Orbits and Perturbations
Given a large spherical gravitating body of mass M and a small test particle at a distance r, the Newtonian equations of motion imply that the test particle undergoes an acceleration of magnitude M/r2 in the direction of the gravitating body, and no acceleration in the perpendicular direction. (We are using units such that the gravitational constant and the speed of light are both unity.) The test particle will be confined to a single plane, so its position as a function of time can be expressed in terms of the radial magnitude r(t) and an angular position q(t) as
The second derivatives of these coordinates are
Since the absolute value of q is arbitrary, these equations are equivalent to the conditions
Multiplying through the right hand equation by r, we have
and therefore the quantity in parentheses is constant, i.e.,
This represents the conservation of (specific) angular momentum, and it applies to any central force law, because such a force imposes no torque on the system. The constancy of this quantity also accounts for Kepler's second law, because the incremental area swept out by the position vector in an incremental time is dA = (1/2)r2dq.
Making the substitution dq/dt = h/r2 into the left hand equation (1) gives
Notice that for a circular orbit all the derivatives of r vanish, and this equation reduces to h2 = rM. Making the substitution h = r2w where w = dq/dt, we have w2r3 = M, in accord with Kepler's third law.
We can also use the relation dq/dt = h/r2 to express the derivative of r with respect to time in terms of the derivatives of r with respect to the angular position q. We have
Inserting this expression for the second derivative of r into equation (2) and simplifying gives
Notice that the quantity in parentheses is just the negative of the derivative of 1/r with respect to q. Therefore, letting u = 1/r, we have the simple harmonic equation
In general the solution of an equation of the form
for constants W and p can be written in the form
where k is a constant of integration. In the present case we have W = 1 and p = h2/M. Recalling that r = 1/u, the path of the test particle in the gravitational field of a spherical body of mass M is
If the magnitude of k is less than 1, this is the polar equation of an ellipse with the origin at one focus (Kepler's first law), and with semi-latus rectum p = h2/M. Now, the total area A of an ellipse with the major and minor semi-diameters a and b is pab, and this area can be expressed in terms of the integral
where we have made use of the fact that r2dq = hdt. Hence we have
which of course is Kepler's third law, M = w2a3. Thus we've reproduced Kepler's three laws for stable elliptical orbits of test particles around a perfectly spherical gravitating mass.
However, in actual physical situations, the gravitating body may not be exactly spherical. For example, if the central body is spinning about its axis, it will be slightly oblate. In such a case, the Newtonian gravitational field is not spherically symmetrical, and the force exerted on a test particle at a distance r is not exactly proportional to M/r2. As a result, the actual orbit of a test particle in such a case will not be exactly Keplerian. To see how the effects of a non-spherical source, suppose the mass M is split into two spherical objects on the x axis separated by a distance 2d. The acceleration of a test particle located (also on the x axis) at a distance r from the center of mass of these two bodies is
Since d is small compared with r, we can expand the expression in square brackets to gives
Thus the predominant term in the acceleration is still inversely proportional to the square of r, but we now have a component inversely proportional to the fourth power of r, and another inversely proportional to the sixth power of r, and so on. The non-Keplerian effects for a slightly non-spherical source can be inferred from just the first two terms. Instead of (2), the radial equation of motion is now of the form
where J is a constant related to the oblateness of the gravitating body. Repeating the previous transformations, we arrive at
This is a non-linear equation, but can treat it as a quadratic in u of the form
Solving this algebraically for u gives
The second term inside the square root is much less than 1, so we can closely approximate the solution using just the first couple of terms of the expansion, so we have
Simplifying and re-arranging terms, we get
The last term on the right hand side is negligible small, so we essentially have an equation of the form (3) with
Hence the orbital path is described by the relation
This again is the equation of an ellipse, except that the period of the radial function is not exactly equal to the period of the angular position q. The angular travel necessary to go from one apogee to the next (for example) is not 2p, but rather 2p(1+PQ). Hence the ellipse precesses by the amount 2pPQ radians per revolution. In the case of our oblate gravitating body we have P = MJ/h2 and Q = M/h2, so the orbit precesses by 2p(M/h2)2J radians per revolution.
