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:

$latex 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:

$latex 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 “$latex 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:

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

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:

$latex 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 ($latex 3.5 \times 10^{27}$ 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:

$latex 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:

$latex 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:

$latex 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 $latex 8.6 \times 10^{-11} $ C/kg. In which case you would only need about $latex 5 \times 10^{14} $ C worth of charge on the Earth, but $latex 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 $latex 9 \times 10^{34} $C², in which case that single extra electron would need to be balanced by a charge of $latex 6 \times 10^{53} $ C, or about $latex 4 \times 10^{72} $ electrons. That’s a mass of $latex 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.