## Wednesday, November 3, 2010

### Gobbledygook Pt. 1: Geophysics Without Fear

It was another object lesson in science's ability to obfuscate, intimidate and make you scratch your head till it bleeds. While I was researching the post before this one, I came across this passage in Geodynamics: Applications of Continuum Physics to Geological Problems:
"The gravitational potential anomaly [ΔU] due to a shallow, long wavelength isostatic density distribution is proportional to the dipole moment of the density distribution beneath the point of measurement."
For the layperson, this is the very model of a "What the fuck????" gob of indecipherable science babble. Let's break it down [DON'T BE SCARED, IT'S ONLY SCIENCE]:

Gravitational potential -- If you pick up a brick, hold it above the ground, and then let it go, it won't just stay suspended in mid-air. By lifting it up, against the pull of gravity, you gave it a certain amount of stored energy, called potential energy. And since that energy -- which is equal to the energy the brick will have when it falls -- comes from doing work on the brick against the earth's gravitational pull (and don't forget, the brick is pulling on the earth too), we call that energy gravitational potential.

Gravitational potential anomaly -- General science tip: whenever you see a delta ("Δ") in an equation, that usually refers to some kind of change or difference between two things. "ΔU" represents the gravitational potential anomaly and refers to how different the measured gravitational potential is from a standard reference potential. The difference is due to the fact that the ground beneath us does not have a uniform density, so a reference potential is used to get an idea of how different the pull of gravity is at any given location.

Isostatic density distribution -- You can take two columns of earth, for example one that starts at the top of a mountain and one that starts on the ocean floor, and by the time you burrow down and reach the creamy filling (i.e. the big mass of molten lava that all land masses rest on) you'll have two columns of earth that weigh pretty much the same, but you'll find the mountain column will be less dense than the ocean floor column. The principle of isostatic density distribution tells us this will be true for any two columns of earth we might want to consider. The "long wavelength" part just means you're talking about something massive enough to make a dent in the lithosphere (which consists of the earth's crust and the top part of the mantle layer).

Dipole moment of the density distribution -- This one really threw me at first. I knew about dipole moments as they related to electric charges and magnetic fields, but didn't know what they had to do with rocks and dirt. If a gravitational anomaly is detected, that's an indicator the density of the earth in that area has sort of adjusted itself, compensated to achieve normal isostatic density distribution. Think of the dipole moment as a measure of the extent of the self-adjusting that occurred to get the right density distribution. Plus, it's proportional to the gravitational potential anomaly (as the equation at the beginning shows... trust me, that's what it says).
Moho -- Beneath the earth's crust, but before you get to the mantle layer, there is a boundary called the Mohorovičić discontinuity, or Moho for short. It was named after a Croatian seismologist and marks the depth at which there are notable changes in the earth's chemical composition compared to the crust above it. This has nothing to do with anything, I just like the word "Moho."
So here's what the bizarre, scary passage at the beginning was saying:

Geophysicists, folks who might be looking for ore and petroleum and things underground, start with a standard measure of what gravity is like for the whole planet. But the earth doesn't have a uniform density, so the gravity is actually a little different depending on where on earth you are. This is why, in some places, you weigh a little less and in other places you weigh a little more (there can be other factors, but here we're concerned with the effects of density). The difference in how much you weigh in a particular location compared to the standard measure depends on the nature of the difference in the density of the earth where you happen to be standing. You can take measurements and calculate the difference because there is a mathematical relationship between how the earth's density at your location changes and how the gravity changes in that same place. Different things alter density in different ways, some of which are known. That's why knowing about gravitational potential anomalies is useful: measuring what the gravity is like at a particular location gives us a hint of what might be underground.

There, you see? Was that so hard? Don't we all feel better now?

Moho.

Sources
• Geodynamics: Applications of Continuum Physics to Geological Problems. ©1982, John Wiley & Sons, Inc.
• Geophysical Methods in Geology, 2nd Ed. P.V. Sharma. ©1986, Elsevier Science Publishing Co., Inc.
• U.S. Geological Survey, http://www.usgs.gov.

### Wrong, Wrong, Wrong!

A certain incident has been nibbling away at my consciousness for ages like a kind of psychic termite. It happened in my freshman physics class when the teacher was introducing Newton's law of gravitation. In order to protect the reputation of said teacher, I shall refer to him here as Professor X.

Professor X presented the equation for finding the force exerted on an object by the earth's gravity. It's a classic, lovely little equation...
Me = the mass of the earth, about 1.3 x 1025 lbs or 5.97 x 1024 kilograms. (Do we remember our metric conversions and our scientific notation?)
m = the mass, in kilograms, of some object.
r = how far that object is from the center of the earth.
G = the universal gravitation constant, a sort of cosmic fudge factor, equal to
6.673 x 10-11 N•m2/kg2
(the "N" stands for "Newtons", the unit of measure for force; read as "Newton meters squared per kilograms squared").

A student asked a question: "So, if r = 0, the force is infinite?"

Knowing that we had been taught in our math classes that anytime you divide a number by 0 the answer is , Professor X said, "Yes, if r is 0, Fg is infinitely large."

There are a few reasons why the professor might have responded to the student's question with such an incredibly wrong answer: