- #1

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- 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?

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?