I have a few problems with your description of tides. Now I have to admit that I am a little hazy about this myself, but there are parts of your description that just seem wrong.

You make quite clear that the apparent absense of force in free fall is an illusion. The best example of this is if you jump out of a plane with scales glued to your feet, if you ignore air resistance you will appear weightless, but you are still being accelerated towards the ground.

You then go onto say that the centre of the Earth is in freefall. I am not sure how one part of a solid body can be in free fall without the rest of it also being in freefall. But this is not important, the bad bit is you claim that if you measure the gravitional force of the moon relative to the centre of the Earth and then define the centre of the Earth as having zero force then you get a negative force on the far side. This is a bit like claiming that if I define the gravitational force of the Earth at ground level as zero, I should be feeling a force away from the Earth on the first floor. Gravity always acts towards the centre of mass, it doesn't matter where you stand on the Earth the moons gravitational force is always acting towards the moon.

There is however another 'force' involved here, which you mention when talking about the effect of the Sun. That is the centrifugal 'force'. I know as another post has already pointed out in another post that this is not realy a force, but in this case it is helpful to think of it as such. If you include the centrilfugal force much of your description begins to make more sense. At the centre of the Earth, the gravitational force and the centrifugal force balance and it could be said that a body here has no force acting on it. Now a person standing under the moon will be subject to the same centrilfugal force, but because they are closer to the moon the gravitational force will be stronger than at the centre of the Earth and so they will be pulled towards the moon. Where as the person on the far side will again have the same centrilfugal force but a lower gravitational force and will feel a force pushing them away from the moon.

I realize I have gone on a bit here and I hope I have made myself clear.

Simon

Gravity does not always act towards the center of mass. For instance, if you were to draw a straight line down from Chicago, you'd miss the center of the Earth by about twenty kilometers. If you subtract the effect of the rotation of the Earth, you'd still miss the center by ten km.

The centrifugal force due to the Earth and moon's mutual orbit about each other is constant on the surface of the Earth--it is essentially the same magnitude and direction everywhere on the Earth. It doesn't change from one side of the Earth to the other. What does change, whether the Earth is orbiting the moon or not, is the differential of the strength of gravity from one side to the other--just as BA says.

Here's an alternative way to think about it. It's easier to use the Moon as the example, but the same thing applies to the Earth (which is in orbit around the barycenter of the Earth-Moon system).

The Moon is in orbit. But it is not a point mass; it has extent (3400km). Now, point mass at a distance from Earth corresponding to the Moon's near side would naturally orbit at a different speed than the Moon does (a slower linear speed). Likewise, an object at distance of the far side of the Moon would naturally orbit at a faster linear speed. This is simply an application of Kepler's laws.

So, the only things that keep the near side of the Moon from shearing off and going into a slightly different orbit are the Moon's own self-gravity, and the tensile strength of the rock. Likewise for the far side.

But one can easily see that the Moon undergoes stress because of these forces. If you work out the magnitude and direction of the forces over the whole Moon, the net effect is a "stretching" force along the Moon-Earth axis: the near side is stretched toward the Earth, while the far side is stretched away. In a sense, the Moon is trying to tear itself apart and spread out into a ring... which it would do, if it were so close to the Earth that the tidal stretching could overcome its gravity and tensile strength.

This stretching force is entirely due to the difference in gravity from the near side of the Moon to the far side, just as the difference in orbital speed of satellites at different altitudes is due to the difference in gravity at those altitudes.

(Note that I'm referring to orbital speed, not orbital period. Lower orbits are smaller in circumfrence, so they complete sooner even though they are slower. The space shuttle completes an orbit in some 90 minutes, while a geosynchronous satellite requires 24 hours; yet the latter is moving much faster than the former.)

For the Moon, this argument is relatively straightforward, because the E-M barycenter is (from the Moon's perspective) pretty close to the Earth itself. The case of the Earth is trickier, because the barycenter is actually inside the planet; but the same argument holds nevertheless. (For you mathophiles, to do the integral of the forces on the Earth you must include parts of the planet that extend past the barycenter, so the sign of the vector changes in those regions. However, the net result is the same as for the Moon, and for the same reasons.)

