strong?"
    It is very strong and this is surprising too because it can work over an infinite range. To give you an idea, the electromagnetic force is approximately 10^36 times stronger than the earth's gravitational field!

    Electromagnetism - the force we know best - is 10 to the 36th power stronger than Gravity.

    And last, and contextually the best:
    Phyisics Forums (Yay CWatter!)
    The gravitational attraction between two electrons is only 8.22*10^−37 of the electrostatic force of repulsion at the same separation.

    equivalent to 1.22E+36


    Btw, did anyone else work out the density of electrons to blow up the earth?
    I come up with ≈3000 electrons per cubic millimeter. Which means that if you could see electrons with your magnifying glass, you would see about 14 of them per millimeter.
    I was just trying to visualize these big numbers, at a human scale.
    Last edited: Mar 15, 2017
  13. Mar 16, 2017 #12
    Fair enough, but you still have to state what you are comparing the forces between for it to be really meaningful. The electromagnetic force between me and the earth is not 1036 times the gravitational force. The charge/mass ratio of a proton is ca. 108 C/kg, while we calculated 8.6 x 10-11 C/kg to blow up the earth (or at least prevent it coalescing).

    I agree with that, given my answer of 5.38 x 108 electrons/kg, and an average earth density of 5510 kg/m3.
  14. Mar 16, 2017 #13


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    It would appear that we are once again in agreement.

    I actually like aftershock's number the best:
    I hear all the time how the electric force is so much stronger than gravity.
    I understand both forces are inversely proportional to the distances squared, and that the gravitational constant is roughly 10^20 times greater than the coulomb constant.

    as it compares things on a base SI unit scale: kg vs coulomb

    I also calculated the linear ratio of atoms to electrons.
    In between each electron, there will be about 462,000 (neutral) atoms*.

    From this I concluded that atoms are really tiny. But I think everyone already knew this.

    * Somewhere I found on the internets, the number of atoms in the earth.
    I did some suspicious maths (I averaged the authors exponents: 10^50.5 = 3.16E50) and came up with: 6.6 million atoms per linear mm.
    (6,600,000 atoms/mm) / (14.4 electrons/mm) ≈ 462,000 atoms/electron
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