moon pulls the earth from both sides?

[rockinreel]

The moon causes a bulge in the earth from the moon side and the opposite side, thus creating our tides. What's confusing about this is how does the moon create a bulge on the opposite side of the earth. Since the sun's gravity also pulls earth's waters, it makes since if the Sun and the Moon are on opposite sides of the earth (full moon). If they are on the same sides of the earth, aligned, (new moon) what causes the side of earth opposite of the moon and sun to bulge?

[01101001]

Check out Bad Astronomy Tides, the Earth, the Moon, and why our days are getting longer


I'm sure you can find some topic or topics discussing it on this forum. I'd point you to them, but the Search function is malfunctioning for me right now.

[Jens]

Actually, this site has a pretty easy-to-understand description.

http://csep10.phys.utk.edu/astr161/lect/time/tides.html

I wasn't sure myself, so thanks for the question. Basically, it says you can visualize the middle part of the earth being pulled toward the moon at a greater rate than the ocean on the far side.

I was thinking more along the lines of a water balloon, where if you pull one side, the middle will flatten and the far side will also bulge. I'm not sure if that has anything to do with it, but in any case, the explanation given on that site seems pretty easy to understand.

[rockinreel]

Vector addition has me convinced it's better described as a magic wand.

It is described below from 01101001's recommended site:

This part is tricky, and is the hardest part of this explanation to understand. A drawing of these forces looks like this:


--> ----> ------->
far center near
side of Earth side

where the arrows represent the force (and direction) of the Moon's gravity on these three points of the Earth. Now, we measure the gravity of the Earth relative to the center of the Earth; everywhere on the Earth, the center is "down". In a sense, we see the center of the Earth as "at rest". It is mathematically correct to then subtract the force of the Moon on the center of the Earth from the force felt on the near and far sides. This is called vector addition. If we do that, our diagram will look like this:


<- X ->
far center near
side of Earth side


Now that means on the farside, the water (and earth) is being pushed into a bulge, pushed into a high tide. True? Magic?

Jen the earth being pulled from the ocean on the farside seems the water would all pull to one side of the planet. (on new moon's).

[hhEb09'1]

Vector addition has me convinced it's better described as a magic wand.

Arthur C. Clarke said "any sufficiently advanced technology is indistinguishable from magic." He meant, indistinguishable by those who do not understand it. And it doesn't have to be just technology

Now that means on the farside, the water (and earth) is being pushed into a bulge, pushed into a high tide. True?

That's the way it appears to us. It's relative.

Magic?

Yes, you want to see some more?

Jen the earth being pulled from the ocean on the farside seems the water would all pull to one side of the planet. (on new moon's).

Why? and which side?

[Robert A.]

The key to understanding the two high tides is to remember that the Earth and Moon are not sitting still in space - they are rotating about a common center of gravity. That common center of gravity lies between the two high tide bulges (in fact, about 1000 miles from the center of the Earth in the direction of the Moon), and so it's as if the "outer" bulge were being slung away from the Earth, just as you would be slung away from the center of a merry go round if you hung on at the outside.

Don't let anyone tell you centrifugal force is not real. It is a very real force in a non-inertial (spinning) reference frame (though not in an inertial reference frame). For that matter, gravity does not exist in a freefalling reference frame, but no one will argue that it is not a "real" force!

[hhEb09'1]

The key to understanding the two high tides is to remember that the Earth and Moon are not sitting still in space - they are rotating about a common center of gravity. That common center of gravity lies between the two high tide bulges (in fact, about 1000 miles from the center of the Earth in the direction of the Moon), and so it's as if the "outer" bulge were being slung away from the Earth, just as you would be slung away from the center of a merry go round if you hung on at the outside.

While that seems to be an intuitively easy way to accept the concept, it's wrong. Centrifugal force, in that reference frame, due to the revolution about the earth/moon barycenter, does not contribute to the tidal effect--as the lack of it in the BA's explanation indicates.

That particular reference frame is valid, but you have to be careful in the actual calculations. Once you subtract the effects of the rotation of the earth, which clearly are not what we consider "tidal", then there is no significant centrifugal effect left. Here (and here) is an old thread where this was discussed (note the last post to the first thread, by mik sawicki, and his link to his webpage about the tides--the BA includes Mik as a reference in his book).

