# Torque and Precession

1. Dec 4, 2008

### mdewdude

1. The problem statement, all variables and given/known data

The earth's axis of spin precesses just like that of a top or a gyroscope. (This is why the North Star moves away from the north over many centuries.) The cause is the torque on the earth due to the gravitational pull of the moon and the sun on the earth's equatorial bulge. The time for a complete cycle of this precession is 2.60×104 years.

How much combined torque do the sun and moon exert on the earth to cause the observed precession?

2. Relevant equations
Torque=r*f

3.

To find the combined torque, i had to find the force that the sun exerts on the earth. I did this by finding the gravitational force which was

6.674*10^-11 ((1.9891*10^31 kg)(5.97*10^24kg))/ (149600000 km*1000)^2 the force should equal 3.53 *10^22 N the moon equals out to be 1.98 *10^20

Now i have the forces so i plug it into the equation t=r*f

t= 3.53 *10^22 N *1.49*10^11 m=5.28*10^33

Moon: t= 1.98*10^20 N * 384403000m=7.6*10^28

The combined torque would be 7.6*10^28 + 5.28*10^33 = 5.28*10^33

Am i using the wrong equation to find the force of the moon and sun exerted on the earth?????

2. Dec 4, 2008

### jambaugh

Yes. The formula you are using is for a force f applied off center by a distance r.

Your approach is wrong in that figuring the torque from tidal force is too difficult a problem.
Rather knowing that the torque is inducing a precession i.e. effecting a time rate of change in the angular momentum.

The relevant equation would then be:
$$\frac{d\vec{L}}{dt} = \vec{\tau}$$ where $$\vec{L}$$ is the vector angular momentum and $$\vec{\tau}$$ is the vector torque.

Calculate the angular momentum of the Earth using its angular velocity and the moment of inertia for a spherical mass. Then recall that this is a vector parallel to the axis of rotation. Figure (assuming the angular speed doesn't change) the time rate of change of this vector due to its change of direction. You will need more information, in particular, the angle of the cone in which the Earth's axis precessed.

Otherwise you'll have a nasty vector integration problem using the Moon's and Sun's $$k/r^2$$ force integrated throughout the earth's volume as an ellipsoid. You'll need the amount of bulge and the angle off of the solar and lunar orbit planes and also you'll need to time average over orbits. Just a nasty nasty calculation!

Hmmm... well you can look up the earth's mass quadrapole moments and the sun's and moons tidal forces to get a good approximation... then time average over the respective orbital periods. Still as the problem reads I would think the best method is to directly figure out the time rate of change of angular momentum.

3. Dec 4, 2008

### mdewdude

The angular momentum is 2.663 * 10 ^40 kgm2s-1 of the earth and the sun and the angular momentum of the earth and moon is 2.88E+34. The angle of the tilt of the earth is 23.8 degrees. and when i convert the years into seconds i got 8.19*10^11 sec. So after i found the angular momentum., the sun would be (2.663*10 ^40)/8.19 *10^11 sec. That would equal to 3.25 *10^28 so when i add the moon i get a sum value of 3.25 *10^28. I guess im confused at plugging the proper numbers into the equation