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SimonCB
2002-Mar-28, 01:42 PM
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

GrapesOfWrath
2002-Mar-28, 04:35 PM
On 2002-03-28 08:42, SimonCB wrote:
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.

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.

Donnie B.
2002-Mar-28, 09:29 PM
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.)

Kaptain K
2002-Mar-29, 06:00 AM
...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...
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.

SimonCB
2002-Mar-29, 10:26 AM
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

GrapesOfWrath
2002-Mar-29, 12:56 PM
On 2002-03-29 05:26, SimonCB wrote:
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 key word is "relative." That makes it correct.


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.

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.


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,

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.

<font size=-1>[Fixed BBcode]</font>

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

SimonCB
2002-Mar-29, 03:07 PM
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

GrapesOfWrath
2002-Mar-29, 03:19 PM
On 2002-03-29 10:07, SimonCB wrote:
Thanks again.

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.

SimonCB
2002-Mar-30, 08:27 AM
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

johnwitts
2002-Mar-30, 11:13 PM
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.

SimonCB
2002-Apr-01, 08:29 AM
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

GrapesOfWrath
2002-Apr-01, 11:09 AM
On 2002-04-01 03:29, SimonCB wrote:
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.

This example is...


[Explanation of in-orbit example] I don't see anyway to explain this without including inertia (or the centrifugal force) in the explaination.

...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.

SimonCB
2002-Apr-02, 07:43 AM
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

GrapesOfWrath
2002-Apr-02, 08:46 AM
On 2002-04-02 02:43, SimonCB wrote:
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.

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?

SimonCB
2002-Apr-02, 09:02 AM
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

SimonCB
2002-Apr-02, 11:27 AM
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]

/phpBB/images/smiles/icon_smile.gif 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

GrapesOfWrath
2002-Apr-02, 11:52 AM
LOL! Do you need a lozenge?

You can create a link to that other thread (http://www.badastronomy.com/phpBB/viewtopic.php?topic=878&forum=9&41) using BBcode (http://www.badastronomy.com/phpBB/faq.php#bbcode) "url" commands. The slashes are forward slashes.



On 2002-04-02 06:27, SimonCB wrote:
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.

The original post in that other thread (http://www.badastronomy.com/phpBB/viewtopic.php?topic=878&forum=9&41) 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.

SimonCB
2002-Apr-02, 12:09 PM
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.



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



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.



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)

/phpBB/images/smiles/icon_eek.gifops

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>

GrapesOfWrath
2002-Apr-02, 12:46 PM
On 2002-04-02 07:09, SimonCB wrote:
I know but its nice to a have quote from Feynman agreeing with what you have been saying.

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.

SimonCB
2002-Apr-02, 01:27 PM
I don't think Feynman would agree with you there at all. Physicists often shift reference frames in doing calculations.



I know, and I hadn't fully grasped in my head what the problem was and so I expressed it badly. You do get a negative force relative to the centre of the Earth, but this does not explain why the far side of the Earth moves away from the moon, this is what I had meant by a negative force, and I admit this was careless use of language (plus the fact I had't realy thought about the tidal effect on a falling body).
The way I read the argument was a man at the top of the tower feels less force from the Earth than a man in the middle of the tower. So relative to the man at the middle of the tower the man at the bottom feels a negative force, so the man at the top moves away from the man in the middle and from the Earth.

Now if the tower disapeared I accept that the man at the top will move away from the man in the middle. You have to explain if the man at the middle doesn't move, why does the man at the top get further away, both from the man in the middle and the Earth. (The far side of the Earth moves away from both the Centre of the Earth and the moon)

I am going to be away from computers for the next few days, so I will not be able to contribute again until next week.

Simon

GrapesOfWrath
2002-Apr-02, 01:47 PM
On 2002-04-02 08:27, SimonCB wrote:
Now if the tower disapeared I accept that the man at the top will move away from the man in the middle. You have to explain if the man at the middle doesn't move, why does the man at the top get further away, both from the man in the middle and the Earth. (The far side of the Earth moves away from both the Centre of the Earth and the moon)

If the tower is rigid, they don't move at all, althought they might feel a slight difference in force. If the earth were completely rigid, it wouldn't move at all either. In fact, the more rigid part of the Earth does not experience as much tidal movement as the oceans, which of course are fluid and not rigid at all.

