I was asked a quite reasonable question on my post about tides. Azmat Hussain posted the question to my blog and on LinkedIn. Here is what he had to say on LinkedIn:
"I liked your writing style, and but it could have used some mathematics and some quantification. Like how much is the force on the body of water on this side vs the other side, and is the difference significant?"
I like it when people try to keep me honest!
I looked on Wikipedia to get some rough numbers. The perigee and apogee of the Moon (closest and farthest points from the Earth) are 362,570 km and 405,410 km. Let's just say that the Moon is 380,000 km away from the Earth when we happen to be looking at the tide. The Earth has a radius of 6,371 km. I'm going to round that to 6,000 km just for ease of reading.
So, here are the numbers that I am going to play with: 374,000 km from the Moon to the closest point on Earth, 380,000 km to the center of the Earth, and 386,000 km to the far side of the Earth. [1]
Next, we throw in out good friend the inverse square law. Someone wrote a very entertaining blog that had something to do with that. It says that the pull of gravity falls off as the square of the distance between two objects. Based on that, I have assembled this table that shows the relative strength of the Moon's pull.
Location
|
Distance
|
Relative pull
|
Side closest to Moon
|
374,000 km
|
103.23%
|
Center of Earth
|
380,000 km
|
100%
|
Side farthest from Moon
|
386,000 km
|
96.92%
|
So... we have a total swing in gravitational pull of 6%. Wow. I was actually expecting a smaller number! I think this proves this is a huge effect, right? [2] And it also explains the whole monthly weight gain problem that women complain about. It's not bloating, ladies. It's the gravitational effect of the Moon!
But, then again, this isn't the whole story. I am talking about the gravitational pull of the Moon. Isn't that tiny? Who cares if it is varies, if it is too small to measure?
Hmmm... If I am standing on the Earth, how does the gravitational pull of the Moon compare to the gravitational pull of the Earth? The gravitational pull goes as the square of the distance. I am 6,000 km from the center of the Earth and 380,000 km from the Moon, so we are talking (6000 / 380,000) squared, which is about 0.025%.
Wait!! That's only a part of the answer, since gravitational pull also goes as the product of the masses, that is, the product of my mass with that of the Earth, and the product of my mass with that of the Moon. My mass is the same, so relatively speaking, we can just look at the relative masses of the Moon and Earth.
The Moon is about 7 X 10^22 kg, and the Earth is about 6 X 10^24 kg. The ratio here is about 1.2%.
So, I conclude that the gravitational pull of the Moon is on the order of 0.025% X 1.2% of the pull of the Earth. If I am using my calculator correctly, I cipher this out to about 3 parts in a million.
I have gone through the whole calculation below.
Location
|
Distance
|
Relative pull
|
Pull relative
to the Earth |
Side closest to Moon
|
374,000 km
|
103.23%
|
3.00 X 10-6
|
Center of Earth
|
380,000 km
|
100%
|
2.91 X 10-6
|
Side farthest from Moon
|
386,000 km
|
96.92%
|
2.82 X 10-6
|
Wow. I was actually expecting a larger number! I think this proves that the Moon has a negligible effect, right?
Then again, this effect is tiny, but we are talking about tides of a few meters... just a tiny distance compared with 380,000 km, about 5 parts per billion.
Azmat has asked a very reasonable question. At this point, I am afraid I must shrug my shoulders and say "flies walk on the ceiling". I say this whenever I am confronted with a problem that is orders of magnitude outside of where I normally live.When I make this comment about flies, I mean that the glue on the foot of a fly that holds it to the ceiling is tiny in my experience, and big for a fly. Alternately, gravity is a pretty big deal for me, but flies seem to be able to take it or leave it.
My next inclination would be to put together an experiment to test the hypothesis. I would measure some tides with the Moon in place where it is, and then remove the Moon and measure the tides again. Simple experiment, really. Maybe I would repeat it a few times. I have applied for a grant to the NSF. Right now, it's hung up on the "romantic impact assessment". Some silly folks are concerned that the removal of the Moon might have a negative impact on the mating habits of homo sapiens. Darn tree huggers! I'll let you know when I hear back from the NSF.
Seriously, my next approach would be a computer simulation in which I modeled the Earth as a thousand little balls. Each ball would be given an initial position and direction vector. They would move through space under the effects of momentum and gravity, and subject to the constraint of non-compenetrability [they can't occupy the same space). Simply enough to write the code, right? (John rolls his eyes.)
In the mean time, I think that Azmat has provided a significant question about insignificance.
------------------
[1] A thought occurred to me when I was figgering those numbers. I have made the assumption that perigee and apogee mean the distance between the centers of the Earth and whatever satellite we are talking about. I wasn't sure about that, so I went to look it up on Wikipedia:
"An apsis ... is the point of greatest or least distance of a body from one of the foci of its elliptical orbit. In modern celestial mechanics this focus is also the center of attraction, which is usually the center of mass of the system. Historically, in geocentric systems, apsides were measured from the center of the Earth."
Uh-oh. Things are getting complicated. I'm gonna stick with the geocentric model.
[2] Rule number 1 about reading John the Math Guy blogs. When I end a statement with "right?" it means that it's not right.
[3] This footnote is not referred to in the main text, but I thought I should put it in for completeness.
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