Don't let anyone tell you centrifugal force is not real. It is a very real force in a non-inertial (spinning) reference frame (though not in an inertial reference frame). For that matter, gravity does not exist in a freefalling reference frame, but no one will argue that it is not a "real" force!

Not true. I've said as much for a long time:

With general relativity, gravity is the shape of spacetime. In other words, it's another fictitious force, like coriolis or centrifugal.

[grant hutchison]

1) The late, great Isaac Asimov had an analogy he used to show how a force acting in a single direction could make an object bulge in two opposed directions.
He likened the force of gravity to the psychological "force" pushing a runner towards the finishing tape of a race. Then he pointed out how a bunch of runners will separate under the action of that "force" into leaders, the pack, and trailers: the leaders and trailers are two tidal bulges formed from the central pack of runners, under the differential action of a single force.
It's an imperfect analogy that doesn't bear too much examination (as Asimov of course understood), but I find it at least gets people over the initial hurdle of the single force and the two bulges.

2)

With general relativity, gravity is the shape of spacetime. In other words, it's another fictitious force, like coriolis or centrifugal.

Gravity shares a property with Coriolis and centrifugal "forces", in that the force it exerts on an object varies with its mass: so it's measured in N/kg (units of acceleration), rather than newtons alone.
I have no idea if this is a deep connection of some sort, or merely a chance resemblance.
Anyone?

Grant Hutchison

[hhEb09'1]

2)Gravity shares a property with Coriolis and centrifugal "forces", in that the force it exerts on an object varies with its mass: so it's measured in N/kg (units of acceleration), rather than newtons alone.
I have no idea if this is a deep connection of some sort, or merely a chance resemblance.
Anyone?

That's basically the Equivalence Principle--which has been fruitful

[grant hutchison]

That's basically the Equivalence Principle--which has been fruitful

Exposure to too many popsci diagrams of rockets and elevators seems to have inextricably linked the Equivalence Principle to a constant acceleration vector in my brain.
But of course: rotating reference frame = accelerating reference frame.

Grant Hutchison

[Robert A.]

While that seems to be an intuitively easy way to accept the concept, it's wrong.

Thanks for pointing that out. I've used it as a way of helping students feel comfortable with the idea (which of course has nothing to do with truth), but I've never bothered to do the math. I will make sure students know that it is only an analogy.

Gravity shares a property with Coriolis and centrifugal "forces", in that the force it exerts on an object varies with its mass: so it's measured in N/kg (units of acceleration), rather than newtons alone.

I have studied a lot of physics (Ph.D., 2001), but I have never heard or seen anything like that. If you are talking about the force of gravity, it makes no sense to use any units other than force. The fact that it depends on the mass of the object is already incorporated into the equation (I am refering to Newton's law of gravity here - I have not studied the mathematics of GR, and doubt I will ever have the time to master differential geometry). If everything I have learned and taught is wrong, please explain!

[Robert A.]

That's basically the Equivalence Principle--which has been fruitful

The Equivalence Principle says that any affects that occur in an accelerating reference frame will also occur in a gravitational field that has the same acceleration due to gravity. But that doesn't mean the force of gravity is measured in units of acceleration - the force will be mg, and the units are still Newtons.

Am I missing your point?

[grant hutchison]

If you are talking about the force of gravity, it makes no sense to use any units other than force.

I'm talking about the "acceleration due to gravity", which is the way that gravity is usually quantified: we'd normally say that the Moon has a surface gravity of one sixth of a g, for instance, which is a measure of acceleration, not force.
You can't quantify gravity purely as a force without knowing the mass of the object acted upon; that's different from (say) the force applied by a compressed spring. Whereas you can state the acceleration gravity will cause without knowing the mass of the object acted upon* - again, different from the case of the compressed spring.
This strikes me as an intriguing and important distinction. But perhaps you can say why it's trivial ... I've already had one such revelation today, and am quite prepared for another.

Grant Hutchison

Edit: *For M>>m, of course.

[hhEb09'1]

This strikes me as an intriguing and important distinction. But perhaps you can say why it's trivial ... I've already had one such revelation today, and am quite prepared for another.

A force due to gravity is of course measured in newtons, that is the unit of force in the SI. But you are right, gravity is usually measured in N/kg, which is the same as m/s^2, which is also a derived unit called the galileo of course, sometimes called the gal. One gal is 1 centimeter per second squared, so gravity on the surface of the earth is around 980 gal. Variations in gravity (due to latitude, or altitude, or local mass densities) are expressed in milligals.

