Calculating tidal forces

[newts]

I have had difficulty finding webpages that accurately describe the tide-raising force, and that show how to calculate it. So I have now produced my own. http://squishtheory.wordpress.com/the-tides/?preview=true&preview_id=210&preview_nonce=69cba02793

 

I show how to calculate the tide-raising force, both by differential gravity, and also by the difference between centrifugal forces and gravitational potentials.

 

Once I had figured out exactly what needed to be calculated, the maths was quite straightforward. It would be helpful if anybody who has struggled with the concept of the tides, would like to tell me if my explanations are sufficiently clear; and also if any tidal experts would like to check it for errors.

Edited January 28, 2012 by newts

[ewmon]

Sea water has a density of 1.025 g/cm, so applying a force if 1.025 g for every cm² of sea surface will raise the water 1 cm, and the work performed is 1.025 g × 1 cm for every cm² of surface raised. Physics is not one of my main strengths; am I right?

[newts]

Sea water has a density of 1.025 g/cm, so applying a force if 1.025 g for every cm² of sea surface will raise the water 1 cm, and the work performed is 1.025 g × 1 cm for every cm² of surface raised. Physics is not one of my main strengths; am I right?

Judging by your excellent post on the 'how radio works' thread, I presume your expertise lies in electrical engineering.

 

I think your question might be about how atmospheric conditions affect tide heights. That would presumably be the case, but it is not something I know much about. My page is about calculating the basic force behind the tides, whose heights can be calculated hundreds of years into the past and future, based on the position of the moon and sun.

 

The mass of the water is not usually included in the calculations, as the basis of the tide-raising force is the difference in the force of gravity on different parts of the earth; and as Galileo said 'all bodies fall at the same rate', or as modern physicists say, gravitational and inertial mass are equivalent.

 

However once a tidal bulge forms, the gravitational attraction of the bulge itself increases its own size. Because water is only about 1/5.5 times the average density of the earth, this factor is not very large, and would perhaps increase tide heights by about 1/10. But if the seas were made of mercury, this would be a major factor, and if my calculations are right then these mercury tides would be enormous.

[D H]

However once a tidal bulge forms, the gravitational attraction of the bulge itself increases its own size. Because water is only about 1/5.5 times the average density of the earth, this factor is not very large, and would perhaps increase tide heights by about 1/10. But if the seas were made of mercury, this would be a major factor, and if my calculations are right then these mercury tides would be enormous.

Uh, no. If the oceans were made of mercury the tides would be smaller. Mercury is more viscous than water and is more dense than water. The greater viscosity results in a greater lag and a smaller amplitude. The greater density results in a greater self-gravitation, which acts against the tidal potential from some third body.

 

Think of it this way: The Earth tides would be much larger than the ocean tides were your analysis correct. It's the other way around; the Earth tides are smaller than the ocean tides.

[imatfaal]

Uh, no. If the oceans were made of mercury the tides would be smaller. Mercury is more viscous than water and is more dense than water. The greater viscosity results in a greater lag and a smaller amplitude. The greater density results in a greater self-gravitation, which acts against the tidal potential from some third body.

 

Think of it this way: The Earth tides would be much larger than the ocean tides were your analysis correct. It's the other way around; the Earth tides are smaller than the ocean tides.

 

Taking it to the other extreme - can we measure the change in the atmosphere, ie does the atmosphere have tides as well? I presume it must have, I just wonder whether we can define the edge well enough to be able to notice.

[D H]

Taking it to the other extreme - can we measure the change in the atmosphere, ie does the atmosphere have tides as well? I presume it must have, I just wonder whether we can define the edge well enough to be able to notice.

The atmosphere also has tides.

 

Defining tides a bit more loosely as any regular variations (i.e., after removing effects such as weather) in behavior, the atmospheric tides have thermal and gravitational components. The thermal tides dominate over gravitational tides in the case of the atmosphere. The upper atmosphere has a diurnal bulge. Density at a given altitude varies with local time of day, with a maximum at about 2 PM local time and a minimum at about 4 AM. These density variations affect the behaviors of vehicles in low Earth orbit (hence my interest in the tides).

 

The atmospheric thermal tides result because the atmosphere is a compressible fluid and thus is highly sensitive to temperature. The atmosphere is heated mostly from the ground and in the ozone layer. It's the heating in the ozone layer that is the primary driver of upper atmosphere thermal tides. Thermal tides also exist in the lower atmosphere, but they're a bit more complex than the upper atmospheric thermal tides.

 

Gravitational tides in the atmosphere can be seen after one removes the weather and the thermal tides. They are very small.

[newts]

Uh, no. If the oceans were made of mercury the tides would be smaller. Mercury is more viscous than water and is more dense than water. The greater viscosity results in a greater lag and a smaller amplitude. The greater density results in a greater self-gravitation, which acts against the tidal potential from some third body.

