Replacing Gravity

I’m teaching introductory E&M this term, so it’s kind of fun to play around with silly applications of Coulomb’s Law. For example, let’s imagine that gravity suddenly switched off, but we wanted to keep the Earth in its orbit. How much charge would we need to move from the Earth to the Sun for the electrostatic attraction to take the place of gravity?

The key here is to set the gravitational force, which we can reasonably approximate by Newton’s Law of Universal Gravitation:

F_{grav}=G\frac{M_1 m_2}{R^2}

(where the M‘s are the masses, R is the distance between them, and G is a constant to get the units right) equal to the electrostatic force, which we can reasonably describe by Coulomb’s Law:

F_{elec}=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{R^2}

(where the q’s are the charges (one positive, the other negative) and the constant to get the units right is “1/4\pi\epsilon_0 ” for historical reasons that really don’t matter).

The nice thing here is that both of these depend on the distance in exactly the same way, so I don’t even need to look up the radius of Earth’s orbit. With a bit of algebra, we can re-arrange this to collect constants of nature on one side of the equation:

\frac{q_1 q_2}{M_1 m_2} = \frac{G}{1/4\pi \epsilon_0}

This says that you can replace gravity with electromagnetism provided these ratios are equal. There’s a bit of ambiguity as to how you achieve this, though– all that matters is the product of the charges and the product of the masses, so the individual values can vary wildly. You could hold the Earth in orbit with a single extra electron, provided you put shitloads– metric shitloads– of extra protons on the Sun.

Of course, that’s not very practical. If you were going to do this for real, you would move some number of electrons from the Sun to the Earth, leaving behind a positive charge of the same magnitude as the negative charge the Earth acquired. In which case, you would have the same thing for both q‘s, and then you can easily solve for q:

q = \sqrt{\frac{G}{1/ 4 \pi \epsilon_0}} \sqrt{M_1 m_2}

So the charge you would need depends on the geometric mean of the two masses (3.5 \times 10^{27} kg), which is kind of cute. If you look in the back of your favorite physics textbook (you do have a favorite physics textbook, right?), you can find numbers for all these things, and get a value of:

q = 3 \times 10^{17} C

That’s a really big number– for reference, if you shuffle your feet on the carpet and shock your younger sibling, the total amount of charge you’ve picked up to make that spark might be in the 10-6C range.

Of course, since this is all a silly exercise, we can ask how many electrons that would be. The answer, again from plugging in values from the back of your favorite book, is:

N = 2 \times 10^{36}

That’s enough that you might reasonably wonder whether you would be skewing the mass by enough to throw off the acceleration needed to keep the orbit going. Not to worry, though, because electrons are really light. The total mass involved is:

m_{elec} = 2 \times 10^6 kg

So by shifting just about 2000 metric tons from the Sun to the Earth, you could replace gravity. Provided that whole mass was in electrons, and you had some magic way to keep all that charge in one place. And, for that matter, to keep the Sun burning, because if you switched gravity off, all that incandescent gas would just fly away in short order…

Another, arguably more physicist-y way to go about this would be to note that you could get the same effect by taking a star and a planet with a constant charge-to-mass ratio of around 8.6 \times 10^{-11} C/kg. In which case you would only need about 5 \times 10^{14} C worth of charge on the Earth, but 2 \times 10^{20} C on the Sun.

And, just for giggles, what would you need to keep the Earth in orbit with a single extra electron? Well, you need the product of the charges to be about 9 \times 10^{34} C2, in which case that single extra electron would need to be balanced by a charge of 6 \times 10^{53} C, or about 4 \times 10^{72} electrons. That’s a mass of 3 \times 10^{42} kg, or around a trillion times the mass of the Sun.

So, clearly, that’s just completely crazy. As opposed to the other solutions, which are totally reasonable…

For homework, using the equal-charge figure above, calculate the strength of the magnetic field that would be produced due to the extra charge on the Earth acting like a current loop. Send your answers to Rhett for grading, and make sure you show all your work.

Comments

  1. #1 Uncle Al
    January 7, 2014

    metric shitloads” merde-loads, “8^>) Gravitation does not spark (yeah well – black holes) , thus being more stable though less exciting. Sunspot AR1944 is loaded and positioned to EM blast the Earth in historic proportions. Ending electricity and communication ends Global Warming. It renormalizes armed conflict to depopulating economically disenfranchised young males. Go Maxwell!

  2. #2 Ricky
    January 7, 2014

    It’s everyone’s favorite crazy Uncle! You deserve your own show, Uncle Al.

  3. […] Replacing Gravity – Uncertain Principles. […]

  4. #4 Sili
    January 7, 2014

    “metric shitloads”

    kilopoops.

  5. #5 Max Erwengh
    January 7, 2014

    Right now I’m wondering about the energy loos per year due to accelerated charges (synchroton radiation). It shouldn’t be too big. Well let’s calculate it: del E = 2,2 * 10^22 J.

    Ok maybe its not that accurate but this is about the energy you need to accelerate to earth by 0.10 m/s^2.

  6. #6 D. C. Sessions
    January 7, 2014

    Another in that line is the question of what kind of magnetic field would be required to (per Velikovsky) stop the Earth’s rotation, assuming the maximum surface charge consistent with the breakdown field of air.

    I worked that on back in the 60s as a reply to an article on Velikovsky’s attempt to rationalize Joshua.

  7. #7 Joey
    January 8, 2014

    Eli5

    What would happen in your calculations if we increased the earths rotation?

  8. #8 Robert
    January 9, 2014

    If we assume that the charge is equally distributed over the whole mass of the earth (including it’s inhabitants) you could calculate the acceleration you would feel due to the electrorepulsive force the earth is exerting on your charged body.

    And now the fun starts, if my very quick back of the envelope calculation is correct, all the E&M terms would cancel and this would be:

    G M_sun / R_earth^2

    in other words equal to the gravitational acceleration you would feel if you compressed the sun to the size of the earth and were standing on it’s surface.

    I suspect this is strong enough to overcome atomic bonds, possibly even nuclear ones. If we charged the earth that much it would quickly transform into a rapidly expanding cloud of elementary particles.

  9. #9 Larry
    Atlanta
    January 19, 2014

    To emphasize the relative weakness of the gravity force, Richard Feynman calculated that if two people were at arm’s length, and each had one percent more electrons than protons, the electric repulsion between the two would be large enough to lift the “weight” of the entire Earth. That is, it would exceed the mass of the Earth times g.