If the solid Earth moved as much as the oceans, we would hardly even notice the tides, as the solid and fluid parts would be moving together.

johnwitts
2002-Apr-02, 11:22 PM
So what about the 'slingshot' effect? This is not just a gravitational effect, because everything is also in motion in circles. The Moon circles the Earth, and the Earth consequently 'wobbles' about the common Earth/Moon centre of mass. They both revolve around the Earth/Moon/Sun Centre of Mass. Centrifugal force (inertia, or whatever) has to play a part because it's all moving. Take a long piece of elastic, put three markers on it, then swing it round your head. The markers move apart. Nothing to do with gravity, everything to do with centrifugal force (inertia, or whatever).

GrapesOfWrath
2002-Apr-03, 01:18 PM
Do you mean the slingshot effect where a spaceprobe can steal some of the momentum of a planet to boost its orbit?

johnwitts
2002-Apr-03, 09:53 PM
No, I mean the forces contained in a real slingshot. Spin a weight on elastic round your head and the elastic stretches. It stretches to a length that counteracts the tendency for the weight to want to go off in a straight line. Gravity applies a force just like the elastic supplies a force (except the force of the elastic get stronger as it's stretched whereas gravity gets weaker with distance). A slingshot stores and accumulates rotational energy until it is let go. This accumulated energy is then enought to kill Goliath. Spin any elastic substance and it will want to stretch in response to the 'centrifugal' forces.

GrapesOfWrath
2002-Apr-04, 11:30 AM
On 2002-04-03 16:53, johnwitts wrote:
Spin any elastic substance and it will want to stretch in response to the 'centrifugal' forces.

That is an interesting example. Forces in and of themselves don't stretch things, though. Just pushing or pulling on an elastic doesn't stretch it significantly. You have to pull on one end in one direction, and pull on the other end in the opposite direction. In your example, the weight on the end is pulling the elastic with centrifugal force, and you are holding it at the center.

You can stretch the elastic by pulling it on one end only, if you don't also pull on the other end in the same direction, since the inertia of the elastic at the other end will tend to stretch it. However, if you pull on both ends in the same direction, it won't stretch.

In the case of the tides, the centrifugal force on the near side of the Earth is the same as the centrifugal force on the far side. That is not immediately obvious, but it can be easily shown that every point of a non-rotating body revolves in the same size circle when it orbits in a circle around another mass. Thus, as above, there is no stretching due to the centrifugal force. (The centrifugal force does keep the body in orbit, and it will balance the gravitational force near the center.)

johnwitts
2002-Apr-04, 09:00 PM
In the case of the tides, the centrifugal force on the near side of the Earth is the same as the centrifugal force on the far side. That is not immediately obvious, but it can be easily shown that every point of a non-rotating body revolves in the same size circle when it orbits in a circle around another mass. Thus, as above, there is no stretching due to the centrifugal force. (The centrifugal force does keep the body in orbit, and it will balance the gravitational force near the center.)

Every point of the Earth is following the same 'orbit' around the Earth/Moon centre of mass, but is each part following the orbit it would like to occupy if it wasn't being held by gravity? The water on the side of the Earth farthest from the Moon would like to be further away from the centre of the rotation because it's travelling too fast for the orbit it is in. That is, it's travelling at the average orbital velocity for the whole body (the Earth), which is too fast for it. Imagine a satellite in orbit. What happens when we increase it's orbital velocity? It raises it's orbit to a higher one. That's what the water is trying to do. On the 'Nearside', the water is travelling too slowly to maintain the orbit it is in, therefore tries to fall into a lower orbit. Gravity stops it all getting out of hand.

GrapesOfWrath
2002-Apr-05, 09:19 AM
On 2002-04-04 16:00, johnwitts wrote:
Every point of the Earth is following the same 'orbit' around the Earth/Moon centre of mass, but is each part following the orbit it would like to occupy if it wasn't being held by gravity?

Only the center of the Earth orbits the center of mass of the Earth/moon, the other parts follow paths that are offset some.

OTOH, the inclusion of centrifugal force in the description of tides is not wrong, of itself. There are often many ways of looking at at a problem--I guess, in another context, that is what relativity is all about.

johnwitts
2002-Apr-05, 09:52 PM
Only the center of the Earth orbits the center of mass of the Earth/moon, the other parts follow paths that are offset some.

The centre of the Earth orbits outside the centre of Mass of the Earth/Moon system. The actual centre of the system is some 1600 km below the surface of the Earth on the side nearest the Moon. Essentially, they orbit round each other.

