# tidal force diagram

For the initial consideration the earth’s axial rotation is ignored and just the mutual revolution of both earth and moon about the E-M barycentre is considered.

Initially the much slower movement of the E-M couple about the sun will also be left out. This will be considered as an inertial reference frame, with the moon revolving in the direction indicated.

The solid earth will revolve around the barycentre with radius re determined by the mass ration of earth and moon.

Since the earth is not rotating each point on the earth will describe the same motion on a parallel path.

Two equal quantities of water at points on the surface of the ocean diametrically opposite and on the instantaneous E-M axis are considered. The gravitational attraction at each point will not be equal. They will be respectively greater and less than the value at the centre of the earth.

F2 > Fe > F1

These instantaneous values of the forces are the centripetal force that will cause the revolution along the given paths.

Since the angular velocity of each point must be the same, the instantaneous circular motions due to the centripetal acceleration will be inversely proportional to these forces.

F = m2.w/r

r2 < re < r1

Thus the water closest to the moon will be in equilibrium at a smaller radius and the far-side will have a greater radius. Similar arguments will apply to a reducing degree, further away from the axis, as the distance to the moon approaches that of the centre of the earth. Hence the typical description of two bulges of water, one near side and one far side, of roughly equal form is found. (Small deviations due to the small convergence of the lines of force and non-linearity being ignored at this stage).

One quarter of a lunar revolution ( 27.2 / 4 days ) later the moon will be below in this diagram. The earth will have revolved to be above the barycentre ( point B ).

Since the earth is not rotating about B, point 1 will still be to the left thus now perpendicular to the line of E-M axis. It will experience the same gravity as the solid earth and will have a lower height and a radius of movement re. Similarly for point 2. Water will have flowed towards and away from the moon around points 1 and 2 allowing them to attain ‘low tide’ level.

Point e will now be the far-side bulge.

Another quarter cycle later and the initial case will be reversed:

r2 > re > r1

with point 2 being the far-side with the larger instantaneous radius of motion.

Since this is an inertial frame of reference this gives a simplistic but complete description and there are no fictitious forces such as centrifugal or Coriolis to consider.

The effect is solely a result of the gradient of the gravitational field.

Simplifications:

The bulges would be symmetrical if the force line were all parallel. In fact they are slightly convergent towards the centre of mass of the moon. The tangential forces are not at points like point on a section through the earth’s centre but closer to the moon.

If account is taken of this convergence the near-side bulge become more pointed and the far-side bulge slightly broader and of lesser height. The result is a slight egg-shaped deformation.

The earth-moon couple is also revovling around the sun (it is perhaps better described as a binary planet in solar orbit since the lunar path is always convex, ie. it does not have little loops in its path around the sun.)

Similar arguments can be applied to orbit around the sun giving rise to two more independant “bulges” that periodically align with the lunar ones. This is the cause of the neap and spring tides.