Calculation of tide-raising force using differential gravity

Calculation of tide-raising force by finding the difference between centrifugal forces and gravitational potentials

One final calculation

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It is interesting how many webpages contain inaccurate information about the tides. What is even more fascinating, is the number that accuse others of getting it wrong, and then still contain errors themselves.  Probably the moral of the story, is that if one cannot use a theory to calculate anything, then one does not properly understand it, and usually those webpages with the worst errors, do not contain calculations.  Therefore the main point of this page, is to show ways of calculating the tidal force; so that readers can judge for themselves the best description of what is really happening.

The cause of the tides, along with the maths, was first identified by Newton around 350 years ago.  We are indeed fortunate that Newton did this, because if the task had been left to modern physicists, then in addition to gluons and gravitons, we might well have to contend with tideons, and presumably also anti-tideons to explain the bulge on the far side from the tide-raising body.

The reason tides exist, is that the gravitational force exerted by a body such as the moon, differs slightly over the vast expanse of the oceans. Imagine you had a large spherical jelly, and you decided to move it by attaching pieces of string to points all over it. If you were to pull each string equally hard, and in exactly the same direction, then the jelly would move without distorting. But as soon as you pull harder on one string than another, or you pull one string in a slightly different direction, then the jelly will lose its spherical shape.

So the tides are caused by the fact that the gravity of the sun and moon, pulls with different strengths, and at different angles, on the particles that make up the earth, and in particular on the water particles.

In Newtonian physics, the force of gravity depends only on a body’s position, not on how it is moving. So the instantaneous tide-raising force on the earth, would be the same, whether the earth was moving towards the moon, away from it, orbiting it, or was momentarily motionless. So if we can calculate the tide-raising force for the stationary earth, at the correct distance from the moon and sun, then that result will also apply to the earth in orbit.

If we start with a spherical earth, here represented by just 5 independent particles, and allow them to fall a small distance towards a source of gravity, we get the following effect:

The red circle represents the original spherical earth, the blue ellipse a vast exaggeration of the earth’s shape after it has fallen. The effect on the particles above and below the page, is the same as the particles on the left and right; so in 3 dimensions the earth becomes slightly lemon-shaped; that is to say it has a circular axis that is smaller than its elliptical axis.

If we deduct the motion of the earth towards the source of gravity, and just consider how the tidal forces affect the shape of the earth; we get this, where the length of the arrows indicate the strength of the force:

Calculating actual tide heights is not practical, as they depend on the shape of the land and oceans. So a calculation of the tide-raising force, actually involves working out what the shape of the ocean would be, if the earth was not spinning and was covered by a uniform thickness of water. Or more specifically the difference between the circular diameter of the resulting lemon-shape and its length.


Calculation of tide-raising force using differential gravity


One way to do this, is to imagine tunnels drilled to the centre of the earth. We can then resolve forces in two directions, and add these together.

We could think of the problem in terms of balancing the water pressure in the two tunnels. But it is perhaps easier to consider dropping a ball down one tunnel, deflecting it at the centre, then seeing how high above the surface it rises when it reaches the top of the second tunnel.

What we are calculating, is a tidal force times distance, which gives us a certain amount of energy; and then working out how high this amount of energy will allow a ball or a water-drop, to rise above the surface of the earth, against the force of gravity at the earth‘s surface.

In order to simplify the calculations, I will take the mass of the earth as 1, its radius as 1, and therefore the force of gravity at its surface will also be 1. This means that the mass of the moon becomes 1/80, and the distance between the centre of the earth and the centre of the moon = 60On the descent, the tidal force results from the angle of pull. At the surface, the gravitational force of the moon, is almost the same magnitude as at the centre = 1/80 ´ 1/60² . The component of the force is about 1/60. So the actual tidal force at the surface is roughly 1/80 ´ 1/60² ´ 1/60 = 1/80 ´ 1/60³ .

Halfway down the force is half that, so the average force over the whole tunnel = ½ ´ 1/80 ´ 1/60³ .  The energy is that times the distance, which is the radius of the earth, which is 1. An energy value of ½ ´ 1/80 ´ 1/60³ ; is enough to raise a drop of water that proportion of the earth’s diameter, against the force of gravity at the earth’s surface. So it creates a tide of ½ ´ 1/80 ´ 1/60³ ´ 6,400,000 m = 0.19m.

