# Tidal force, torque and precession

1. Jun 26, 2006

### tim_lou

imagine a sun and a planet, both rigid and perfectly spherical. the distance between the center of mass of these two bodies is much greater than the radius of both bodies.

considering newton's inverse square law of gravity:
now, due to the non-uniform nature of this gravity, different parts of the planet experiences different forces. let's say the planet is spinning around a axis with a non-zero inclination angle. the tidal forces would actually result in a torque even though the planet is perfectly spherical and it would result in precession. i have worked out some basic intrgrations and approximation regarding this problem. perhaps someone can tell me if it has been discovered before and information about it? i know tidal force... but what about "tidal torque"?

2. Jun 26, 2006

### Staff: Mentor

3. Jun 26, 2006

### tim_lou

is there any actual calculation/derivation on these things? i couldnt find any online.

4. Jun 26, 2006

### Tide

I'm not sure you'll have any tidal effects if both objects are "rigid and perfectly spherical."

5. Jun 28, 2006

### Meir Achuz

There would be no torque if the planet were "perfectly spherical".
The earth's precession is due to its slightly uneven shape.

6. Jun 28, 2006

### Andrew Mason

Yes. If the center of mass is not the physical centre of the sphere the sphere will experience a tidal torque from a single gravitational source. In the earth's case, however, I think the reason for precession has more to do with the gravitational forces being supplied by two sources, the moon and earth.

AM

7. Jun 28, 2006

### tim_lou

that is i think why the idea is catchy. there is torque if the planet is spinning on a tiled axis.

the center of mass does not always coincide with the "center of gravity" in a non-uniform gravitional field. hence there will be a torque....

i do not know to what extent this torque affects the precession. i do know that it is quite significant.
it is approximately proportional to radius^2/distance^3 and sin(angel of inclination*2)
(if my math works out are correct... i did some simple integrations)

i dont konw how much the moon contributes to the torque... maybe someone knows? i couldn't find it on google.com or anything...

8. Jun 29, 2006

### pervect

Staff Emeritus
I agree. As far as the request for sources go, I would suggest that the OP look it up in "Classical Mechanics" by Goldstein. In the second edition, Goldstein ssay:

pg 225

A useful formula here is MacCullagah's formula is derived on pg 228, one can find various online sources as well that also cover some of this material, for instance

http://scienceworld.wolfram.com/physics/MacCullaghsFormula.html
http://geophysics.ou.edu/solid_earth/notes/maccullagh/maccullagh.html

This formula gives the potential function for a point mass orbiting a non-spherical planet. What is important are the moments of inertia of the planet around it's three principle axis (A,B, and C on the scienceworld webpage).

It's common to assume A=B (see the bottom page of the scienceworld url), then one can see that the potential function for a point mass consists of the usual potential plus a multi-pole potential proportional to (C-A).

In some cases (for instance the tidal torque on the moon), one has to deal with the fact that the Earth isn't a perfect figure of revolution (i.e. A is not equal to B), but usually it's not necessary, one usually only has to worry about the fact that C is not equal to A.

Sometimes the factor J2 is used instead to represent the oblateness parameter (see the bottom of the second webpage above).

9. Jun 29, 2006

### pinestone

10. Jun 29, 2006

### tim_lou

thx man, never know that these ideas and formulae existed until you pointed it out. too many names, too little explicit explainations on formulae until you learn them... hmm gotta take more time to digest it.

i have to disagree on this one.... i cant explain very well...so i'll post a diagram.

the "center of gravity" does not coincide with the center of mass... (i know "center of gravity" is not a very good word in non-uniform gravitional field...but i dont know how else i can explain it).

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11. Jun 29, 2006

### tim_lou

if you rotate the planet such that the z-axis is the axis of rotation(the planet is spinning around it),
theata is the angle of inclination.
the net torque would be....
I'm pretty sure that the integral isn't zero. there is no symmetry, and i did some approximation which does not really equal zero.

correct me if i'm wrong.

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12. Jun 29, 2006

### tim_lou

i used a lot of approximations and the integral turned out to be approximately,
$\tau=\frac{3GMmr^2sin{2\theta}}{5d^3}$

i will explain my results if anyone thinks that my logic is correct.... i gave this project to my physics teacher and he didnt have much to say...

if my idea is correct, then i'll continue to work on and see what the precession period will be solely based on this idea (for a planet in a spherical orbit). if this is significantly smaller than other theories predicted... we might have something here. (i'll need a lot of help from you guys though)

Last edited: Jun 29, 2006
13. Jun 29, 2006

### pervect

Staff Emeritus
The Earth is basically very close to an equipotential surface (with minor variations, such as mountain ranges and valleys).

Because it is rotating, it has a distinct equatorial bulge

Without this equatorial bulge, there will be no torques on the Earth. The rotation doesn't matter to the force in Newtonian gravity. The force between two perfect spheres acts as if the force were at the center of mass - it doesn't matter whether or not they are rotating. It is only imperfections in the spheres (due to equatorial bulges or tidal bulges) that can cause torque.

I'd have to see more of your derivation to find your mistake - but as I said, Goldstein works thorugh a lot of this stuff, and he introduces the chapter in which he does so with the remark that I quoted, about how no planet could produce a torque on the Earth if it were perfectly spherical.

