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Tides

Started by walkingcarcass May 23, 2003 at 7:31 AM 15 replies 1k views

Original Post

May 23, 2003 at 7:31 AM

I''m sure many of you could tell me why the oceans bulge towards the moon AND away from it on the opposite side, but I started thinking about the way the water moves over time after seing a SHM simplification of tide depth. BTW I''m not asking you to solve this! Take a perfectly spherical stationary planet radius R mass M in isolation with a uniform liquid of density d and depth D all over it. This liquid flows instantly into equilibrium shape. Add to this system a satelite mass m at a distance r from the centre of M in a circular orbit with period T. Take a cross-sectional view of the planet in the plane of orbit. Taking the origin of reference to be the centre of the planet, initially the satelite is seen at (r,0) and moves counter-clockwise. What is the shape of the ocean surface? An ellipsoid? After a given amount of time, what is the depth of water at (x,y,z)? (or in terms of latitude/longitude/whatever) ******** A Problem Worthy of Attack Proves It''s Worth by Fighting Back

vanillacoke

May 23, 2003 at 9:48 PM

Oh, come on, that's easy! All you have to know is fluid dynamics and three-variable second order differential equations.

Honestly, though, I don't even know how to solve for r(t) for a planet in orbit. I wonder how (or if) the planet positions affect the fluid systems of the sun.

It would probably be significantly easier to figure out the shape for a stationary mass m (just have the bodies separated by an infinitely rigid, infinitely thin, massless rod). I might even be able to figure that out... but I doubt it.

edit: because I'm insane

[edited by - vanillacoke on May 23, 2003 10:53:13 PM]

GaulerTheGoat

May 23, 2003 at 11:06 PM

Hmmm... Good question! I stink at physics, though, so don''t expect any help from me. But, according to my encyclopedia, "Toward the moon, the lunar attraction is added to a comparitively small centrifugal force; in the opposite direction it is subtracted from a much larger centrifugal force." The reason is this: The center of gravity of the Earth and Moon lies on a line inbetween them. As the Earth and moon rotate, it causes a centrifugal force on the side of the Earth away from the moon. That causes one tide and the Moon''s gravity causes the other! Cool, I guess...

BTW, the equation of the planet r (t) = r because it has a fixed radius according to the problem. But the polar coordinates would be (r, 2pi t/T) or x = r cos (2pi t/T), y = r sin (2pi t/T).

no wise fish would go anywhere without a porpoise - The Mock Turtle

walkingcarcass

Author

walkingcarcass

May 24, 2003 at 9:49 AM

Centrifugal force doesn''t exist.

What i''m wondering is what kind of problem is this, one that can''t be solved except by numerical approximation perhaps.

If the tide bulges out the other side because of the sun''s gravity, what happens when the sun, moon and earth are not colinear?

********

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Proves It''s Worth by Fighting Back

vanillacoke

May 24, 2003 at 10:01 AM

No, by r(t) I meant the elliptical orbit of a single planet around a single sun. In an ideal situation, the ellipse becomes a circle, and that''s easy. For an elliptical orbit, I don''t know how to get r(t) solely from the law of gravitation, but I know that it''s possible.

Punty50

May 24, 2003 at 10:19 AM

Each mass will cause it's own tide, and the actual tide will be the superposition of the various tides.

Brendan

[EDIT]
Vanilla- Look up the conic sections formula on wolfram. Those are in terms of theta, and I am sure that you can integrate your radial velocity and acceleration to get theta in terms of t.

[edited by - Punty50 on May 24, 2003 11:20:45 AM]

GaulerTheGoat

May 24, 2003 at 4:14 PM

quote:
walkingcarcass
Centrifugal force doesn''t exist.