[Robert A.]

I'm talking about the "acceleration due to gravity", which is the way that gravity is usually quantified

Thanks for clarifying. I misunderstood your original post. In my classes, I try to be very clear about the distinction: I try to say "acceleration due to gravity" or "the force of gravity", because that very point can be confusing to students.

Whereas you can state the acceleration gravity will cause without knowing the mass of the object acted upon*

Edit: *For M>>m, of course.

Actually it works for M<<m, too. The acceleration due to gravity g is simply everything in Newton's Law of Gravity other than the little m:
g = GM/(d^2). Of course, if M<<m, the acceleration of the mass m may be so small as to be immeasurable.

This strikes me as an intriguing and important distinction.

This is most definitely an important point. It's the equivalence between inertial mass and gravitational mass, which probably leads to the idea Einstein came up with called the "Equivalence Principle" (EP). This principle was a key insight in his development of GR.

I say "probably" because I'm not entirely sure. Experiments to show the equivalence of inertial mass and gravitational mass continued on long after Einstein published his general theory - I assume they are still going on today (as are lots of tests of GR). So I don't really know if Einstein assumed it and it lead him to the EP, or if the EP just came to him as a flash of insight.

Anyway, thanks for the clarification.

[grant hutchison]

Actually it works for M<<m, too. The acceleration due to gravity g is simply everything in Newton's Law of Gravity other than the little m:
g = GM/(d^2). Of course, if M<<m, the acceleration of the mass m may be so small as to be immeasurable.

My sloppy language is fostering confusion again, I think. Here's what was in the background to my afterthought footnote.
Using N/kg = m/(s^2):

F/m = GM/(d^2)

F/M = Gm/(d^2)

If M>>m then F/M is negligible, so mass M can be said to have a characteristic acceleration due to gravity at distance d (for instance, its "surface gravity"), which is insensitive to variations in m. But once the two masses are of the same order of magnitude, then they have an acceleration towards each other of (F/m+F/M), and so the magnitude of m becomes important.

I'm sure you're already well aware of all of the above.
But certainly for the sake of clarity I should have stipulated that my footnote applied to the relative acceleration of the bodies, rather than the action of one upon the other. My excuse is that I was thinking specifically of the surface acceleration due to gravity of M, and was anxious to note that the concept rather broke down when m was of comparable mass.

Grant Hutchison

[DALeffler]

Now that means on the farside, the water (and earth) is being pushed into a bulge, pushed into a high tide. True? Magic?

Imagine 3 identical objects in free fall toward the sun.

Object 1 is denoted by “(“
Object 2 is denoted by “-“
Object 3 is denoted by “)”

These objects are identical, not connected to each other, not in orbit around each other, and are not in orbit around the Sun. They are simply moving towards the Sun in a straight line under no forces other than that caused by the Suns gravity and are currently equidistant from each other. Thus:

(-)              --->               (Sun)    T+0

After any time interval, all three objects will have moved towards the sun such that object 3 will have moved closer to the sun than object 1 or 2. Object 2 will have moved closer to the sun than object 1 but less than object 3, and object 1 will have moved towards the sun the least amount. The reason for this is because the gravity of the sun pulls harder on objects that are close to it and less hard on objects that are further away. Thus:

  (  -  )        --->               (Sun)    T+1

If we take a viewpoint from the surface of object 2, we would see that both objects 1 and 3 are moving away from us in opposite directions caused by “some force”. All three objects are still falling toward the Sun, but to us on object 2, something seems to be “pulling” object 1 and 3 away from us in different directions. This is the suns’ gravitational force acting over distance causing an apparent (and very real) “tidal” force.

Connecting objects 1, 2, and 3 together with rock, dirt, magma, etc., doesn’t change the tidal force: It only limits the effect of the tidal force to the elasticity of the “mud ball” and to the size (the distance between points 1 and 3) of the mud ball. Fluid parts of the planet will show the tidal force much more readily.

(http://www.jal.cc.il.us/~mikolajsawicki/tides_new2.pdf)

You are correct in asserting that an “ideal” planet should be stretched more “to one side” for a planet’s cross-section. But with distances between real planets, massive stars, and extremely short distances between the near and far side of a planet compared to its distance from the primary, the stretching caused by tidal forces is essentially equal on both sides of the planet (down to like the 11th decimal place – even when talking about the Earth/Moon system).

