Clarification about Earth's Precession

[bznm]

I haven't a textbook where I can study Earth Precession and I'm not sure to have correctly understood... Could you tell me if I do mistakes?

I consider two points P and Q on the equatorial bulge. The torques of the gravitation forces between P-sun and Q-sun (calculated wrt the Earth center) are not equal and opposite, so I have a $\tau_{tot} \neq 0$ that generates $\Delta L$ ($\tau=dL/dt$).

$\vec{\tau_{tot}}=\vec{\tau_p}-\vec{\tau_q}=\vec{R}\times\vec{F_p}-\vec{R}\times\vec{F_q}=\vec{R}\times\vec{\Delta F}$

$\Delta F=\displaystyle \frac{4G mMR^2 cos \theta}{a^3}$

so $\tau_{tot}=\displaystyle \frac{4G mMR^2 cos \theta}{a^3} Rsin \theta$

but $\tau_{tot}=\displaystyle\vec{\Omega_p}\times{L}=\Omega_p L sin \theta$ . From this last relations, I obtain:

$\displaystyle \frac{4G mMR^2 cos \theta}{a^3} Rsin \theta=\Omega_p L sin \theta$ -> $\Omega_p=\displaystyle\frac{4GmMR^2 cos \theta}{a^3 I \omega_{rot} }$

Is it correct? My colleague wrote on his blocknotes $|\Omega_p||L|=|\tau_{tot}|$ so $\Omega_p=\frac{2GmMR^2 sin2 \theta}{a^3 I \omega_{rot} }$

 

[ORF]

Hello

I'm not sure, but maybe it's that the angle between L and Ω_p is 90deg, and not θ. Drawing the vectors may help.

Greetings

 

[bznm]

Mmmmh. $\theta$ is the angle between the Earth rotation axis and the Earth direction axis that we would have if the Earth wasn't "crooked"...

 

[ORF]

Hello

I'm not completely sure, but if nobody answers... I think that the Ω_p is always perpendicular to the ω_{rot}; the tricky point here is to see that the direction of Ω_p is not constant, but its module is (ie, the frequency of precession is constant).

So the angle between Ω_p and L would be 90deg, and not theta, and you will keep the sin(theta) in the cross product, in order to get 2sin(theta)cos(theta) = sin(2theta), as your classmate. I still think that a drawing is useful (I'm not native English speaker, and it would be easier also for me ;) ).

Another point: I think there is another mistake in the second equation; maybe it's just R not R squared (magically it desappears in the next steps ;) ).

Greetings! :D

 

[bznm]

The setting should be this:

I can't understand the reason why $\Omega_p$ should be perpendicular to $w_r$.. see also

http://oceanworld.tamu.edu/students/iceage/images/precession_1.jpg

 

[ORF]

Hello

After thinking carefully, it's pretty probable I were wrong.

I found a very complete explanation about this problem:

http://astro.cornell.edu/academics/courses/astro6570/Precession_Free_and_Forced.pdf

In the page 7/19 it's shown the same formula that you proposed, the frequency of precession depends on cos(theta), and not on sin(2*theta).

Thank you for making me think in this issue ;)

Greetings