Roy Batty
2002-Apr-05, 10:01 PM
Err didnt you both just say the same thing.. or do I have this really confused?



<font size=-1>[ This Message was edited by: Roy Batty on 2002-04-05 17:09 ]</font>

johnwitts
2002-Apr-05, 10:11 PM
Yes we did, but in different ways. We're both trying to describe the same thing really, just from different viewpoints or frames of reference. I my view, the tides are caused by the combination of more than a few factors, adding together to produce the effect. We're probably both right, at least with parts of the puzzle. Now if we could get together, we'd probably become an awsome force to be reckoned with...
/phpBB/images/smiles/icon_smile.gif

SimonCB
2002-Apr-06, 01:45 PM
Hello, back again good to see this discussion carrying on without me.

To go back to my tower example, what would happen if you increased the Earths rotation rate such that the centrifugal force on the man in the middle matched the gravitional attraction. The man in the middle now feels equal and opposite forces and if the tower disapeared he wouldn't move, (he is in orbit around the Earth). The man at the top of the tower however feels a slightly smaller gravitational pull (and in this case a slightly larger centrifugal force) so he starts to rise.

This is what happens with the far side of the Earth, at the centre of the Earth gravity and the centrifugal force are balanced so the centre of the Earth moves in a circle around the barycentre. The far side of the Earth feels the same centrifugal force, but a slightly lower gravitational force and thus feels a net force away from both the centre of the Earth and the moon. Though this is of course tiny compared to the self gravity of the Earth.


Simon

GrapesOfWrath
2002-Apr-06, 05:41 PM
On 2002-04-05 17:01, Roy Batty wrote:
Err didnt you both just say the same thing.. or do I have this really confused?

As near as I can tell, johnwitts agrees with what I said there, but my objections were against other things. When he said "Every point of the Earth is following the same 'orbit' around the Earth/Moon centre of mass," I pointed out that every point of the Earth does not orbit around the Earth/moon center of mass. Positions on the surface of the Earth orbit points that are offset by 6000 kilometers from the Earth/moon center of mass.

Also, I do understand the explanations of the tides that include centrifugal force in their analysis--I just do not think that it is necessary to include centrifugal force. That is why I defend the BA's choice to not.

johnwitts
2002-Apr-06, 08:12 PM
So what causes a spacecraft to increase it's orbital height when you increase it's orbital velocity, or indeed to lower it's height when 'retro firing' along the orbital path.

GrapesOfWrath
2002-Apr-07, 03:07 AM
On 2002-04-06 15:12, johnwitts wrote:
So what causes a spacecraft to increase it's orbital height when you increase it's orbital velocity, or indeed to lower it's height when 'retro firing' along the orbital path.


That is also an interesting subject (http://www.badastronomy.com/phpBB/viewtopic.php?mode=viewtopic&topic=744&forum=2&start=53), but when you think about it, it's clear: you have to fire a rocket to get into space and then into orbit, but when that rocket rises, it loses some of its energy, and velocity.

SimonCB
2002-Apr-08, 08:05 AM
I also have no doubt that if you calculate all the forces relative to the Centre of the Earth, you get the right answer but I don't think it good way to understanding the forces involved.

To give another example it is realtively simple to reformulate the equations of motion of the planets to be relative to the centre of the Earth and this would be just as accurate as the equations relative to the Sun or the Barycentre. In this frame you would see something very similar to Ptolemy's epicycles.

But although this is accurate it is not a good way of describing the fundemental physics involved in the system.

Similarly with the Earth centred discription of the tides. The fundemental thing it seems to me in understanding the tides, is to understand why the tidal force exerted by the moon causes part of the Earth to move away from the moon. Though I have no doubt this will fall out of the maths in the Earth centred case, (I have been desperatly trying to avoid digging out my old textbooks and going through it). I think the best way of understanding what is going on is with the centrifugal force (or inertia).

On an other matter, you don't need rockets to change you distance from a object you are orbiting, it simple a matter of velocity. The simplest example is the eliptical orbit, the distance is constantly changing but not a rocket motor insight.


Simon

GrapesOfWrath
2002-Apr-08, 09:09 AM
If you're concerned with the best (or even better) way of describing the tidal effect, then none of the ways described in this thread fit the bill. A far simpler and more complete analysis would calculate the form of the equipotential surface around the Earth. But that's not the point.

Clearly, with the "Earth-centered" version, you do not have epicycles falling out of it, or any other complications, so there really isn't any problem with using it.