On the ascent back up the second tunnel, the tidal force results not from the angle of pull, but from the differing strength of the moon’s gravity over the length of the tunnel. At the centre of the earth, the gravitational force = 1/80 ´ 1/60² , whilst at the point on the earth’s surface furthest from the moon, it is only 1/80 ´ 1/61² . (1/80 ´ 1/60² ) (1/80 ´ 1/61² ) is very close to 2/80 ´ 1/60³ . Halfway down the tunnel, the force is roughly half that, so the average force over the tunnel is 1/80 ´ 1/60³ , and the length of the tunnel is again 1. Therefore the energy of the ascent, is twice that of the descent; so the total tidal energy of the whole journey is enough to raise the water to a height of 3 ´ 0.19 = 0.57 metres. Allowing for my approximations of mass and distance, that agrees with the value given on Wikipedia.

The formula I have derived, shows that the tidal difference between the circular radius of the lemon-shape, and half its length, is proportional to the mass of the tide-raising body divided by the distance between the two bodies cubed. Since the sun is 333,000 ´ 80 times as heavy as the moon, and since it is 390 times as far away, its tidal effect will be (333,000 ´ 80) / 390³ = 0.45 that of the moon. So if the earth, sun and moon are arranged in a straight line, the tide-raising force is 1.45 times that of the moon alone, whilst if they are arranged such that the force of the sun is working against the moon, then the force is only .55.  Which explains why tide heights vary so much over the lunar month.

The general equation for the height of the bulge created on a non-rotating body of mass M(e), by the tidal force of a body of mass M(t), is given by the equation:

3/2 ´ M(t)/M(e) ´ (R/D)³ ´ R

Where R is the radius of the body being squashed by tidal forces, and D is the distance between the centres of the two bodies.


Calculation of tide-raising force by finding the difference between centrifugal forces and gravitational potentials


There is often debate amongst tidal theorists, as to whether or not the tides are caused by centrifugal forces. As so often in these cases, it is less a question of fact, and more one of definition. I calculated the tide-raising force, without mentioning centrifugal force, because the value of the tide-raising force does not depend on how the earth is moving.

However, realistically, the earth cannot just hover in space; the only way that heavenly bodies remain a constant distance apart, is by orbiting each other, or more precisely orbiting the centre of gravity of the system, known as the barycentre.

The gravitational force attracting two orbiting bodies is known as the centripetal force. To explain why the bodies do not move towards each other under the effect of this force, an equal and opposite centrifugal force is postulated. However this centrifugal force is not a force in the same sense as gravity or electric forces, whose values can properly be described in terms of how much energy is released when objects move closer together or further apart. Despite this, the centrifugal effect can be usefully modelled as though it is a force, it is just a pity that physicists do not instead call it ‘centrifugal inertia’.

In order to show that Newtonian mechanics is consistent, I will now do some tidal calculations taking centrifugal forces into account. Let us start with the earth and the moon solidly fixed together with a stick of the correct length. Since the system is not rotating, there will be a tendency for the water to flow from the far side, to the side nearest the moon, under the effect of gravity, as shown below.

The size of the bulge can be calculated by drilling a hole through the centre of the earth, so that we can resolve forces in a straight line. The gravitational force of the moon on the far side is 1/80 ´ 1/61² , at the near side it is 1/80 ´ 1/59² , so the average force over the distance of 2 is roughly 1/80 ´ 1/60² . This provides enough energy to raise the water 2/80 ´ 1/60² of the earth’s radius of 6,400,000 metres = 44 metres.

So I have calculated that if the earth and the moon were held rigidly together with a stick, but were not rotating; then there would be a bulge of 2/80 ´ 1/60² on the side nearest the moon. Of course as soon as we start rotating the system around its barycentre, there will be a centrifugal force trying to push the water in the opposite direction. If we rotate the system at exactly the speed at which the stick no longer has any effect, then since the tide heights at the points nearest and furthest the moon, are approximately equal; we would expect the centrifugal force to more or less cancel out the 44 metre gravitational bulge.

When the earth and moon are in freefall, and the stick no longer has any effect, the centrifugal force on the centre of the earth, due to its rotation about the barycentre, must exactly balance the gravitational attraction of the moon, which is 1/80 ´ 1/60² . If we were to assume that the barycentre was on the edge of the earth:

Then from the diagram, the average centrifugal force would be exactly {1/80´ 1/60² }, and the distance over which it acts 2. So it would indeed balance out the 44m bulge calculated earlier.