My physics is weak... what is that force that pushes you out the side of a car when you make a turn?

no wise fish would go anywhere without a porpoise - The Mock Turtle

mwtb

May 24, 2003 at 5:24 PM

I''ll probably get this horribly wrong, but as far as I know there is no force pushing you out. All that is happening is that your body is being accelerated as the car turns and the force you feel is that acceleration through whatever is holding you into the car. If you weren''t being held in, no force would be applied and you''d just merrily continue with the same velocity as you had before the car turned without you. So, the "centrifugal force" is actually just Newton''s 1st law of motion in action.

mwtb

Punty50

May 24, 2003 at 9:37 PM

That force is the force of your seat/surroundings forcing you to go in a circular or curved path, which is quite different from the straight line you want to go in. By newtons first law, you want to go in a line, but, because you are turning, there has to be a force (friction) causing you to do so.

Brendan

GaulerTheGoat

May 24, 2003 at 11:00 PM

I don''t like physics. I like vectors... and DEs (especially PDEs...) but I don''t like physics!

(Sorry ''bout this pointless post.)

JohnBolton

132

JohnBolton

May 25, 2003 at 3:41 AM

quote:
Original post by walkingcarcass
Centrifugal force doesn''t exist.

That statement is analagous to this statement:

Gravity doesn''t exist.

Drilian

108

Drilian

May 25, 2003 at 3:45 AM

It's actually "Centrifical force doesn't exist"

Centrifugal force is very much alive and well in the minds of physicists everywhere.

EDIT: Whoops. I got the word "Centrifugal" and "Centripital" mixed up. In retrospect, I don't think there is Centrifugal force after all. But CENTRIPITAL force (regardless of the correct spelling) is very much alive and well in the minds of physicists everywhere :D

[edited by - Drilian on May 25, 2003 11:20:40 AM]

walkingcarcass

Author

walkingcarcass

May 25, 2003 at 4:27 AM

If you''re in a car turning a corner and you feel a force throwing you out, it''s just the car pushing in on you as it feels a centripetal force causing the turn. Centrifugal force is an illusion by virtue of inertia, it''s because of Newton''s 3rd law.

It''s not analagous to "gravity doesn''t exist" but it is analagous to "pulling doesn''t exist".

But this isn''t the point of the thread.

********

A Problem Worthy of Attack
Proves It''s Worth by Fighting Back

Cedric

May 25, 2003 at 9:35 AM

quote:
Original post by walkingcarcass
If you''re in a car turning a corner and you feel a force throwing you out, it''s just the car pushing in on you as it feels a centripetal force causing the turn. Centrifugal force is an illusion by virtue of inertia, it''s because of Newton''s 3rd law.

Centrifugal forces don''t "exist" only in an inertial frame of reference. In the frame of reference of the car, you can say that centrifugal force is pulling you to the side.

In fact, one of the ways to solve the ellipsoid orbit problem is to use an effective potential which is the gravitaionnal potential plus the potential associated to the centrifugal force. This allows you to find the shape of the ellipse. To find r(t), you can use 0.5*r²* dtheta/dt = L / 2m.

Tides is a non-trivial problem. I''m sure that you can find discussions of it around the net.

Cédric

Cedric

May 25, 2003 at 9:55 AM

Ok, given a force field F acting on your sphere and its fluid, what you want is to find an equilibrium position for the fluid (i.e.: a function r(latitude, longitude)). We'll assume that the equilibrium position is always reached, for a given position of the moon. You need a function Pressure(r, F), and you want Pressure to be equal for every lat, long.

Pressure(r, F) = (G * |r| + F Dot r)

This is a guess, and it's probably wrong because it negligeates the component of the force perpendicular to F.

Now, F is pretty easy to find, it's the gravity field associated to the moon. It can probably be treated as constant. F can also include the centrifugal force.

Cédric

EDIT: Changed the pressure formula to what I originally intented; not that it really seems to matter

[edited by - cedricl on May 26, 2003 8:32:48 AM]

walkingcarcass

Author

walkingcarcass

May 26, 2003 at 8:55 AM

Interesting approach, that.

********

A Problem Worthy of Attack
Proves It''s Worth by Fighting Back

vanillacoke

May 27, 2003 at 10:22 AM

Or you could find an equilibrium position for a perfect fluid with the gravity of the moon, earth, and the fluid acting on it. I would try a simulation with 10^30 particles of water.

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