 

Think of it this way: The Earth tides would be much larger than the ocean tides were your analysis correct. It's the other way around; the Earth tides are smaller than the ocean tides.

 

I see the point you are making, and it could be true for the earth rotating once a day. It is not something I would know how to calculate.

 

Certainly the land tides are less than the water tides. But in the case of the equatorial bulge created by the earth’s rotation about its axis, the bulge is nearly twice what it would be if the earth was a solid sphere covered in water. So by the same argument, if the same side of the earth always faced the moon, then the gravity of the tidal bulges at each end would attract more matter towards themselves; and in the case of the mercury, this gravity would theoretically be strong enough to totally drain a line round the circumference of the earth in between.

 

If the earth rotated very slowly, then that argument would still hold. With the earth rotating once a day, you might be right. Or it might depend on the geometry of the oceans. When calculating the tide-raising force, I just assumed a solid spherical earth covered in a uniform depth of water; and then calculated what the difference would be between the high and low tides, if the earth always had the same point facing the moon. With the earth rotating fast, I think tides heights would depend on how the tidal bulges resonate, but that would not be an easy calculation.

 

 

[D H]

Certainly the land tides are less than the water tides. But in the case of the equatorial bulge created by the earth’s rotation about its axis, the bulge is nearly twice what it would be if the earth was a solid sphere covered in water.

That's just wrong. The bulge would be larger (but only slightly) if the earth was covered with water. If you were correct, the equatorial regions would be dry while the poles would be twenty miles underwater.

 

 

 

So by the same argument, if the same side of the earth always faced the moon, then the gravity of the tidal bulges at each end would attract more matter towards themselves; and in the case of the mercury, this gravity would theoretically be strong enough to totally drain a line round the circumference of the earth in between.

That's even more wrong. Ignoring third body effects (tidal gravity) and rotational effects, self-gravitation acts to pull an object into a spherical shape. Always. A spherical shape is the shape that minimizes energy. Add in those other effects and the minimal energy shape will of course be something other than spherical. Increase the density of the primary body and self-gravitation will make the shape more spherical, not less.

 

 

I suggest you look into Lagrangian and Hamiltonian dynamics.

[newts]

That's just wrong. The bulge would be larger (but only slightly) if the earth was covered with water. If you were correct, the equatorial regions would be dry while the poles would be twenty miles underwater.

You misunderstand the mechanism. The equatorial bulge is created by centrifugal forces; but once it exists, the gravity of this bulge attracts more matter towards it. If the solid earth remained spherical, and the bulge was solely composed of water, then the relative lightness of the water would mean a smaller bulge. Because the rock of the earth is deformed, the bulge is denser and creates a larger gravitational effect, and this gravitational effect applies to the oceans as well as the land.

 

If you cannot follow that, you could try calculating the size of the bulge yourself using Lagrangian and Hamiltonian dynamics, or you could visit my page http://squishtheory....uatorial-bulge/.

 

self-gravitation acts to pull an object into a spherical shape. Always. A spherical shape is the shape that minimizes energy.

If a body is of uniform mass then gravity will always want to pull it into a spherical shape. If we have a solid sphere covered in a liquid that is less dense than the sphere, then the same holds true.

 

But imagine a spherical planet made of polystyrene, covered in a uniform layer of mercury. That certainly does not minimise gravitational potentials. In that extreme example the mercury would want to form its own sphere, only slightly distorted by the presence of the attached polystyrene planet.

[D H]

You misunderstand the mechanism.

You've got that exactly backwards.

 

Rather than doing this on your own, perhaps you should pick up a text on geodesy. You aren't going to find these derivations (written well) on the 'net. You can find some sites that do derive the nature of the bulge, but with more than a bit of hand waving.

 

If you insist on doing this on your own, start with a uniform density object. You should find that the flattening is proportional to the square of the rotational rate and inversely proportional to density.

 

 

or you could visit my page http://squishtheory....uatorial-bulge/.

I have. I suggest that you read a book.

[newts]

You aren't going to find these derivations (written well) on the 'net. You can find some sites that do derive the nature of the bulge, but with more than a bit of hand waving.

That is why I created the pages. If there were other pages that showed how to calculate the tides and the equatorial bulge correctly, I would not have bothered. If there are books which do this simply, it is a bit odd that no one of the web appears to have read and understoood them.

 

If you insist on doing this on your own, start with a uniform density object. You should find that the flattening is proportional to the square of the rotational rate and inversely proportional to density.

That is what I did do. But the way I derived the formula quoted on Wikipedia, means that I can easily adapt it, to show that an equatorial bulge caused by water alone on a otherwise solid spherical earth would be much smaller than when the rock also distorts. The formula quoted on Wikipedia, is correct for a planet of uniform density, but on its own it gives no understanding of the mechanism, which is presumably why you incorrectly assumed that a bulge of water alone would be larger.