If the barycentre was between the edge of the earth and the moon, it is not too difficult to see that the average centrifugal force across the earth, would still be the same. In fact even with the barycentre inside the earth’s radius, the same calculation holds true, but I will omit that bit of maths.

Actually the approximate calculation of the gravitational bulge with the stick in place can be done more precisely, using gravitational potentials, as: (1/80) ´ (1/59 1/61) = (1/80) ´ (61 59) ¸ (59 ´ 61) = (1/80) ´ 2/(60² 1). If we deduct from this the centrifugal energy of {2/80 ´ 1/60² }, we get 2/80 ´ [(60² ) (60² 1)] ¸ [(60² 1)(60² )] which is roughly equal to 2/80 ´ 1/604.

The tide height I calculated for the points nearest and furthest from the moon, was roughly 3/2 ´ 1/80 ´ 1/60³ . If we express 2/80 ´ 1/604 as a proportion of that, we get 1/45, which gives us a value for how much larger the tide-raising force towards the point nearest the moon is, than the force towards the point furthest from the moon.

You may have noticed that whilst the tide-raising force depends on the inverse square of the distance between bodies; the difference in tide-raising force between the points nearest and furthest from bodies, depends on the inverse fourth power. So since the sun is 390 times as far from the earth as the moon, its differential tide-raising force will be only 1/390 ´ 1/45.

As the difference of 1/45 in the tide-raising force only applies to the moon, when the earth, sun and moon are lined up, the overall difference will actually be about  [1/45 ¸ 1.45] = 1/65.   However when the sun is effectively cancelling out part of the moon’s tide-raising force, and tides are at their smallest, then the total tide-raising force on the side nearest the moon will be about  [1/45 ¸ 0.55] = 1/25  larger than that on the opposite side, which is something that might be detectable in tide tables.

However it might not be very noticeable, because whilst the general tide heights have two weeks to adjust from the minimal neap tides, to the maximal spring tides; what we are really talking about with the factor of 1/25, is the difference between the two tides, at a particular place, on a particular day. So in theory the tide when the moon is overhead, ought to be up to 1/25 greater than the next one, twelve hours later, when the moon is underfoot. However high tides do not actually occur when the moon is directly overhead or underfoot, as the water takes time to respond to the tidal force, so there are a number of difficulties in finding evidence of the 1/25 factor.

What I now want to do, is calculate the tide-raising force between point P and the point nearest to the moon. 

The difference in gravitational potential = [1/59 1/Ö (60² + 1)] times the mass of the moon. From this we need to deduct the centrifugal energy released in moving a water-drop from the point nearest the moon to point P.

If we were to assume that the barycentre was at the point nearest the moon, then again we would get the right answer for the centrifugal energy. However rather than expect you to accept that on faith, or try to prove it mathematically, I will instead incorrectly assume the barycentre is at a distance of exactly ¾ from the centre of the earth, and show that this does still give exactly the right answer (using my figures, the barycentre would actually be a distance of 61/80 from the centre of the earth).

The centrifugal force of the centre of the earth away from the barycentre, must exactly equal the centripetal force, which is {1/80´ 1/60² }. That means the centrifugal force at a distance of ¾ is {1/80´ 1/60² }. So the centrifugal force at point P = (5/4 ¸ ¾){1/80´ 1/60² } = 5/3 ´ {1/80´ 1/60² }.

The average centrifugal force over a tunnel from the barycentre to point P, will be a half of that. To find the centrifugal energy, we need to multiply force times distance: 5/4 ´ ½ ´ 5/3 ´ {1/80´ 1/60² }= 25/24 ´ {1/80´ 1/60² }.

From this result we still need to deduct the centrifugal energy lost in moving from the point nearest the moon to the barycentre: ½ ´¸ ¾)´ ¼ ´ {1/80´ 1/60² }= 1/24 ´ {1/80´ 1/60² }. So we end up with a final centrifugal energy, in going from the point nearest the moon to point P of {1/80´ 1/60² }.

We now deduct this centrifugal energy of {1/80´ 1/60² }, from the gravitational energy gained in going in the other direction 1/80 times [1/59 1/Ö (60² + 1)], which by calculator gives 1.517 ´ [1/80 ´ 1/60³ ]. This is larger than the originally calculated tide-raising force of 3/2 ´ [1/80 ´ 1/60³ ], by a factor of about 1/90, because the tide-raising force towards the point nearest the moon, is larger than the tide-raising force towards the point furthest from the moon by about 1/45, and the originally calculated value was in a sense an average.