SimonCB
2002-Apr-08, 10:06 AM
There is best as is most complete, but there is also best as in allowing physical understanding.


It is by no means obvious in the Earth centred frame why part of the Earth moves away from the moon. It is not enough to say that the force is less, as the tower example shows. If you do the maths for the Earth centred case, the centrifugal force is implicit as the the Earth-Moon distance remains constant. I think for understanding of what happens the centrifugal force has to be made explicit.

It is not enough to understand the forces acting, there has to be some consideration of the reaction to the force and this is best done in an inertial frame.

I keep coming back to the fact I don't think the description of tides in the book gives good explanation of why part of the Earth moves away from the moon.

Simon

GrapesOfWrath
2002-Apr-08, 10:58 AM
On 2002-04-08 06:06, SimonCB wrote:
It is by no means obvious in the Earth centred frame why part of the Earth moves away from the moon.

It's not necessarily an Earth-centered frame, for one thing. This is not a geocentrist argument. Not that there's anything wrong with that.


It is not enough to say that the force is less, as the tower example shows.

I don't see where the tower example shows that. You may have missed this post (http://www.badastronomy.com/phpBB/viewtopic.php?topic=887&forum=9&start=0&start=20), since it came right after you left for vacation.


If you do the maths for the Earth centred case, the centrifugal force is implicit as the the Earth-Moon distance remains constant. I think for understanding of what happens the centrifugal force has to be made explicit.

We certainly disagree about that. If the Earth were simply falling straight "down" (no centrifugal force), the tidal stretching would still be there. Of course, the Earth is simply falling.

SimonCB
2002-Apr-08, 11:44 AM
It's not necessarily an Earth-centered frame, for one thing. This is not a geocentrist argument. Not that there's anything wrong with that.



In any frame the dstance from the far side of the Earth to the moon increases and this has to be explained.





I don't see where the tower example shows that. You may have missed this post, since it came right after you left for vacation.




I saw it, the point I was getting at is that in the stationary case, despite the difference in force both men remain where they are. But if you spin the system at just the right rate the man in the middle stays put but the man at the top moves up. If the tower is rigid he will find himself pushed against the ceiling.




We certainly disagree about that. If the Earth were simply falling straight "down" (no centrifugal force), the tidal stretching would still be there. Of course, the Earth is simply falling.



In the falling case the tidal force would still be there, but all of the Earth would be moving towards the moon. I come back again it not just the force, but the reaction to the force. If you place two particles at 1AU from the sun, the force they feel is identical, but the way they behave depends on their initial velocity. They could crash into the Sun, orbit or fly off into deep space.

I am not sure I would agree that the Earth was simply falling. Although one way of thinking about an orbit is say that it is falling and missing. This however does have flaws, it does not explain, for example, why an object after passing pericentre moves away from the object it is orbiting. Another way of thinking about an orbit, is that the particle is trying to get away from the object and is always being pulled back, or to put it another way a balance between gravity and inertia.

Of course I wouldn't claim that the second way of thinking about an orbit is anymore 'right' than the first, but I do think it enables a better understanding of the effect of tidal forces.

Simon

GrapesOfWrath
2002-Apr-08, 12:11 PM
On 2002-04-08 07:44, SimonCB wrote:
Of course I wouldn't claim that the second way of thinking about an orbit is anymore 'right' than the first, but I do think it enables a better understanding of the effect of tidal forces.

But that's a personal standpoint--for me, it's not true at all. The point is, BA's presentation is not bad astronomy, from a physics standpoint. As you say, neither is better than the other.

Avoiding use of the term centrifugal force in the discussion because of its perception as a "forbidden term" would be bad, in my opinion, but I don't think that is the case here. Nor is it a matter of insisting upon a geocentrist position to the exclusion of all others.

SimonCB
2002-Apr-08, 12:25 PM
I suspect that we are going to have to agree to disagree on this, but I come back again to the fact part of the Earth moves away from the moon, it gains potential energy. In order to understand tides this has to be explained and differences in force won't do it.


Simon

GrapesOfWrath
2002-Apr-08, 12:44 PM
On 2002-04-08 08:25, SimonCB wrote:
I suspect that we are going to have to agree to disagree on this,

I thought we did agree that both explanations are OK? No disagreement there.

SimonCB
2002-Apr-09, 07:58 AM
Sorry I meant disagree about the best way of explaining it, rather than which is true or not.

