Torque imposed by the Moon and Sun on the Earth given its precession

[ago01]

Homework Statement

What torque, in Nm, must be applied by the sun and moon to create the observed precession of 26,000 years?

Relevant Equations

Angular momentum, precession, torque

From the givens:

Approximate Earth as a sphere:

$I_e = \frac{2}{5}MR^2 = \frac{2}{5}(5.97x10^{24})(6.371x10^6)^2 = 9.69x10^{37} kg*m^2$

$\omega_e = 7.29x10^{-5} \frac{rad}{s}$

To calculate the rate of precession of the disk the Earth precesses around (1 revolution every 26,000 years):

$\frac{1}{8.199x10^{11} s} * 2*\pi = 5.15x10^{-10} \frac{rad}{s}$

Given the equation of angular precession: $\omega_p = \frac{rMg}{I\omega}$

$rMg = I\omega\omega_p = (9.69x10^{37})(7.29x10^{-5})(5.15x10^{-10}) = 3.638 × 10^{24} N*m$

But it appears this answer is incorrect. I had thought that since in deriving $\omega_p$ we cancel $\theta$ the angle between the planets wouldn't matter and $rMg$ would be our torque. However, this seems not to be the case.

What have I done wrong?

 

[haruspex]

It's very hard to follow your reasoning when it's all in numbers. Please post a purely algebraic version.

 

[ago01]

It's very hard to follow your reasoning when it's all in numbers. Please post a purely algebraic version.

To be honest I'm not sure what else I can do to make it more algebraic. But I'll take a shot.

Moment of inertia of the Earth (approximated as a solid sphere):

$I_e = \frac{2}{5}MR^2 = \frac{2}{5}(5.97x10^{24})(6.371x10^6)^2 = 9.69x10^{37} kg*m^2$

I used the mass $5.97x10^{24} kg$ and the radius $6.371x10^6m$ from reference.

Angular velocity of the Earth about it's axis (some unit conversions):

The Earth completes 1 rotation about it's axis in 24 hours. So to solve for $\omega_e$ I get:

$\omega_e = \frac{1}{24 hours} * \frac{1}{60 min} * \frac{1}{60 s} * 2\pi = 7.29x10^{-5} \frac{rad}{s}$


From the givens the Earth precesses once every 26,000 years. So this is just another conversion.

Precession of the Earth in seconds (some unit conversion):

$\frac{1}{8.199x10^{11} s} * 2*\pi = 5.15x10^{-10} \frac{rad}{s}$

Where $8.199x10^{11} s$ is 26,000 years in seconds.

Calculating the torque:

From my book the precession angular velocity is equal to (treating the Earth as a gyroscope):

$\omega_p = \frac{rMg}{I\omega_e}$

It's derived by:

$T = rMg\sin{\theta}$
$dL = rMgsin{\theta}dt$
$d\phi = \frac{dL}{Lsin{\theta}} = \frac{rMgsin{\theta}}{Lsin{\theta}}dt = \frac{rMg}{L}dt$

Since $\omega_p = \frac{d\phi}{dt}$

$\omega_p = \frac{rMg}{L}$

or

$\omega_p = \frac{rMg}{I\omega_e}$So, I was lead to believe that the numerator represents a torque and I could solve for it. So moving some variables around:

$rMg = I\omega_e\omega_p$

and

$rMg = I\omega\omega_p = (9.69x10^{37})(7.29x10^{-5})(5.15x10^{-10}) = 3.638 × 10^{24} N*m$

The values of which I derived above. Hope this helps understand my reasoning. It's really hard to determine if this number even makes sense as everything in space is quite large.

 

[Orodruin]

To be honest I'm not sure what else I can do to make it more algebraic.

You don’t insert any numbers until you have the very final algebraic expression.

Edit: I also suggest using \times or \cdot for the multiplication in your TeX code instead of the variable x.

 

[ago01]

You don’t insert any numbers until you have the very final algebraic expression.

Edit: I also suggest using \times or \cdot for the multiplication in your TeX code instead of the variable x.

That's usually what I do. However, the only algebra I did really was the last part where I manipulated the equation (as I did in the original answer). That is why I was confused. The rest was just unit conversions that, admittedly, could be made more clear. Which I tried to do in my response.

 

[haruspex]

treating the Earth as a gyroscope

The formula for precession involves a cross product. How might that affect your equations?

 

[mjc123]

There is a mistake in your arithmetic. I calculate the rate of precession as 7.66e-12 rad/s, not 5.15e-10.