In Case Of Sudden Failure Of Planet Earth

Have you ever figured out just exactly what you would do if the Earth were to suddenly disappear from the universe, leaving just you and whatever’s around to fall towards whatever the nearest heavenly bodies are? No, me neither. Asked to improvise one, I suppose I’d suffocate within minutes and then everything else becomes not so interesting to me, although possibly my heirs might be interested, if they’re somewhere.

My mother accidentally got me thinking about this fate. She’s taking a course about the science and world view of the Ancient Egyptians. (In talking about it she asked if I knew anything about the science of the Ancient Egyptians, and I tried to say I didn’t really know more than the average lay reader might, although when I got to mentioning that I knew of their awareness of the Sothic Cycle and where we get the name “Sothic Cycle” I realized that I can’t really call myself ignorant about the science of the Ancient Egyptians.) But the class had reached the point where astrological beliefs of the ancients were under discussion, and apparently some of the students insisted that astrology really and truly works, and my mother wanted me to figure out how strong the force of gravity of the Moon is, compared to the force of gravity of another person in the same room. This would allow her to go into class armed with numbers which have never dissuaded anyone from believing in astrology, but, it’s fun for someone.

I did double-check, though, that she meant the gravitational pull of the Moon, rather than its tidal pull. The shorthand reason for this is that arguments for astrology having some physical basis tend to run along the lines of, the Moon creates the tides (the Sun does too, but smaller ones), tides are made of water (rock moves, too, although much less), human bodies are mostly water (I don’t know what the fluid properties of cytoplasm are, but I’m almost curious enough to look them up), so there must be something tide-like in human bodies too (so there). The gravitational pull of the Moon, meanwhile, doesn’t really mean much: the Moon is going to accelerate the Earth and the people standing on it by just about the same amount. The force of gravity between two objects grows with the two objects’ masses, and the Earth is more massive than any person on it. But this means the Earth feels a greater force pulling it towards the Moon, and the acceleration works out tobe just the same. The force of gravity between two objects falls off as the square of the distance between them, and the people on the surface of the Earth are a little bit closer or a little bit farther away from the Moon than the center of the Earth is, but that’s not very different considering just how far away the Moon is. We spend all our lives falling into the Moon, as fast as we possibly can, and we are falling into the Moon as fast as the Earth is.

So, each of us is accelerating towards the Moon by a certain amount, but the Earth is accelerating by the same amount in the same direction. The net effect is that the Earth isn’t falling out from under or pressing up into our feet no matter where the Moon is.

But, it’s what she wanted, and I was curious, so now let me, nearly two months in, finally put some equations into this math blog. Equations are said to be killers of reader interest; worse, the web has never figured out how to present equations in anything resembling a graceful manner. But I’ll try anyway. The equations will start innocently enough. The force of gravity F_{1, 2} felt by one object, which I’ll label as object 1, exerted by a second object, which I’ll label as object 2, is:

F_{1, 2} = G * (m₁) * (m₂) / (d_{1, 2}²)

where m₁ is the mass of object 1, m₂ is the mass of object 2, and d_{1, 2} is the distance between objects 1 and 2, and we square that because that’s the gravitational formula we know going back to Newton and Euler and folks like that. The equation may not look very pretty, at least rendered in text like that, but it’s important to remember there are no good ways on web pages to render mathematical equations. Little images are maybe the least bad approach, and that only barely.

G, meanwhile, is the gravitational constant of the universe, one of those physical parameters describing how the universe is put together. In the metric system, this has a value of something like 6.673 times 10^-11 meters-cubed per kilogram-seconds, which seems incredibly tiny, but only because it is. Gravity is surprisingly weak considering it’s the only force in the universe that works at large distances. But then if it were enormously stronger we would have an uncomfortably tiny universe.

But let me consider this: imagine there are three objects in the universe. Number one is some person. Number two is some other person, elsewhere in the room. Number three is the Moon, out wherever it is the Moon is supposed to be. What my mother wanted to know is how strong the force due to the other person in the room, F_{1, 2}, is, compared to the force caused by the Moon, F_{1, 3}. That ratio is F_{1, 2} divided by F_{1, 3}, and will be a little nicer to work with:

F_{1, 2} / F_{1, 3} = (G * (m₁) * (m₂) / (d_{1, 2}²) ) / ( G (m₁) * (m₃) / (d_{1, 3}²) )

This seems to make a liar of me in talking about things being nicer to work with. Fractions divided by fractions are never popular. But we have a familiar old rule we sort of remember to work with: (a/b) / (c/d) is equal to (a/b) * (d/c), and so, flipping over the big denominator and changing this into one fraction times another gives us:

