dark matter (today), Final review (Monday). No class on Wednesday. Final exam on Friday, 1:20 to 3:50 in room 177 (same room as lecture).
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, rmin = 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
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, &phimax = ±Acos(1/ε). The shape of the orbit is an hyperbola, (x-δ)²/α² - y²/β² = 1.
| eccentricity | energy | orbit |
|---|---|---|
| ε=0 | E<0 | circle |
| 0<ε<1 | E<0 | ellipse |
| ε=1 | E=0 | parabola |
| ε>1 | E>0 | hyperbola |
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:
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]
Please read this example of a manuever to change from one circular orbit to a second of larger radius.
The orbit of known planets are perturbed by other planets, especially Jupiter and Saturn. These perturbations were known, and approximations were used to calculate them. The calculations were in good agreement with the observed orbits for all the planets except Uranus, the outermost of the known planets. A number of possible explanations were considered, for instance:
Newtonian mechanics and the law of Universal Gravitation has been successfully applied to the orbits of more and more distant objects: orbits of Neptune, Pluto, and comets; spacecraft; binary and multiple star systems; and finally within and between galaxies. A strange effect occurred when the motion of stars and gas around the center of the Milky Way and other galaxies was studied. First, we need to determine what we would expect for the motion of a star due to the gravitational force from all other objects in the galaxy.
One of the hurdles Newton had to overcome with his law of Universal Gravitation was to prove that the gravitational force due to a spherical distribution of mass was equivalent to the force from a point object of total mass equal to the mass distribution, located at the center of the sphere. He, and many texts, do this with the force vector. We will take advantage of our knowledge that the 1/r² force is conservative and can therefore be computed from a potential function,
It is generally much easier to sum (or integrate) the scalar potential energy functions than to sum (or integrate) the vector forces. let's begin by finding the potential for a spherical shell of radius R, total mass M, and uniform mass per unit area ρ, at all points inside and outside of the shell. At a point a distance r from the center of the sphere, call s the distance from an infinitesimal mass to the point. Then the total potential can be written as
Use spherical coordinates (R (fixed), θ, φ) to integrate over the shell. Choose the polar (z) axis to run through the center of the sphere and the point. Then for fixed R and θ all points on the shell are the same distance s from the point r, independent of φ. So the integral for U becomes
The total mass of the shell is density times area, M = 4πρR². From the geometry of the problem we have r² + R² -2rRcosθ = s². Holding r and R constant, the differentials of both sides are 2rR sinθdθ = 2s ds. Using these, we can substitute for sinθdθ to get
If the point is outside the sphere, the integral over ds runs from r-R to r+R and the potential energy is
If the point is within the shell, the integral over ds runs from R-r to R+r and the potential energy is
This result says that the potential due to a uniform shell of material is equivalent to that from a point object of the same total mass, located at the center of the shell for points outside the shell, and constant for points inside the shell. At the shell, the potential function is continuous, but the derivative (force) is discontinuous, going to zero inside the shell. It is straightforward to now integrate over a stack of such shells to find the potential due to a solid sphere (with density that can vary with r, but not θ or φ) and get the result that
for points outside the sphere. We have been using this result all along. If the point lies within the sphere, only those with radii less than r contribute to the force, those with larger radii contribute a constant to the potential but nothing to the force.
We will now use this result to understand the speed of stars rotating around the center of our galaxy, or any other galaxy. Our galaxy is not shaped like a sphere, it is more like a disk with a spherical bulge in the center and spiral arms, but this result has the basic qualities we need. It shows the result, generally true, that an object rotating around the center of a mass distribution feels a force that depends on the distance from the center of the distribution, and the amount of mass in a sphere (or disk) of the same radius centered on the distribution. Mass outside this sphere doesn't effect the object.
As an example, let's assume that the mass distribution is uniform so that M(R) = (4/3)πρR³ and the potential energy varies like U(R) = -(4/3)πρkmR². We can use the virial theorem to relate the kinetic energy of a object at distance R from the center to the potential energy and find T = -½U, so that the tangential velocity of the object is v = (2/√3)(πρk)½ R. Within a spherical region where the mass density is approximately constant, the tangential velocity (also called the rotational speed) increases like the distance from the center.
At a point where there is little additional material, like near the edge of a galaxy, the mass within the sphere is approximately constant, so that U(R) = -kMm/R. Again using the virial theorem, we can find the rotation speed v = sqrt(KM/R). Far from the center, where the mass density is much smaller than near the center, the rotational speed decreases like the inverse square root of the distance from the center.
These results are approximate, but give a rough idea of the behavior we expect to see if we measure the rotation speed of objects as a function of their distance from the center of a galaxy. The objects measured are stars and clouds of gas. The expectation is that the rotation speeds will increase linearly near the center of the galaxy, and then turn over and decrease as we move further from the center. In constrast, the measured rotational speeds remain approximately constant to a great distance from the center of the galaxy. [sketch figure from Fowles & Cassidy.]