Charging the Earth

Teaching Physics 201 has me digging out some of my old favorite concept-y problems. Nothing dramatic in the mathematics, but at the 201 level you can't even assume knowledge of derivatives. But you can try to catch their minds with interesting examples. Here's a classic one:

You've got the earth and the moon. They have mass and so they attract each other with gravity. Both the earth and the moon are pretty large, and so the attraction is considerable. On the scale of earthbound undergraduate lab equipment however, gravity from anything but the earth is pretty hard to measure. Pretty much impossible, in fact. We can do a little better with electrical forces. You can charge a glass rod by rubbing it with fur, and observe the effect of the charge on little sheets of foil or similar. These electrical forces are very, very tiny. But even so we're able to see them, which is more than we can say for gravity.

So, if you sprinkled some extra electrons on the earth and the same number of electrons on the moon, how much charge would you have to add in order to completely cancel the gravitational attraction? Let's figure it out. Call the charge q, and we'll put the gravitational force on the right and the electrical force on the left

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Of course Me of the mass of the earth, and similarly for the moon. The q2 really is the product of the charge on the earth and the charge on the moon, but we'll assume they're equal for conceptual convenience. Nicely, the distance r will cancel and solving for q we get:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

Plug in the appropriate numbers to find the required charge on each body. I find that the answer is 5.71 x 1013 coulombs. That's a pretty huge amount of charge in total. But for the earth it's only about 10-11 coulombs per kilogram. And that itself is only sixty million or so excess electrons per kilogram. And considering each kilogram of earth contains trillions of trillions of electrons, only an absolutely tiny percentage of the total atoms in the earth would actually need to be ionized.

We often think about gravity being the strongest force in our daily experience. It isn't. Electrical forces just have the decency to cancel themselves out to very high precision. Gravity builds up on itself without limit. Which in the final analysis is a pretty lucky thing for us - it's nice having these solid planets which hold themselves together and solar systems which stay mostly stable over quite long time periods.

Still, next time lint clings to your shirt think about how amazing it is that electrical charge can produce forces strong enough to have visible effects from something so small. Gravity couldn't hold lint to your shirt in a million years.

More like this

The best description I ever heard for the difference in magnitude between gravity and electromagnetism is that if you jump off a 100-storey building gravity will accelerate you all the way down, but electromagnetism will stop you in a fraction of a millimetre.

http://www.seismo.unr.edu/ftp/pub/louie/class/plate/composition.html

Average FW ~ 34.3. Earth masses 5.9723x10^27 g. A mole of electrons is a Faraday, 96,485 C. 5.71 x 10^13 coulombs to be dispersed.

1.74x19^26 g-moles of atoms, 5.92x10^8 moles of electrons. Each atom averages 3.4x10^(-18) extra electron. 34.3 grams of atoms get 2.05 million electrons, a kilogram gets 59.7 million electrons in excess amidst its nominal 1.1x10^24 electrons vs. protons. Fair enough, originally estimated 60 million electrons added to a trillion trillion electrons/kg.

By Anonymous (not verified) on 04 Sep 2008 #permalink

The various planets and sun all carry charges. This is due to the fact that the proton and electron weigh such different amounts. At any given temperature, electrons escape easier than protons (or positively charged nuclei), leaving the earth positively charged.

Of course the result is that the earth builds up a charge sufficient to balance the loss rates so the charge is constant.

By Carl Brannen (not verified) on 04 Sep 2008 #permalink

I still think comparing gravitational to coulombic force gets the message across better.

F=Gmm/r^2 vs. F = Kqq/r^2 that's 10^-12 * 10^-31*10^-31 / r^2 = 10^-74/r^2 vs. 10^9*10^-19*10^-19/r^2 = 10^-29/r^2. 45 orders difference.

By erik Remkus (not verified) on 04 Sep 2008 #permalink

Actually, 1+1=2, water is wet, and most adhesive tapes will easily remove lint from your shirt. Science... pfft.

By Time Bomb (not verified) on 04 Sep 2008 #permalink

Gravity sucks.

As I understand it, mass bends space-time. Unless acted upon by another agency objects continue to travel in the direction they were put into motion. So when an object appears to change direction near a more massive object it is actually the direction it is proceeding in that has changed direction. Thus there is no need for a gravitational force.

Furthermore, all objects with mass curve space-time. Therefor, not only does the Earth bend space-time, so does the Moon. Which means that not only is the direction the Moon is traveling in changed by the presence of the Earth, the direction the Earth is traveling in is changed by the presence of the Moon. Giving us an object with a shifting center of mass and a sort of quivering gravitational field we call the Earth-Moon system.

Thats about 320 kg of electrons!I was gonna say, don't forget the mass of all those elctrons, but its much less than the precision you used.
I wonder how strong a box it'd take to contain the pressure from all that repulsive force, anyone know how to calculate it?
Also, Thats also enough electrons to make about 2 trillion lightning bolts.

" but at the 201 level you can't even assume knowledge of derivatives"

Ap, Op, Up (various other sounds of flabergation) .....

What!? My daughter is in year 9 and she's learning calculus (pretty much the same time as I first learnt it).

And you're teaching 2nd year physics????? Sorry, I'm a bit stunned, what level of knowledge are you talking about here?

Physics 201 isn't a second-year physics class, it's the intro non-calculus-based physics for non-science majors. The "2" just denotes that there's prerequisites that typically are taken during the 1st year.

Matt #11 - ok, thanks for clarifying that - my mistake.

By Anonymous (not verified) on 11 Sep 2008 #permalink