Incidentally, if J = 3h2 (corresponding to a term equal to h2/r3 in the potential) this gives the same precession as general relativity predicts for a perfectly spherical gravitating body. Of course, the quantity h is specific to the particular orbit in question, whereas J is a fixed characteristic of the gravitating body, so it is not possible to replicate the relativistic precessions of more than one planet by hypothesizing solar oblateness. In order to duplicate the relativistic prediction it would be necessary to hypothesize a gravitational potential that is dependent on the angular velocity of the test particle, not just on its position.
It's worth noting that although the actual oblateness of the Sun is believed to be negligibly small, the relativistic precession is still just a small fraction of the overall precession for a planet such as Mercury. This is because the simple two-body Keplerian orbit is disturbed by the presence of gravitating matter outside the orbit. For example, the orbit of Mercury precesses by about 526.7 seconds of arc per century due to the disturbing effects of the other planets, primarily Venus, Earth, and Jupiter. The French astronomer Urban LeVerrier in 1859 computed the following values for the contributions of the various planets to the advance of Mercury's perihelion. Even though these effects are strictly Newtonian, the calculations are far from trivial, especially for the effect of Venus. However, with certain simplifying assumptions we can derive a simple formula that is fairly accurate for the effects of the outer planets, as well as for the asteroid belt (which LeVerrier seems not to have included).
The orbital periods of the planets are not synchronized, so over the course of many orbits their angular relationships will be averaged out. On this basis, we could make a rough estimate of the perturbing effect of an outer planet (like Jupiter) on the precession of Mercury's orbit by imagining that the mass m of Jupiter is distributed uniformly in a ring around the Sun with a radius equal to the radius R of Jupiter's actual orbit. As discussed in Gravity of a Torus, the gravitational acceleration inside such a massive ring for a test particle (in the plane of the ring) at a distance r from the center is
directed radially outward from the center. Hence the lowest order effect is an outward acceleration directly proportional to r, superimposed on the inverse-square acceleration inward toward the central gravitating body. The radial equation of motion is therefore
Transforming to the u variable as we did previously leads to the equation
Unfortunately this can't be solved in the same manner as the previous equations, because of the u3 in the denominator of the last term. However, for small eccentricities the range of r values that must be covered is fairly small, and we can represent the right hand side of (5) by a more tractable form. Letting r1 and r2 denote the minimum and maximum radial distances of Mercury from the Sun, we seek constants A and B such that
Solving this for A and B, letting r0 denote the mean orbital radius of the test particle, and assuming very small eccentricity, we can re-write equation (5) approximately as
which can also be written as
Transforming back to the u parameter as before, this becomes
Using the fact that h2 is approximately equal to Mr0, and expanding the coefficient of the leading term, we see that the angular factor is
Subtracting this from 1 and multiplying by 2p, we get the approximate orbital precession per revolution due to a uniform ring of mass m and radius R. For a concentrated mass we expect that the precession would be some fraction of this value, since the mass would be sometimes located so as to give no precession at all. We might plausibly apply a factor of 1/2 to the precession for a ring of the same mass. On this basis we arrive at the formula
We expect this to apply to the outer planets, but not nearly as well to the effect of the inner planets. This is confirmed by the table below, which shows the exact contributions to the precession of Mercury's orbit due to each of the next six planets as computed by LeVerrier, and as given by the above formula. The table also includes the contributions of the asteroid belt and the planet Neptune, neither of which were included in LeVerrier's analysis (presumably because he considered them negligible). The values listed are in units of arc seconds per century.
Oddly enough, the precession for Venus is very close to the value for a ring of uniformly distributed mass, i.e., twice the value given by equation (6). For the Earth the effect is somewhere in between the full ring and half the ring value, whereas for the remainder of the planets the effect is very nearly equal to half the ring value. It's also interesting to note that the crossover point where equation (6) transitions from an under-estimate to an over-estimate is at the asteroid belt, which actually is a somewhat uniformly distributed ring of mass. For that entry in the table we have taken the full ring value, omitting the factor of 1/2 in equation (6).
Presumably LeVerrier's calculation of the Earth's contribution included the mass of the Moon, which is about 1.2% of the Earth's mass. This represents roughly 1 second of arc per century. Apparently his calculations did not include the asteroids, which contribute roughly 0.26 seconds of arc per century. There are also a number of "Earth-crossing" asteroids that would have been omitted from Leverrier's calculation, but the total mass of these may be negligible.