There has been a long animated discussion about tides at Wattsupwiththat which highlighted a confusion about both the causes of tides and their strength. Several people are adamant that tides are a universal phenomena experienced by any object near a large gravitational source. They argue that tides are caused by the gradient of the 1/r^2 field.
While an extended object falling into a star or black hole will experience a transient tidal force, it is only bodies in orbit around each other that experience long term tidal forces acting on the surface. The earth is in orbit about the earth-moon barycenter. It is not in an inertial frame. It is in an accelerating frame of reference, similar to the way that the rotation of the earth about its axis causes a reduction in g at the equator.
So what is the real cause of tides on earth and why is the lunar tide larger than the solar tide ? Here is my derivation for the formulae for tides.
For 2 bodies in orbit the “centrifugal force” must balance the gravitational force. The centrifugal force is constant within a solid body in the same way that it is for a plate spinning on a stick.
Therefore the centrifugal acceleration on body M is
Now consider the net force per unit mass acting on point a). Assuming that = 0 we get
assuming that r>>R we can do a binomial expansion to get
This derives the approximate formula for the tidal force. There are two bulges centered on
Now we can do the full calculation where the angle is no longer zero and thereby identify how the tidal force acquires a vertical component.
The distance a-m is by Pythagorus
Gravity acting on point a) is therefore =
Net tidal force now has 2 components
Now let’s compare the two solutions by calculating the effective tidal acceleration. Figure 3 shows the approximate formula in red and the two components of the exact result in blue.
On earth the gravitational acceleration ‘g’ = 9.8 m/s^2 This can be compared with the above “tidal” accelerations of ~ 10^-6 m/s^2. So the moon’s tidal force is 10 million times less than the earth’s gravitational force at the surface. This is not going to do any direct heavy lifting of the oceans! Instead it is the tractional component of the tidal force parallel to the surface which moves vast quantities of ocean. Now with the addition of the vertical component Fy the vector diagrams of tidal forces looks more like this:
Note that the largest tractional forces are at larger theta angles. This drag of water currents throughout the depth of the ocean results in both tidal bulges. Figure 5 shows the tractional north-south force parallel to the earth’s surface which is unaffected by the earth’s gravity and therefore moves water from outside the bulge towards the centre.
The position and strength of these tractional forces is constantly changing during the lunar month, during the 18.6 year precession cycle and during longer time scale astronomical cycles. So the way I like to understand tides is that both the moon and the sun exert a horizontal drag force on the oceans and atmosphere. Twice a month we get spring tides when the sun and moon line up at new moon and the full moon, whose strength depends on the coincidence with both orbital perigees. Long term astronomical variations must have an effect on climate as orbital parameters slowly change increasing or decreasing ocean mixing and atmospheric dynamics.
There is a new argument about whether centrifugal forces play any role at all in tides. In some sense both sides in this argument are correct. You don’t need to use the centrifugal force to derive the formula for tides. This is because there is a perfect balance between the centrifugal force and the gravitational force at the center of the earth when in orbit around the earth-moon barycenter. This balance also determines the strength of tides on earth. When in doubt see what Feynman says.
What do we mean by “balanced”? What balances? If the moon pulls the whole earth toward it, why doesn’t the earth fall right “up” to the moon? Because the earth does the same trick as the moon, it goes in a circle around a point which is inside the earth but not at its center. The moon does not just go around the earth, the earth and the moon both go around a central position, each falling toward this common position. This motion around the common center is what balances the fall of each. So the earth is not going in a straight line either; it travels in a circle. The water on the far side is “unbalanced” because the moon’s attraction there is weaker than it is at the center of the earth, where it just balances the “centrifugal force.” The result of this imbalance is that the water rises up, away from the center of the earth. On the near side, the attraction from the moon is stronger, and the imbalance is in the opposite direction in space, but again away from the center of the earth. The net result is that we get two tidal bulges.
A body in free fall into the sun experiences ever increasing tidal forces until it is torn apart. A body in orbit eperiences varying tidal forces depending on the eccentricity of the orbit. So on a purely logical basis Willis and Greg are correct because centrifugal forces don’t enter into the formula for tides. However in order to calculate the variations of tides on earth you need to include orbital dynamics because they change the earth-moon distance.