Now I was adding personal knowledge, but I read the description as saying, because the far side of the Earth feels a negative force relative to the centre of the Earth, it moves away from the centre of the Earth. Therefore as the distance between the centre of the Earth and the Moon is fixed the far side of the Earth moves away from the Moon.

Now thinking about it a bit more I realize that the fact the Earth-Moon distance is fixed, must imply a centrifugal force and that force must be proportional to the mass of the moon and hence the tidal force, so if I could be bothered to do the maths everything would come out right.

But, and in my mind at least it remains a big but, I don't think this is the best way of gaining physical understanding of what is going on. The discription implies that movement of the tides is simple due to the diferential gravity and this is not the whole answer. There has to be motion for the tidal force to have any effect, either falling or spinning. I know that the decription makes clear that the Earth is in freefall, but I don't think it makes clear enough how important the motion is.


On the otherhand the book and the discussion here have greatly expanded my understanding of tides. The last time I have looked at it was as a small part of a solar system course eight years ago that I knew I wasn't going to be examined on and to be honest I was more worried about Lagrange points and the disturbing function. So I have to thank you GrapesofWrath (can I call you Grapes) and all the others who have taken part in this discusion.

Simon

GrapesOfWrath
2002-Apr-09, 11:13 AM
Well, I'm thinking of changing my nom anyway, to make it easier to type.



On 2002-04-09 03:58, SimonCB wrote:Sorry I meant disagree about the best way of explaining it, rather than which is true or not.

Different styles are not really bad astronomy, though.


Now I was adding personal knowledge, but I read the description as saying, because the far side of the Earth feels a negative force relative to the centre of the Earth, it moves away from the centre of the Earth. Therefore as the distance between the centre of the Earth and the Moon is fixed the far side of the Earth moves away from the Moon.

Well, the distance is not fixed--but the important part, from the perspective of the Earth, is that two opposite sides move away from each other.

SimonCB
2002-Apr-09, 11:42 AM
Different styles are not really bad astronomy, though.



No.



Well, the distance is not fixed


No, by assuming a circular orbit has no real effect on the basic argument.



but the important part, from the perspective of the Earth, is that two opposite sides move away from each other.


He is the root of my problem, I think the important bit is that it moves away from the moon. The effect of the tidal force is move part of the body away from the object exerting the force. This is so counter intuitive I think it needs explanation and I don't think the book does this.

Simon

2002-Apr-09, 12:01 PM
<a name="20020409.5:49"> page 20020409.5:49 A.M.5:49 aka "_ALWAYS_"
On 2002-03-28 11:35, GrapesOfWrath wrote:


On 2002-03-28 08:42, SimonCB wrote:
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.

Gravity does not always act towards
HUb' 5:51 A.M. AGREE absolutly [NOT]!
maybe examples later I've been on way to long today
[/quote] Math Just takes time {no way around it}

GrapesOfWrath
2002-Apr-09, 01:30 PM
On 2002-04-09 07:42, SimonCB wrote:
This is so counter intuitive I think it needs explanation and I don't think the book does this.

OK, there is where I guess we disagree.



On 2002-04-09 08:01, HUb' wrote:
Gravity does not always act towards
HUb' 5:51 A.M. AGREE absolutly [NOT]!

I think I gave an example earlier in this thread. Just look at the Earth/moon system. For us, standing on the Earth, gravity seems to point a couple tens of kilometers away from the center of mass of the Earth, but it misses the center of mass of the Earth/moon system by thousands of kilometers.

2003-Mar-20, 03:59 AM
<a name="20030319.p24"> page 20030319.p24 aka push 2 look 4
March 19, 2003 8:02 P.M.
in an effort to find a path to actual
data for Pacific Northwest Tides
i've pushed this thead up..
:::50:::::::::::**********************:::::::::
:::30:::::**?***::::::::::::::::::::::******:::
:::10:::**:|::160+200*cos(t),:50*sin(t):::::**:
::::0::*-(Sun)----25yr/*--in AU units---------*
:::20:::**:|::HUb's ORBIT of planet X:::::::**:
:::40:::::*******::::::::::::::::::::*******:::
:::50::::::::::::********************::::::::::
::::::50:::0:::50:::100::150:200::250::300::350
to look for the eXpected delay then advance time's of
the actuall hi tides compared to Astronomical calculations

mik sawicki
2003-Mar-25, 08:03 PM
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.
...
Simon


Here's the link to an extensive discussion of tides:
http://www.jal.cc.il.us/~mikolajsawicki/gravity_and_tides.html
Enjoy
MS