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

In summary: This affects your equations because the torque (which is rMg) is multiplied by the angular velocity (which is I\omega_e).
  • #1
ago01
46
8
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?
 
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  • #2
It's very hard to follow your reasoning when it's all in numbers. Please post a purely algebraic version.
 
  • #3
haruspex said:
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.
 
  • #4
ago01 said:
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.
 
  • #5
Orodruin said:
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.
 
  • #6
ago01 said:
treating the Earth as a gyroscope
The formula for precession involves a cross product. How might that affect your equations?
 
  • #7
There is a mistake in your arithmetic. I calculate the rate of precession as 7.66e-12 rad/s, not 5.15e-10.
 

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

1. How do the Moon and Sun affect the Earth's precession?

The Moon and Sun both exert a gravitational force on the Earth, causing it to bulge slightly towards them. This bulge creates a torque on the Earth's axis, which in turn affects its precession.

2. What is the magnitude of the torque imposed by the Moon and Sun on the Earth?

The magnitude of the torque depends on the distance between the Earth and the Moon/Sun, as well as their masses. On average, the Moon's torque is about 2.2 times stronger than the Sun's, due to its closer proximity to the Earth.

3. Does the torque from the Moon and Sun cause the Earth's precession to speed up or slow down?

The torque from the Moon and Sun causes the Earth's precession to speed up. This is because the Earth's axis is tilted at an angle, so the torque from the Moon and Sun is not aligned with the axis and therefore creates a rotational force.

4. How does the Earth's precession affect its climate?

The Earth's precession has a small but significant impact on its climate. It affects the distribution of sunlight on the Earth's surface, which can lead to changes in weather patterns and the amount of solar radiation received by different regions.

5. Is the torque from the Moon and Sun the only factor affecting the Earth's precession?

No, there are other factors that can also affect the Earth's precession, such as the gravitational pull of other planets and the Earth's own shape and rotation. However, the torque from the Moon and Sun is the most significant factor in causing the Earth's precession.

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