You got that backwards. The nearside of the moon is trying to orbit faster than the center of mass and the farside is trying to orbit slower than the center.

The point I was trying to get across is that the tides are an ineraction between the strength of gravity of the moon and the centrifugal force (or inertia) of the different parts of the Earth.

At the bottem of page 68 the book says:

It seems paradoxical that gravity can act in such away as make something feel a force away from an object, but in this case it's because we are measuring the force relative to the centre of the Earth. When you do that, then you do indeed get a force pointing away from the moon on the far side of the Earth.

This is simple wrong the gravitational force does only act towards the moon and it doesn't matter where you measure it relative to. The force which pushs the far side out is the centrifugal force (or inertia) which is not mentioned until the effect of the Sun is included, were it says that the tides are due to the effect of both centrifugal forces and gravity.

The point I was trying to make with the force always acting towards the centre of mass was that it always acts 'downwards' and was not able to push things apart, (Well at least on the scales we are talking about I am aware that some cosmologists are suggesting that the cosmological constant should be reintroduced) rather than define the exact direction of the force.

I hope I clarified my point and now I can get back to reading the rest of the very enjoyable book.

Simon

The key word is "relative." That makes it correct.

But, if the moon were falling straight towards the Earth, and there were no centrifugal force, there would still be a tidal force acting on the moon. The reason is the same, that there is a differential between the farside and the nearside. There is no difference in the centrifugal force from one side to the other, whether it is falling or in orbit.

As I mention above, that is wrong--it is able to tear things apart. And although the force of gravity always acts 'downwards,' that is a result of the definition--the direction is along the gradient. Gravity rarely acts towards the center of mass. Imagine you were on the near side of the moon of the Earth--would the force you feel be towards the center of mass of the Earth and moon? No, it would be in the opposite direction.

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<font size=-1>[ This Message was edited by: GrapesOfWrath on 2002-03-29 08:06 ]</font>

Thanks for your replies, I have never really sat down and thought about tides before, I usually just think of planets as point masses.

You got me on the centre of mass, it was a very poor use of language.

I think I have sorted some stuff out in my head. I started to think about astronauts, actually I started to think about a hole to the centre of the Earth openning up underneath me, but the problems of how to deal the mass above me made me think about astronauts instead. If you had three astronauts falling straight towards the Earth in a line, they would seperate with the lowest one being subject to a slightly large force than the middle one and so on. Relative to the lowest one the other two are moving up and thus could be said to be subject to a force away from the Earth.

I think this is where my problem lay, I find it hard to think of a astronaut plumetting towards the Earth as being subject to a force away from it.

I still have a few issues about the use of freefall to define the centre of the Earth as the zero point, but I will try and think these through and may get back to you with some more questions.

Thanks again.

Simon

And thank you for reminding me about my manners. I apologize for not welcoming you to the board! I didn't notice your post count, sorry.

Welcome to Phil's Wonderland.

Thanks, in retrospect my first post was a little strident.

Simon

The way I understood tides was like this. Imagine for a moment that the Earth is orbiting the Moon (in a way, it is). Therefore the water on the side nearest the Moon is travelling too slowly to orbit at that distance, so it tends to try and fall to a lower orbit, just as a spacecraft will fall to a lower orbit when it fires it's retros. The water on the side of the Earth opposite the Moon is travelling too fast to be in the orbit it is in, so it tries to 'climb' to a higher orbit, just as a spacecraft will increase it's orbital height when firing it's rockets along the line of orbit. The same thing happens in a sling shot. The sling's angular velocity at the inner end is way too fast for the outer end, so it swings out away from the centre.
Put it another way. Hammer throwers in athletics are swinging a weight around that is much less massive than themselves yet they end up leaning backwards before they let go of the hammer. Same effect, different scale.

I still have a problem, with the gravity only explantion.

In the case when the Earth (astronauts) are falling towards the moon, I can accept that relative to the the centre the far point feels a force away from the centre and hence away from the moon.

However in the case of the orbiting system, the distance between the centre of the Earth and the moon is fixed (well it would be if the orbits were circular). This means that when the point on the far side of the Earth moves away from the centre it is also moving away from the Moon, it is going up the gradiant, out of the potential well, how ever you want to describe the force of gravity, it is going the wrong way. It is feeling a force away from the moon. I don't see anyway to explain this without including inertia (or the centrifugal force) in the explaination.

