Okay, I don’t know how stupid this question is, but please bear with me.
One day after my astronomy class I thought of the following scenerio:
Suppose the Earth is perfectly cut in half vertically, north pole to south pole, along zero longitude. The division occurs instantly (as if the Earth suddenly became two pieces.) Geological and oceananic effects are ignored.
Would these two halves stay together because of gravity, or would they drift appart due to rotation?
Fire away.
I think that by far the most significant force acting on these two halves would certainly be the attraction between two giantic masses just inches away. The earth would snap back together and all hell would probably break loose.
Two things: First of all, if the Earth were spinning fast enough for the halves to separate, we wouldn’t be able to stay on the surface, either. In fact, we’d fly off before the halves even separated, because we’re farther from the center, and gravity decreases with distance, while centrifugal force increases.
Secondly, if the Earth were spinning that fast, it wouldn’t matter if you cut it. Rock isn’t all that strong, on planetary scales, and the Earth (and other planets) is held together pretty much entirely by gravity.
Don’t worry about it. If you come back again and ask the same question next week, then that would be a stupid question. You’ve got to find things out somehow :).
I’ve worked this out - with a few simplifying assumptions. As I understand it, we’re assuming that this rift in the earth occurred magically, leaving two hemispheres (trying to) continue rotating about one another. Would the gravitational attraction provide the necessary centripetal (no, I don’t mean “centrifugal”) force, or not?
To greatly simplify my life, I’m going to assume that the earth is a sphere of uniform density. This is clearly not true, but I’m going to run with it anyway. I’m also going to assume that the rotational period of the earth is precisely 24 hours (it’s not), and I’m going to ignore the fact that the earth is rotating around the sun.
Mass of earth, m = 5.97x10[sup]24[/sup] kg.
Angular velocity, [sym]W[/sym] = 2[sym]p[/sym] radians/day = 7.27x10[sup]-5[/sup] rad/sec
Radius of earth, R = 6.378x10[sup]5[/sup] m
When the two halves split, they’ll each have a mass of m[sub]H[/sub] = 2.99x10[sup]24[/sup] kg. We also need to know the center of mass of each half. This is where the assumption of uniform density comes in handy.
This page is kind enough to derive the center of mass of a uniform hemisphere for us - it’s a distance of (3/8)R out from the center of the original sphere. What this means is that once the split occurs, we can treat it as two bodies of mass m[sub]H[/sub] a distance of (3/4)R apart. Call this separation D[sub]H[/sub] = 0.75R = 4.78x10[sup]5[/sup] m. Let R[sub]H[/sub], half of that, be the distance of each body from the center of rotation. R[sub]H[/sub] = 2.39x10[sup]5[/sup] m.
What is the magnitude of the gravitational attraction between the two halves? This is given by:
G m[sub]H[/sub][sup]2[/sup]
F[sub]G[/sub] = ----- = 2.6x10[sup]27[/sup] N
D[sub]H[/sub][sup]2[/sup]
where G, the universal gravitational constant, has a value of 6.673x10[sup]-11[/sup] N m[sup]2[/sup] kg[sup]-2[/sup].
OK, great. Is that sufficient to act as a centripetal force to keep a hemisphere orbiting about the original center of the earth? The force required to do that is given by:
m[sub]H[/sub]([sym]W[/sym]R[sub]H[/sub])[sup]2[/sup]
F[sub]C[/sub] = -------- = m[sub]H[/sub][sym]W[/sym][sup]2[/sup]R[sub]H[/sub] = 3.78x10[sup]21[/sup] kg m/s[sup]2[/sup] = 3.78x10[sup]21[/sup] N
R[sub]H[/sub]
The gravitational attraction is about a million times stronger than that, so it will easily be enough to hold the two halves together.
How fast would the earth need to be spinning so that it would fly apart under these conditions? We need F[sub]C[/sub] > F[sub]G[/sub], or
Gm[sub]H[/sub]
[sym]W[/sym][sup]2[/sup] > ------ = 3.65x10[sup]-3[/sup] s[sup]-2[/sup]
4R[sub]H[/sub][sup]3[/sup]
which is [sym]W[/sym] = 6.04x10[sup]-2[/sup] radians/sec, or about 830 revolutions per day.
I think it’s safe to say that Chronos is quite right - if the earth were spinning that quickly, we’d have all sorts of problems to worry about. In fact, I’d say that if there were that much angular momentum floating around, the earth would never have formed in the first place, since I believe planetary accretion is predominantly gravitaional forces slowly overcoming rotational ones.
I think it’s worth noting that the earth originally formed as a coalescence of scattered material. Given that the earth formed due to gravity, you’d need something more than a cut down the middle to split it apart.
This isn’t as convincing as brad_d’s math, but it’s a good back-of-the-envelope argument.
[nitpick]
It’s not quite true that you can treat the two hemispheres as point masses, each centered at their respective centers of mass. This only works for spherically symmetric things. But this should still be the dominant effect, and it gives the right general idea.
[/nitpick]
As I half-expected, I just found an error in the math I did above. Specifically, my value for the radius of the earth is off by a factor of ten. Here is where I got it from, but I converted from the diameter (given in kilometers) erroneously. The actual value of R is 6.378x10[sup]6[/sup] meters.
This makes F[sub]G[/sub] = 2.6x10[sup]25[/sup] N, and F[sub]C[/sub] = 3.8x10[sup]22[/sup] N. The general idea is the same, but the difference is less dramatic.
On top of that, g8rguy is quite right. I’d forgotten about that completely, but I just checked up and re-educated myself. Thanks for pointing it out.
brad_d, I hope that those calculations didn’t take too much time out of your day. Thanks for your info.
This really started as a “I wonder what would happen if…” kind of question, with an interresting mental image.
(as you can see, I’m not the kind of person who just pops in, asks a question, and never returns. It’s a habbit from another MB I used to frequent.)
You are the kind of person we like around here. You’ve even got a good username. Welcome to the SDMB. 