Throw the planet into orbit around the Sun and the tidal effect is still there: While the planet isn’t falling into the Sun, it’s falling around the Sun, and the side of the planet nearest the sun is still closer than the far side of the planet is to the sun.

Spin the planet on its axis while orbiting the Sun, throw in a moon or two, add a few billion years, and that’s where the neat stuff is!

hth,

Doug.

PS: Don’t forget the tidal squeeze on the poles of the planet…

[worzel]

Here's another way to think about tidal forces. In a circular orbit the orbital velocity is faster the lowe the altitude. Now image we have an object representing the earth in a perfectly circular orbit about the sun. Now imagine two other objects orbiting with the earth at the same speed but one at a lower altitude and one at a higher altitude - these represent the sea. The lower one will be going to slow for its smaller orbit and will lose altitude. The higher one will be going to fast for its larger orbit and will gain altitude.

The fact that the major tidal force experienced on earth is from the moon and that it orbits us more than we orbit it makes it a bit more difficult to see, but it is the same effect.

[hhEb09'1]

Here's another way to think about tidal forces. In a circular orbit the orbital velocity is faster the lowe the altitude. Now image we have an object representing the earth in a perfectly circular orbit about the sun. Now imagine two other objects orbiting with the earth at the same speed but one at a lower altitude and one at a higher altitude - these represent the sea. The lower one will be going to slow for its smaller orbit and will lose altitude. The higher one will be going to fast for its larger orbit and will gain altitude.

The fact that the major tidal force experienced on earth is from the moon and that it orbits us more than we orbit it makes it a bit more difficult to see, but it is the same effect.

This is an even better expression of the example that Robert A. gives, and I've seen it a lot. Most people consider it easy to grasp, and the two tidal bulges seem easier to understand.

However, you have to be very careful in actually applying the model. For circular motion of a body around a central point, every position on the body actually follows a path that has the same diameter as every other position on the body, if the body is not rotating. Here is an illustration of that.

The example uses the idea of inertial force, and that is where it runs into problems in the calculation--the values due to centrifugal force just disappear.

[worzel]

However, you have to be very careful in actually applying the model. For circular motion of a body around a central point, every position on the body actually follows a path that has the same diameter as every other position on the body, if the body is not rotating. Here is an illustration of that.

That's a very good point, which I missed when thinking about it.

The example uses the idea of inertial force, and that is where it runs into problems in the calculation--the values due to centrifugal force just disappear.

I don't understand. With a non-rotating body, the tidal force will be experienced as I described, right? If spinning, the centrifugal force will be even all round the circumference, so the difference in force at different points on the circumference will be the same as if it were non-rotating, wouldn't it?

[hhEb09'1]

I don't understand. With a non-rotating body, the tidal force will be experienced as I described, right?

You didn't quantify the effect, so I'm not sure. My point was just that you mentioned an effect of inertia--and there is no significant inertial effect in the production of the tidal effect, at the close and far points of the "bulges".

If spinning, the centrifugal force will be even all round the circumference, so the difference in force at different points on the circumference will be the same as if it were non-rotating, wouldn't it?

I don't think so. The centrifugal forces are in opposite directions, for opposite points on the surface. The difference would be non-zero, wouldn't it? Or is that what you meant?

[worzel]

You didn't quantify the effect, so I'm not sure. My point was just that you mentioned an effect of inertia--and there is no significant inertial effect in the production of the tidal effect, at the close and far points of the "bulges".

I just meant that it would qualatively be as I described. If you stood a person on the point of the earth closet to, and furthest from, the sun, stopped the earth from spinning, and switched off gravity, then both men will float off following their intertial paths. Is that what you meant by inertial?

I don't think so. The centrifugal forces are in opposite directions, for opposite points on the surface. The difference would be non-zero, wouldn't it? Or is that what you meant?

I meant as measured from the center of rotation, the outward centrifugal force would be the same everywhere on the circumference - so any difference due to tidal effects around the circumference would be the same whether it was spinning or not.

[hhEb09'1]

I just meant that it would qualatively be as I described.