To go back to my astronauts and trajectories where I feel more comfortable. If you start the three astronauts in a vertical line and give them all the same velocity (that required to but the middle astronaut into a circular orbit). Then the bottom astronaut will start to move ahead and down relative to the middle astronaut, as he starts on an eliptical orbit with his start point as apogee, the top astronaut on the other hand will move up and behind as he starts an eliptical orbit with his start point as perigee. Of course this can't happen with the Earth as the self gravity is far stronger than the tidal forces, but the example does describe the tidal forces involved.

To sum up my problem with the description, part of the Earth actually increases its distance from the moon due to tidal forces, I don't see any way of explaining this without using inertia (well you could probably do it with energy conservation, but thats a complication).


Simon

This example is...

...the same as that example, though. The strength of gravity is proportional to 1/r^2, and the tidal "force" (yes, it is a fictitious force just like centrifugal and coriolis) is the derivative of that force, proportional to 1/r^3. The computation is the same.

The tidal force exerted is the same, but the fact remains that part of the Earth moves away from the moon (and the centre of the Earth) it gains potential energy that energy has to come from somewhere, in the case of the astronaut it comes from his kinetic energy.

I am not claiming that the motion of the body has any effect on the forces felt only that it has an effect on the responce to the force, in this case to move away from the moon.

Simon

Gaining potential energy relative to which body? In the case of your three vertical astronauts, I guess it would be relative to the middle astronaut?

In the case of the Earth, relative to both the centre of the Earth and the moon. In the case of the Astronaut relative to the Earth, the top astronaut is not effected in any way by the other two.

Simon

I am not sure of the proticol of quoting between topic but from the centrifugal force topic:

[quote]

Still, he uses the term centrifugal force on page 7-5, to explain the tides and the motion of the Earth, and he says that the moon's attraction on the Earth is balanced by the centrifugal force, just as the BA has done. If you accept Feynman as the ultimate authority, I think you'd have to accept BA's use of the term.

[quote]

[img]/phpBB/images/smiles/icon_smile.gif[/img] I wish there was a coughing smiley.

BA doesn't mention centrifugal force in the explanation the tides on the Earth, due to the moon, which is the problem I have with the explanation.

Simon

LOL! Do you need a lozenge?

You can create a link to that other thread using BBcode "url" commands. The slashes are forward slashes.

The original post in that other thread mentions that BA says "The Earth feels a gravitational pull toward the Sun and a centrifugal force away away from it." That is what my words, in your quotation, is referring to.

As we've shown with the falling astronauts, it is not necessary to have a centrifugal force to have a tidal force. So, any explanation can include it, or not. The centrifugal force on the Earth is the same on the near side as on the far side as well as on the center--so once you make the center of the Earth the reference point, all centrifugal forces are cancelled.

I know but its nice to a have quote from Feynman agreeing with what you have been saying.

The centrifugal force is constant, but the tidal force isn't constant and it is the balance of these two 'forces' that defines the behavour. In case when the Earth is falling straight towards the moon the far side is trying to fall a little slower than the near side.

But in the case with the Earth in orbit, the far side moves way from both, the centre of the Earth, and the moon and this is due to the centrifugal force. The tidal force exerted is the same, but the effect is different, (I know its the same in the Earth centre frame of reference, but I think ultimatly you want to explain the effects in an inertial frame)

[img]/phpBB/images/smiles/icon_eek.gif[/img]ops

Thats wrong even in the Earth centre frame you can measure the distance between the far side and the Moon. In the falling case this will dropping quickly and in the orbiting case it will slightly larger than the distance between the Earth centre and the moon plus the radius of the Earth.

Simon


<font size=-1>[ This Message was edited by: SimonCB on 2002-04-02 07:17 ]</font>

Your original criticism in this thread was "the bad bit is you claim that if you measure the gravitional force of the moon relative to the centre of the Earth and then define the centre of the Earth as having zero force then you get a negative force on the far side."

I don't think Feynman would agree with you there at all. Physicists often shift reference frames in doing calculations.