PHY5200 F06

Chapter 8: Two-Body Central Force Problems

Reading

Taylor 8.7-8.8 (today) and dark matter on (Friday).

Recall:

The Kepler Orbits

Derivation of Kepler's Third Law

Let's derive Kepler's third law from Newton's law. Start with the result we obtained for Kepler's second law, dA/dt = l/2μ. We can integrate this over one period to find

The total area of an ellipse with semimajor axis a and semiminor axis b is A = πab. So the integral over the area becomes

Now, square both sides, solve for τ², and express all lengths in terms of the semi-major axis, a. Use the relations b² = a²(1-ε²)

We still haven't exposed all the a dependence. The square of the angular momentum depends on a by l² = ckμ = a(1-\epsilon²)kμ yielding

Finally, because Kepler's third law holds that the same constant of proportionality works for all planets, we must eliminate the dependence on the mass of the planet. This occurs in the factors μ and k, but k = GMμ ≈ GM_sμ, where we leave the factor of μ since it will cancel out, and we approximate M = M_s + m by just M_s, finally yielding

This form clearly has all the a dependence exposed, and the constant is the same for all planets. This is Kepler's third law. The additional advantage of this derivation is that we know have the constant of proportionality.

Relation Between Energy and Eccentricity

Recall that bounded orbits have E<0, and 0<ε<1. The fact that both quantities are bounded suggests that they are related, and they are. To see this, begin with the fact that at the minimum radius, the total energy is equal to the effective potential,

From the equation for the orbit, r_min = c/(1+ε) = l²/[kμ(1+&epsilon)]. Inserting this in the energy equation yields

This result has the expected behavior that negative energies correspond to &epsilon<1, E=0 when ε=1, and E>0 for ε>1. Since the expression involves ε², this result is equally true for negative values of &epsilon

The Unbounded Kepler Orbits

Now let's tackle the unbounded orbits, that is, the cases when ε≥1. When ε=1, r goes to infinity for φ = ±π. The shape of the orbit is a parabola -- can be verified with algebra, as done for the elliptic orbit -- y² = c² -2cx. When ε>1, r goes to infinity for some value of &phi, &phi_max = ±Acos(1/ε). The shape of the orbit is an hyperbola, (x-δ)²/α² - y²/β² = 1.

Summary of Kepler Orbits

Changes of Orbit

With these results for orbits, we can now talk about some interesting spacecraft maneuvers. Spacecraft travel on orbits just like planets and comets, but spacecraft can generate thrust with rockets and change their orbit, an important ability for space exploration. To investigate how this works, begin with the general orbit, including the phase angle, since we can't guarantee that the initial and final orbits will have the same phase.

A common technique for chaning orbits is for the spacecraft to briefly fire its rockets, changing the craft's velocity. We will treat this as an impulse, that is, an almost instantaneous change of velocity. The craft will then move from an orbit with energy E1, angular momentum l1, and orbital parameters c1, ε1, and δ1 to an orbit with E2, l2, c2, ε2, and δ2. At the instant that this occurs, the initial and final orbits must coincide:

A Tangential Thrust at Perigee

To make this problem tractable, we'll specialize to a tangential thrust occurring at perigee (this is for a craft orbiting earth, but can easily be generalized to an orbit about the sun or any other object). We can align the initial orbit so that δ1 = 0. Perigee on the initial orbit then occurs for φ0 = 0. A tangential thrust will increase (or decrease) the velocity, but it will still be perpendicular to the radius, so this point will also be perigee (or possibly apogee, see later) of the final orbit, meaning we can set δ2=0. Then the orbit matching condition becomes

Again for simplicity, let's assume that the thrust changes the velocity by a factor λ such that v2 = λv1, and that the mass of the craft doesn't change appreciably. If λ>1, then the craft speeds up and moves into a "higher" orbit. If 0<lambda;<1, then the craft slows and moves into a "lower" orbit. We won't discuss the possible case that λ<0 where the craft changes direction; such a situation rarely occurs in practice.

The initial and final angular momenta are also related by λ. When the craft is at perigee, its momentum is perpendicular to the radius and the angular momentum is simply l = μrv. Since μ and r don't change between the initial and final orbits at perigee (no change in mass and orbit matching condition), then the initial and final angular momenta are in the same relation as the velocities, l₂ = λl₁. Since c = l²/kμ, the two c's are related by c₂ = λ²c₁. Using this in the orbit matching condition we find a relation between the eccentricities,

If λ>1, then the eccentricity of the final orbit will be larger than the eccentricity of the initial orbit, ε2>ε1. It the thrust is large enough, then ε2 can become greater than or equal to 1, and the orbit becomes unbounded, that is, the spacecraft leaves earth. If 0<λ<1, then the eccentricity becomes smaller. If λ = 1/sqrt(ε1 + 1) then ε2 = 0 and the final orbit is circular. If λ is smaller than 1/sqrt(ε1 + 1), then ε2 is negative. We ran into this situation earlier; a negative ε simply means that the orbit is shifted by a phase of π. In this case, it means that the point where the rockets fired is perigee for the initial orbit and apogee for the final orbit. [sketches will help here]