F_{1, 2} / F_{1, 3} = (G * (m₁) * (m₂) / (d_{1, 2}²) ) * ( (d_{1, 3}²) / (G (m₁) * (m₂)))

So far, perhaps, this is a lateral move. But, now, remember the associative and the commutative rules: we can turn (a * b * c)/(d * e * f) into (a/d) * (b/f) * (c/e), or otherwise shuffle things around, so they’re convenient. For example, move the G in the numerator and denominator way over to the left; and group the mass terms together; and group the distance terms together. This gives us:

F_{1, 2} / F_{1, 3} = (G / G) * (((m₁) * (m₂)) / (m₁) * (m₃)) * (d_{1, 3}² / d_1,2²)

Now things are starting to look better. G/G is one, and m₁ divided by m₁ is one, and we can stop writing them. So the ratio of forces is now:

F_{1, 2} / F_{1, 3} = (m₂ / m₃) * (d_{1, 3}² / d_{1, 2}²)

That last line is a particularly nice one to work with, since it’s gotten rid of that pesky G. It’s also got rid of the mass of the first person, but that’s fair. If the person were twice as massive, the force of gravity the person feels would be twice as large, toward other people as well as toward the Moon.

Here’s another nice thing about this form, hidden behind all these ratios. At some point we need to plug in numbers about how massive people and Moons are, and how far away both are. What units should we work in? Do we use the Metric system or the Imperial system, or do we wait for someone to pipe in with an argument that the United States does not actually use the Imperial system but rather one that just looks like it except for some fuss about liquid ounces and tons that we never keep straight?

It doesn’t matter, as long as we use the same units consistently. I’m partial to the metric numbers, since it’s very easy to find scientific terms in them, but if you want to run them with pounds and feet instead of kilograms and meters that’s fine. If you want to argue that kilograms are units of mass and pounds aren’t, go ahead; if you want to argue that slugs are the proper unit of mass in the Imperial Or Whatever system, you’re on your own as the only people who ever talk of slugs-the-unit-of-mass are people trying to bog down a perfectly good conversation about weights and masses so as to look like they know something the rest of us don’t. But we’ve gotten rid of the physical dimensions; the final number doesn’t depend on what system we use. Dimensionless units are quite nice this way.

How much mass does a person have? That’s hard to say, but if we aren’t going to insist on personal measures for every person in the world, we can say they’re around 75 kilograms, around 165 pounds, and while we’ll rarely be exactly right, we won’t be too horribly wrong. 75 will be our m₁. For the Moon, there’s less room for debate. It has a mass of about 73,600,000,000,000,000,000,000 kilograms, and a touch over twice that in pounds. 73,600,etc, will be our m₂.

How far away is another person? That’s extremely variable, but we started out thinking they were somewhere in the same room. Let’s say they’re about one meter, three feet, away; nearly any room will tolerate being that far away. 1 will be our d_{1, 2}. The Moon’s distance varies more than you might think, but it averages around 384,000,000 meters away from the center of the Earth, and we can say that’s about as far away as it is from us on the surface without being too far off. So 384,000,000 is our d_{1, 3}.

We put those in to the ratio and see something which startled me, at least:

F_{1, 2} / F_{1, 3} = (75 / 73,600,000,000,000,000,000,000) * (384,000,000² / 1²)

Big numbers, but, a calculator saves us by telling us this ratio is about 0.00015. That is, you’re accelerating toward the Moon about 6,600 times as strongly as you are towards a person standing in the room. I was startled, too; accustomed as I am to thinking of the tidal forces I’d expected the nearby person to be a larger factor.

Knowing their relative strengths makes it almost irresistible to figure out their absolute strengths: how fast are we falling towards the moon, in meters per second per second? That’s a matter of plugging in the gravitational constant and the masses and distances into that force equation way up above. It turns out that we’re falling towards the Moon with an acceleration of about 0.0025 meters per second per second, about one fortieth of one percent of our acceleration towards the Earth. But the person in our room is attracting us with a tiny 0.000 000 38 meters per second per second, or about three millionths of a percent of the Earth’s acceleration.

What that tells us, by the way, is that if the Earth were to suddenly wink out of existence, we would begin falling towards the Moon, at first, picking up something like two and a half millimeters per second in speed each second. It’s easy to work out how long, at that acceleration, it would take to fall into the Moon, at some 384,000,000,000 millimeters away, although we wouldn’t be falling at that acceleration for the whole trip. And there are many more complications, too.

And now, you should know, I’m interested in describing some of those complications.