Two-body problem

Let us consider the motion of 2 point masses m1 and m2. Let the positions of these 2 point masses be given by r1 and r2. Newton's 3rd law tells us that if m1 exerts a force F21 on m2, m2 will exert a force F12 = -F21 on m1. This implies that:

This means that we can define a center-of-mass vector:

such that dR/dt = const. We can then boost to a frame in which dR/dt = 0 ⟹ R = const. We can also define the separation vector r = r1 and r2 between our 2 masses, implying that:

A central force depends only on the magnitude of r and can be expressed as F = -∇U(|r|). In this case the energy of the system can be expressed as:

where

is the reduced mass of the system. We have thus reduced the 2-body problem for point masses under the influence of a central force to an effective one-body problem for a single point mass μ in the potential U(r).

Angular momentum

The fact that F(r) is a central force implies:

This in turn implies that the motion is restricted to the plane perpendicular to L. In polar coordinates (r, θ) in this plane:

The constancy of the angular momentum implies that the area per unit time swept out by the particle as it travels along its orbit:

is also constant. This is Kepler's 2nd law of planetary motion. The constancy of the angular momentum reduces the 2-body problem to a 1-dimensional problem in the radial coordinate:

Equations of motion

We can solve the above equation for dr/dt and integrate to determine r(t):

Turning points (pericenter and apocenter) occur where the integrand vanishes. If all we are interested in is the trajectory r(θ), we can instead integrate:

Integrating the above expression from pericenter to apocenter yields ½ the angular motion per orbit; an orbit is closed if this is a rational multiple of π and open otherwise.

Alternatively, we can compute the orbit directly from Newton's 2nd law:

Some not particularly exciting algebra and the substitution dθ/dt = L/μr2 shows that this expression can be rewritten as:

This equation might be useful for finding the orbit (particularly for an inverse-square-law force), and in a very straightforward manner gives the force from the orbit r(θ). This might be useful, for example, if astronomers observed a stellar orbit and wanted to reconstruct a galaxy's spherically symmetric gravitational potential (and hence density distribution, including dark matter).

Effective potential

When we consider central-force motion as a one-dimensional problem in the radial coordinate:

we can think of the last 2 terms as the effective potential:

The effective force will therefore be:

The 1st repulsive term is often called the centrifugal force although it is not a fundamental force like gravitation, electromagnetism, etc. Intuitively, we can think of this as the barrier to orbits approaching zero separation. If the central force is gravity, then:

The effective potential is shown below:

We see that a particle with energy E1 is unbound, a particle with energy E2 is on a bound eccentric orbit with pericenter r2 and apocenter r4, and a particle with energy E3 is on a bound circular orbit with radius r3. If we set u = 1/r, we can find the pericenter and apocenter by solving the quadratic equation. These are alternatively called perigee (Moon about the Earth), perihelion (planet about the Sun), peribothron (motion about a black hole).

In general relativity, which is Einstein's theory of gravity, there is an additional term in the effective potential for Schwarzschild black holes:

This last term leads to a host of interesting phenomena such as capture of particles with nonzero L, unstable circular orbits, and the ISCO.

Kepler's problem

We can integrate

to determine the orbit. This is a 1st order ODE, so we need one integration constant which is fixed by choosing r to be minimized at θ = 0. The solution, which we can check by differentiating, is

where α = L2/GMμ2 is the semi-latus rectum and e = √(1 + 2EL2/(GM)2μ3). This equation describes conic sections, curves where the ratio of the distance between a point and a line (the directrix) is fixed.

The value of e determines the shape of the orbit:V = Vmin ⟹ e = 0 ⟹ circleE < 0 ⟹ e < 1 ⟹ ellipseE = 0 ⟹ e = 1 ⟹ parabolaE > 0 ⟹ e > 1 ⟹ hyperbola

Planets, comets, and other objects bound to the Sun move on ellipses of varying eccentricity. The Earth

has e = 0.0167 (closest to Sun in winter), the former planet Pluto has e = 0.249 (closer to the Sun than Neptune from Feb. 7, 1979 to Feb. 11, 1999, won be again till 2227). Halley's comet has e = 0.967 (tiny for a comet, no one quite knows why) and had its last perihelion in 1986, next in 2061. Hopefully we'll all be around to see it!

Elliptical motion can also be described in terms of the semi-major axis a = α/(1-e2) = GMμ/2|E| and semi-minor-axis b = α/√(1-e2):

The pericenter and apocenter are:

To calculate the period of an elliptical orbit, we can return to the previous result:

Integrating both sides, we find:

or

This equation, showing that the square of the orbital period is proportional to the cube of the semi-major axis, is Kepler's 3rd law. This completes Kepler's 3 laws:

1. Planets move on elliptical orbits with the Sun at one focus.2. The area per unit time swept out by a radius vector from the Sun to a planet is constant.3. The square of a planets period is proportional to the cube of its semi-major axis.

Hohmann transfer

The Hohmann transfer is an efficient orbital maneuver to change between circular orbits and nicely illustrates the energy and angular momenta of Keplerian orbits. Consider a rocket in a circular orbit of

radius r1 initially. As shown above, the energy of this orbit is:

If we want to boost the rocket up to an orbit of larger radius r2, we need to change the velocity by an amount Δv such that the new orbit will have pericenter r1 and apocenter r2, implying that the new semi-major axis will be 0.5 x (r1 + r2) and the new energy will be correspondingly:

Solving for Δv, we find:

This boost will put the rocket onto an elliptical orbit with apocenter r2, where its velocity can be determined from:

A second boost parallel to the velocity will be required to circularize the orbit at this radius. This boost Δv' will increase the energy further to:

Solving for Δv', we find:

Although this is the most energy efficient orbital transfer, it is not the shortest trip if you are trying to travel between planets located at a particular phase on these circular orbits.

Slingshot effect

Encounters with 3rd bodies can be used to redirect satellites or boost their energy as was most famously done with the Voyager spacecraft. Although the encounter is hyperbolic and elastic in the rest frame of the 3rd body, in the frame of the central mass it can greatly enhance the spacecraft's speed.

Since α = L2/GMμ2 and e = √(1 + 2EL2/(GM)2μ3), L → 0 ⟹ e → 1 ⟹ δ = 2 cos-1 (-1/e) – π → π but also: