Solar System Dynamics

orbital period = 2 * π * (semimajor axis³ / (G * mass))^1/2

force due to gravity = G * mass₁ * mass₂ / distance²

We would like to see how Newton provided a simple and elegant mathematical framework encompassing Kepler's Laws. So we will apply Newton's Law of Gravity to a simple system of a planet and a star, and see if we can derive Kepler's Laws from them.

Let us first assume that the mass m is orbiting in the gravitational field created by the mass M, and that m is spherical and far enough from M that it can be considered a point mass. Considering the enormous distances between them relative to their sizes, this is not a problem.

Newton's law of gravity is written

F = - G m M / r²

or, using his Second law (force = mass * acceleration), as

m a = - G m M / r²

From this we see immediately that Newton's law of gravity DOES NOT DEPEND ON m .

In Einstein's General Relativity, this is a result of the Equivalence Principle, which states that Gravity acts the same on all masses regardless of their composition: the mass's acceleration does not depend on the mass itself.

Rearranging, we then have

(G M) / r² + a = 0

But there is another piece of information that is hidden in Newton's law of gravity: THE FORCE ACTS ON A LINE BETWEEN THE TWO MASSES.

Since two points define a plane, this is a planar problem, meaning that we have to think in terms of x and y directions. We lose no generality by choosing a coordinate system so that M is at the origin and m is at {x[t], y[t]}. Since acceleration is the rate of change of the rate of change of the position with time, we write a as

{x''[t], y''[t]}

(where each prime indicates "rate of change") and multiply the (G M)/r² term by the direction {x[t], y[t]}/r, where r is √(x²+y²) (using the Pythagorean theorem). Newton's law then becomes a pair of equations:

G M x[t] / √(x[t]² + y[t]²)³ + x''[t]=0,
G M y[t] / √(x[t]² + y[t]²]³ + y''[t]=0

This is a miserable pair of equations to solve, but if we express them in terms of r and the angle θ between the line connecting m and M, and the x axis:

we can attack them more simply.

First recall that x/r = cos θ and y/r = sin θ. Making these substitutions into our pair of equations and using the calculus that Newton invented for the occasion gives us two equivalent equations that are much more tractable:

G M / r[t]² - r[t] θ'[t]² + r''[t]=0,
r[t]² θ'[t] = J

In the second equation, J is a constant called the angular momentum. It measures both the rate of rotation and the bulk of the rotating object. Since the area of a triangle is one half the base times the height, r[t]² θ'[t] is twice the rate at which area is swept out by the path r[t]:

So we have just found Kepler's second law as a result of Newton's law of gravity: the area swept out in any given time is a constant.

With some significant but perfectly legal manipulations (involving more of Newton's calculus), we can make the substitution u=1/r, trade t dependence for θ dependence, and change the other equation to

u''[θ] + u[θ] = G M / J²

This has the solution (after we substitute 1/r for u)

r = ((1 + ε) r₀) / (1 + ε Cos[θ])

Here, we have traded the two physical constants describing the motion (M and J) for two geometrical constants which describe the orbit: (G M)/J² is 1/(r₀(1+ε)), where r₀ is the perihelion and ε is the eccentricity of the ellipse describing the orbit.

As expected, M is at one focus: the origin. So we have found Kepler's first law as well.

Since the rate at which area is swept out is J/2, the time it takes to complete one orbit (the period) is the area of the orbit (π a b) divided by J/2. Now the length of the semimajor axis (a) is (r[0]+r[π])/2, or

(r₀ + r₀ (1 + ε) / (1 - ε)) / 2 = r₀ / (1 - ε)

and the length of the semiminor axis (b) is a √(1 - ε²). The period of the orbit is then

(2 π a b) / J = (2 π r₀² √(1 - ε²)) / (J (1 - ε)²)

Since J is √(G M r₀ (1 + ε)), the period is

2 π √(1 / (G M)) (r₀ / (1 - ε))^(3/2) = 2 π √(a³ / (G M))

and we have just found Kepler's third law: the orbital period is proportional to the square root of the cube of the semimajor axis.

There is something else pretty special here. It turns out (Bertrand proved it) that the inverse square force law is one of only TWO force laws which allow closed orbits. And the inverse square law is a direct consequence of the fact that we live in 3 dimensions.

Now this was all for one planet and one star, but Newton's laws are powerful enough for us to use them for more complex systems; just not in this text.

And it turns out that we have actually accomplished more than we thought we did. All this time you thought (admit it!) that M was the mass of the star. But since the gravitational force is felt by both the star and the planet, the planet will cause the star to wobble, so it cannot be fixed at any point. (This is how we most often detect extrasolar planets.)

What we have actually done is find both the motion of the planet and of the star: each orbits in an ellipse around their common center of mass, and M is the total mass (star plus planet). The origin is the location of the center of mass of the star-planet system. Since the star is so much more massive than the planet, the center of mass is very close to the center of the star, so we observe the elliptical orbit of the star as the wobble we spoke of above.

What about our treatment of the planet as a point mass? Gauss' law tells us that the gravitational force due to a spherically symmetric mass is equivalent to the force due to a point particle of the same mass, located at the center. Therefore we will usually treat planets (and stars) as if all their mass was located at a point at their center.

Using the masses of these bodies, we can compute a number of interesting parameters: their average densities, their surface gravities (by setting "mass₂" to 1 and "distance" to the radius), their orbital angular momenta ("L", using the equation

angular momentum = 2 * π * mass * rotation rate * orbital radius²)

Here is a time lapse movie of tidal basin filling (17.83 Mb), and one of a tidal bore (6.87 Mb) (the turbulence downstream of a narrow opening).

All of the moons in this table are locked in synchronous orbits: their rotational period is equal to their orbital period. This is due to tidal deformations (bulges toward the planet) which hold the same face toward the planet at all times. Mercury is also in a sychronous orbit, but in a more complicated fashion because of its higher eccentricity: it rotates 3 times for every 2 orbits around the Sun.

In addition, it is interesting to compare the tidal accelerations to that of Europa on Io, which is thought to account for much of Io's internal heating, and subsequent active volcanism. Note that values in the final column are all less than 1. If the tidal acceleration were equal to or greater than the surface gravity, the body would be gravitationally unstable and would break up. This happens to bodies lying withing the Roche Limit.

• Table of Contents
• References
• Index

©2010, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

Please send comments or suggestions to the author.