Another thing to consider, is what would happen if your two objects were orbiting at the same distance as the Earth--instead of at the far or near side, they were at the leading and trailing edge of the Earth. (This is a thought experiment that--I think--sam5 first brought up, and it has helped me to understand some of the things going on.) Since the two objects are the same distance, your model has no tidal effect on them, right? But we know that there is a tidal effect, about half the other, in depression, an "anti-bulge", at those positions.

[Eroica]

Another thing to consider, is what would happen if your two objects were orbiting at the same distance as the Earth--instead of at the far or near side, they were at the leading and trailing edge of the Earth. (This is a thought experiment that--I think--sam5 first brought up, and it has helped me to understand some of the things going on.) Since the two objects are the same distance, your model has no tidal effect on them, right? But we know that there is a tidal effect, about half the other, in depression, an &quot;anti-bulge&quot;, at those positions.

But there's only a finite amount of water available. If there are bulges on the far and near sides, there must be anti-bulges somewhere else.

[worzel]

Another thing to consider, is what would happen if your two objects were orbiting at the same distance as the Earth--instead of at the far or near side, they were at the leading and trailing edge of the Earth. (This is a thought experiment that--I think--sam5 first brought up, and it has helped me to understand some of the things going on.) Since the two objects are the same distance, your model has no tidal effect on them, right? But we know that there is a tidal effect, about half the other, in depression, an "anti-bulge", at those positions.

Yeah I thought of that when I was thinking about what you said before. I figured that the squeeze is due to the direction of fall on the extremities being different. But as the object isn't falling (or getting any nearer as it falls) would it feel the squeese, or just the buldge?

[hhEb09'1]

But there's only a finite amount of water available. If there are bulges on the far and near sides, there must be anti-bulges somewhere else.

In this case, we are not concerned with the matter involved. The tidal effect is a deformation of the equipotential surface--which affects even the solid (non-flowing!) earth. There are bulges and depressions in that equipotential, and the displacement of the solid earth is about half that what we see in the water (if the solid earth had less strength, and deformed more easily, it would rise and fall with the water, and we would never even notice the tides.)

PS: I'm not sure, does this help answer your question, worzel?

[worzel]

PS: I'm not sure, does this help answer your question, worzel?

I'm not sure, hhEb09'1, I know the earth-moon tides are a lot more complicated, but I'm trying to understand what just the gravitational tidal forces are for now.

Suppose we have some free particles all moving at exactly the same velocity at t0 so that we can consider their tidal deformation laterally and radially separately.

Whatever their velocity (at t0), they would suffer a radial deformation such that they separate.

If there is a component of their velocity towards the gravity well, they would suffer a lateral contraction.

If there is a component of their velocity away from the gravity well, they would suffer a lateral expansion.

If, at t0, the velocity was tangential to the direction of the gravity well, they wouldn't suffer any lateral deformation.

Have I got that right?

[hhEb09'1]

I know the earth-moon tides are a lot more complicated, but I'm trying to understand what just the gravitational tidal forces are for now.

Yeah, I'm not talking about wobble-induced tides, or changes in the tides due to topography. Are there others that you want to explicitly exclude?

If there is a component of their velocity towards the gravity well,

Is that the gravity well of the Earth? Are those particles in a cloud about the Earth?

[worzel]

Yeah, I'm not talking about wobble-induced tides, or changes in the tides due to topography. Are there others that you want to explicitly exclude?

For now, I want to include only gravitational tidal effects in an idealised situation. I know you know what I mean

Is that the gravity well of the Earth? Are those particles in a cloud about the Earth?

Yes, but it could be any effectively point-like gravity well.

[hhEb09'1]

For now, I want to include only gravitational tidal effects in an idealised situation. I know you know what I mean

I wasn't sure. I thought that was all we've been talking about all along.

Yes, but it could be any effectively point-like gravity well.

I considered the Sun to be a gravity well too, so I wanted to make sure of the directions. I'll go back to your previous post:

Suppose we have some free particles all moving at exactly the same velocity at t0 so that we can consider their tidal deformation laterally and radially separately.

Whatever their velocity (at t0), they would suffer a radial deformation such that they separate.

Argh. The radial deformation is relative to the Sun, isn't it? I mean, as opposed to lateral? If so, why the distinction between radial and lateral? Velocity towards the "gravity well" would be a "contraction" in either case, wouldn't it?