You may have noticed, that at the beginning I was talking about a lemon-shape, but more recently I have been discussing point P. The reason for this, is that in order to make the centrifugal calculations easier, I assumed that the same point on the earth was always nearest the moon, so the earth’s orbit around the barycentre looks like this:

The point nearest the moon then orbits the barycentre in a circle of radius ¼, the centre of the earth orbits with a radius of ¾, and the point furthest from the moon has an orbital radius of 7/4. What should be apparent is that for each orbit of the barycentre, the earth also revolves once relative to absolute space. So one way of understanding why my calculation only applies to point P and another point on the opposite side of the earth, is to say that the centrifugal force due to the earth rotating once about its centre, forces water away from the areas above and below the page, and towards the drawn circle.

I could have done my calculations on the basis that the earth orbited the barycentre, without rotating relative to absolute space, but that would have been even more difficult, as the orbits would have looked like this:Ideally a scientific theory should start with a logical statement in words, for instance Newton’s theory of gravity can be fully described in a short phrase: each unit of matter in the universe instantaneously attracts every other unit directly towards it with a force proportional to the inverse square of the distance between them. Using this assumption, combined with simple mechanics, the whole solar system can be described mathematically.

Unfortunately people like Newton, with the ability to come up with an abstract idea, then do all the maths to extend it to its logical limit, do not come along very often. Usually physical laws start with an observation, and physicists react by trying to come up with an explanation for the observed phenomenon.

For instance Kepler figured out that a planet on an elliptical orbit, always moved faster the nearer it was to the sun. He then came up with the theory, that the sun was pushing the planets around in their orbits. Not only did this theory agree with the evidence, but it also worked mathematically, as a particular planet half the distance from the sun would go twice as fast, which could easily be explained by saying that the sun was pushing twice as hard.

So Kepler had a theory that agreed with the evidence, was supported by the maths, but was completely wrong. Actually the theory never took hold, but that was because the most gullible fools still rejected heliocentrism; whilst the not quite so gullible fools, like Galileo, rejected elliptical orbits because they thought that all heavenly bodies had to move in perfect circles.

So whenever a physicist claims some incomprehensible piece of nonsense must be a proper description of nature, because it is ‘based on evidence’ or ‘proved by the maths’, remember that even a theory that seems to work, and is created by a great physicist like Kepler, can still be rubbish. You might also note that good physics theories like Newton’s gravity, or the theory of atoms, explain much with little, whilst wrong theories tend to explain little with much.


One final calculation


To finish, I will just show one more way in which the tides can be used to show how consistent Newtonian mechanics are. Let us consider the earth orbiting the sun. As part of my calculation of the tidal force, I assumed that the angle of pull of the sun’s gravity squashed the earth, and accounted for 1/3 of its tide raising force.

Now consider this, if two objects are following one behind the other, on the same orbit, then clearly there is no way that the sun’s gravity can really be squeezing them together:

The paradox is resolved, by noting that the squashing force was used on the basis that the earth did not rotate relative to absolute space; however the system of particles following one behind the other, does actually revolve once around its centre for each orbit of the sun.

So to show that Newtonian mechanics is consistent, I will prove that the centrifugal effect of one revolution of the earth per year about its centre, exactly balances the squashing effect of the component of the sun’s gravity.

The force of the sun’s gravity on the earth is 333,000/R² , the angle of pull is 1/R, so the squashing force on point P is 333,000/R³ .

The centrifugal force on the earth as it orbit’s the sun, must exactly balance the centripetal force, which is again 333,000/R² .  A particle on the circumference of the circle drawn, will rotate about the centre of the earth once a year, but since its radius is 1/R times the radius of the earth’s orbit, the centrifugal force will be 1/R times as much, giving us a centrifugal force of 333,000/R² ´ 1/R = 333,000/R³ , which is equal and opposite to the squashing force.

If you want some more maths, then you could visit my equatorial bulge page, which uses similar principles. If you are interested in the theory of everything, you can read the chapters from my book. If you would just like to have those parts of accepted physics which are actually correct, explained in a simple non-mathematical way, the chapters might also be of interest.

© William Newtspeare, 2012.  Unauthorized use and/or duplication of this material without express and written permission from this blog’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to William Newtspeare, with appropriate and specific direction to the original content.