Expert: Paul Klarreich - 3/9/2006

Question
Hi, I'm Wayne. I'm trying to compute the max torque the moon could have toward causing the precession of the earth. I've read the moon provides approximately 2/3 of the torque toward the 25,800 year precession of the earth. I first computed the torque required to cause this precession of the earth as being 2.955 e 19 kg-m squared/sec. .67 of that is 1.98 e 19 kg-m sq/sec. Then I computed the maximum torque the moon could provide as being 3.3 e 17 kg-m squared/sec. (I show how I did that later) I was told I was using the simplified formula when I computed the maximum torque the moon could provide. The longer more accurate formula uses sine and cos. I agree. However, multipling 3.3 e 17 kg-m squared/sec by sine or cos would always result in a smaller number as sine and cos are always less than 1. I don't see anyway mathmatically that the moon supplies much of the torque towards the 25,800 year precession of the
earth. Am I wrong? If so where?

Here's how I arrived at my figures. When the Earth precesses 360 degrees in 25,800 years I figured its precession rate is 7.717 e -12 radians per second. So the torque or
force needed to cause that precession rate is 7.717 e -12 r/s (the precession rate) times 7.292 e -5 r/s (the spin rate) times 3.3 e 35 kg-m squared (the moment of inertia, that is I, of the extra material around the Earth's equator. divided by 2 pi. So the torque or force that causes the Earth's present precession rate is 2.955 e 19 kg-m squared per second. That times
.67 (which is suppose to be the precessional force suppied by moon) is 1.98 e 19 nm squared per second.


To compute the gravational pull of the moon on the Earth at point A, the center of the Earth, I took the gravational constant 6.6739 e -11 times the mass of the moon 7.35 e22 kg times the mass of Earth 5.972 e 24 kg, then divided by the distance seperating them (384,400,000 meters) squared. That tells me the gravational pull of the moon on point A on the earth is 1.9825314 e 20 nm squared per second.


Now, to compute the gravational pull of the moon on the Earth at point B, the part of the Earth, closest to the moon, I divided the distance between the center of the moon and point A on the Earth (384,400,000 m), from the distance between the center of the moon and point B on the Earth (378,022,000 m). I figured the earth's radius as 6,378,000 meters. I squared the answer I got, since gravational attraction becomes four times greater when the seperating distance is halfed. I multiplied that squared answer times 1.98253140277806351994 e 20 newton meters squared. I got 2.0499944 e 20 nm squared per second. I then substracted that answer from the 1.9825314 e 20 nm squared per second answer. That told me the difference in gravational attraction between point A and point B on the
earth is 6.74630 e 18 nm squared per second.


That's way short of the 1.98 e 19 nm squared precessional pull the moon is suppose to provide. Also, it appears if the moon is attempting to pull up the part of the earth's
equaterial bulge closest to it, then the pull on the part of the earth's equaterial bulge furthest from the earth would have to be substracted from that upward pull to measure
the actual precessional torque experienced by the earth.


To compute the gravational pull of the moon of the part of the Earth furthest from the moon, point C, I divided the distance between the moon and the center of the earth (384,400,000 m),
by the distance between the moon and the furthest point on the Earth (390,778,000 m), point C. Again, I squared that answer, since gravational attraction becomes four times greater when the seperating distance is halfed, and got .9676238. I then multiplied that figure times 1.98253140277806351994 e 20 nm squared per second. That told me the gravational attraction of the moon on point C on the earth is 1.918344586 e 20 nm squared per second.
I then substracted that answer from the 1.9825 e 20 nm squared per second answer. That told me the difference in gravational attraction petwen point C and point B on the earth is 6.41868 e 18 nm squared per second.


I substracted the difference in gravational attraction between the closest and furtherest points on the earth to the moon from each other and found out the difference between the pull of the moon on points B and C on the earth, is
3.27618 e 17 nm squared per second.


Again, that 3.27618 e 17 nm squared per second difference in pull, is far short of the 1.98 e 19 nm squared per second precessional torque the moon is suppose to provide.  

Answer
Hi, Wayne,

I'm sorry, but it has been many, many years since college physics and I'm out of my league here.  What I DO remember about precession is that it is caused by an asymmetry in the mass distribution in the earth, combined with the 23-degree tilt.  

If the mass of the earth were uniformly distributed through its volume, the gravity of the sun would act strictly on its center of mass.  However, the slight asymmetry causes a torque on the earth, which is, of course a vector.  The spin of the earth is expressible as a second vector, and the vector product (cross-product) of these is the precession.  As to the numbers involved, sorry.

Paul