14. Jun 29, 2006

### tim_lou

actually the idea came from how spinning ball on a table curve... so i thought why not gravity?

also, from wikipedia:
http://en.wikipedia.org/wiki/Center_of_gravity
"The center of mass is often called the center of gravity because for at least two purposes, any uniform (constant) gravitational field g acts on a system as if the mass were concentrated at the CM"

however, newton's law of gravity is not uniform.
"If the gravitational field acting on a body is not uniform, then the center of mass does not necessarily exhibit these convenient properties concerning gravity."

quote from wikipedia from feyman's lecture on physics:
"The center of mass is sometimes called the center of gravity, for the reason that, in many cases, gravity may be considered uniform. ...In case the object is so large that the nonparallelism of the gravitational forces is significant, then the center where one must apply the balancing force is not simple to describe, and it departs slightly from the center of mass. That is why one must distinguish between the center of mass and the center of gravity."

that's why i thought perhaps if a planet is big enough, the approximation of gravity acting on the center of mass is not enough...

i honestly do not have Goldstein's book... so i do not know any of his argument.

My math is not all that good... and my derivation is somewhat lengthy (10 pages with diagrams), it also uses a different x,y,z axis for integration... i'll have to clean that up before posting it here....

Last edited: Jun 29, 2006
15. Jun 29, 2006

### tim_lou

my math is really messy... the whole thing has more than 10 pages....and i approximate the first integral over and over again so that it can be integrated again... anyway, i cut all the fat in the following article.

the following is a simple "proof" that i put up to prove that the torque is indeed non-zero. i dont know if it is correct, but i checked it many times... although i often make stupid little mathematical mistakes. i has the "proof" but non of the "fat".

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16. Jun 30, 2006

### pervect

Staff Emeritus
I think that the torque should be (x,y,z) x force, that that is where the error is. (You have (x,y,0) x force).

Think about an x-component of force on the top of the sphere, for instance (0,0,zmax).

There will be a torque along the y axis due to this force. You are not including this component of the torque in your analysis of the y-component of the torque.

On a related issue, there should be no difference in the torque produced on a sphere when it is rotating or when it is not rotating. The torque is still radius x force

So you can simplify the problem immensly more than you've done - the rotation doesn't matter to the torque, and there is spherical symmetry to the problem (not just axis symmetry).

Last edited: Jun 30, 2006
17. Jun 30, 2006

### tim_lou

the reason why it is <x,y,0> is because z is the axial of rotation. <x,y,0> is the moment arm, the component of the position vector that is orthogonal to the rotation axial. i have carefully thought about what should happen when it is non-spinning:

lets assume that the planet is not spinning, and my theory is correct. then, it will experience a torque in the y direction. Let's say now it will start to spin and gain angular momentum in the y direction. however, that mean the axial of rotation will the y axis, not the z axis, a contradiction. now, if the axis of rotation is the y-axis, no torque will be produced. the "moment arm" is paraelle to the "net gravitional" force. r cross F=0

however, the torque can exist when rotation exists. the angular momentum will simply go into precession. energy is conserved because the torque is always orthogonal to the L vector.

hmm, <0,0,zmax>, the moment arm is a zero vector<0,0,0>...it is a point at the axial of rotation...r cross F=0. yes, there will be a torque produced when another axial is chosen. but... how would it make sense to pick a rotation axial other than the one that the planet is spinning around? it would lead to contradiction at the end...

18. Jun 30, 2006

### pervect

Staff Emeritus
The spinning planet is like a gyroscope. When you apply a torque to it, it will tend to precess. But this does not change how the torque is computed. One still uses r X F to compute the torque.

It's easiest to see how a x-directed force on the non-spinning north pole of the Earth, spinning about the z axis, causes a torque on the y direction. But the calculation for torque is the same regardless of whether the Earth spins or is stationary. The velocity is not relevant to the caclutation of torque - it is still rXF, even if the point to which the force is being applied by gravity is moving.

What does happen is that

$$\vec{\omega} = \vec{\omega_0} + \vec{\tau} t$$

where $\omega$ is the angular velocity of the Earth (expressed as a vector), and $\tau$ is the torque, expressed as a vector.

This means that if one applies a toruqe along the y axis to an Earth spinning on its z axis that spin axis moves in the y-z plane, as one expects from a vector addition. You can feel this behavior if you apply torque (i.e. try to twist) a spinning gyroscope.

19. Jun 30, 2006

### tim_lou

i see what you are saying, around the y-axis, there will be torque produced by a force in the x-direction at the north pole.

do you agree that there is a non-zero torque produced by the sun around the axis of rotation?
so, you are saying that this torque will somehow be canceled by torque produced in a different axis right? and the "net" torque should be the sum of all torques calculated in all 3 principle axis right?

btw, spot your typo there, hehe, you meant $$\vec{L}=\vec{L_0}+\vec{\tau}t$$ ?

Last edited: Jun 30, 2006
20. Jun 30, 2006

### tim_lou

after much more thinking... i think my mistake is really not taking into account of torques calculated in other principle axis since informations are lost when only one particular axis is chosen for calculation

I guess in 3D free space things are different than normal day situations..... well, at least i learnt it the hard way... 10 pages of "attempting" to derive something incorrect go to waste. Anyway, at least now i understand the idea of rotations and torque much much more clearly (and i had fun).

how do you actually calculate the net torque though? pick an axis then... ? what are the actual equations used to calculate the net torque in 3 dimensional spaces?

i still need to varify that the torque is indeed zero when it is all added up though...

Last edited: Jun 30, 2006