With reference to the answer that appeared in these columns on Feb 29, a discerning student might very well ask: The centrifugal force is not a real force, so what is the real explanation of tides?
It is not difficult to explain tides. The question to ask is: Does not the moon’s gravity pull the rest of the earth, which is under the ocean water, towards it as well?
It does. But the surface directly under the moon is nearer to it than the rest of the earth below, and hence gets pulled more. If this surface is water, being more elastic, it rises in the direction of the moon. The water at the ocean floor ‘remains’ with the earth. The difference between the ocean surface and the floor becomes greater than it normally is. We recognize this as high tide.
Coming now to the opposite side of the earth, the bulk of the earth is nearer to the moon than the surface here and is pulled more. Water at the ocean floor here is ‘carried along’ with the rest of the earth towards the moon, but water on the surface stays where the lesser pull on it dictates. Again the difference between the surface and the floor increases and we see a high tide. When the moon is at the horizon at right angles to the overhead position the ocean and the earth below it are at roughly the same distance from the moon and experience the same pull. The ocean depth is normal. But, because some of the ocean water has flown to where there is a high tide, there is less water here and we see a low tide. Of course, this is a very simple model and the tides are influenced by many other factors, such as latitude and the shape of the continents.