Tidal Force - Mathematical Treatment

Mathematical Treatment

For a given (externally-generated) gravitational field, the tidal acceleration at a point with respect to a body is obtained by vectorially subtracting the gravitational acceleration at the center of the body (due to the given externally-generated field) from the gravitational acceleration (due to the same field) at the given point. Correspondingly, the term tidal force is used to describe the forces due to tidal acceleration. Note that for these purposes the only gravitational field considered is the external one; the gravitational field of the body (as shown in the graphic) is not relevant. (In other words the comparison is with the conditions at the given point as they would be if there were no externally-generated field acting unequally at the given point and at the center of the reference body. The externally-generated field is usually that produced by a perturbing third body, often the Sun or the Moon in the frequent example-cases of points on or above the Earth's surface in a geocentric reference frame.).

Tidal acceleration does not require rotation or orbiting bodies; for example, the body may be freefalling in a straight line under the influence of a gravitational field while still being influenced by (changing) tidal acceleration.

By Newton's law of universal gravitation and laws of motion, a body of mass m a distance R from the center of a sphere of mass M feels a force equivalent to an acceleration, where:

. . ., and . . . . . . ,

where is a unit vector pointing from the body M to the body m (here, acceleration from m towards M has negative sign).

Consider now the acceleration due to the sphere of mass M experienced by a particle in the vicinity of the body of mass m. With R as the distance from the center of M to the center of m, let ∆r be the (relatively small) distance of the particle from the center of the body of mass m. For simplicity, distances are first considered only in the direction pointing towards or away from the sphere of mass M. If the body of mass m is itself a sphere of radius ∆r, then the new particle considered may be located on its surface, at a distance (R ± ∆r) from the centre of the sphere of mass M, and ∆r may be taken as positive where the particle's distance from M is greater than R. Leaving aside whatever gravitational acceleration may be experienced by the particle towards m on account of m's own mass, we have the acceleration on the particle due to gravitational force towards M as:

Pulling out the R2 term from the denominator gives:

The Maclaurin series of 1/(1 + x)2 is 1 – 2x + 3x2 – ..., which gives a series expansion of:

The first term is the gravitational acceleration due to M at the center of the reference body, i.e. at the point where is zero. This term does not affect the observed acceleration of particles on the surface of m because with respect to M, m (and everything on its surface) is in free fall. When the force on the far particle is subtracted from the force on the near particle, this first term cancels, as do all other even-order terms. The remaining (residual) terms represent the difference mentioned above and are tidal force (acceleration) terms. When ∆r is small compared to R, the terms after the first residual term are very small and can be neglected, giving the approximate tidal acceleration (axial) for the distances ∆r considered, along the axis joining the centers of m and M:

(axial)

When calculated in this way for the case where ∆r is a distance along the axis joining the centers of m and M, is directed outwards from to the center of m (where ∆r is zero).

Tidal accelerations can also be calculated away from the axis connecting the bodies m and M, requiring a vector calculation. In the plane perpendicular to that axis, the tidal acceleration is directed inwards (towards the center where ∆r is zero), and its magnitude is (axial) in linear approximation as in Figure 2.

The tidal accelerations at the surface of planets in the Solar System are generally very small. For example, the lunar tidal acceleration at the Earth's surface along the Moon-Earth axis is about 1.1 × 10−7 g, while the solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about 0.52 × 10−7 g, where g is the gravitational acceleration at the Earth's surface. Modern estimates put the size of the tide-raising force (acceleration) due to the Sun at about 45% of that due to the Moon. The solar tidal acceleration at the Earth's surface was first given by Newton in the 'Principia'