1. Part A due Friday March 17 (if you want to give me something to do during spring break) or Monday, March 27 (if YOU want to have something to do over spring break). Part B of this assignment will be due Friday, March 31
   A simple, 2-step derivation of the orbit equation for a 1/r² forceusing the "dynamically" conserved quantity p₀.

   (a) Use the same procedure as in lecture (Eq 8.39) to rewrite Newton's law for the force of Eq 8.44, dp/dt = -(γ/r²)e_r, in terms of φ-derivatives.

   Here er is a unit vector in the r-direction, what is usually called

   ^ r

   
   Put dp/dφ on the LHS and note that the RHS depends only on the direction (not the magnitude) of the position vector r.

   (b) Integrate the equation of (a) over φ to get the general solution for p(φ), noting again that the RHS does not contain the magnitude r -- otherwise the equation would be difficult to integrate because r is an as yet unknown function of φ, whereas you do know how e_r depends on φ. You are advised to integrate the Cartesian (that is, x- and y-) components of this equation, assuming that the orbit and hence p is in the xy plane. You can express the result as a vector equation for p by using the unit vector e_φ. Be sure to include a constant of integration p₀, which will of course be a vector in the xy plane.

   (c) Interpret the solution to say that the endpoint of p, as a function of φ, describes a circle in the (p_x, p_y) plane. What is the radius of this circle?

   (d) Square this equation and substitute for p² from energy conservation. (Since p is the full momentum, not just the radial part, energy conservation involves the potential U, not U_eff.) Call the angle between the constant direction p₀ a new f and solve for 1/r(f). The result is an equation for the orbit. Compare with the equation before 8.49, from which it should differ only by a renaming of the constants.

   By the way, the two steps were (b) and (d). The rest is "trivial" once you get the idea. Also, the Runge-Lenz vector is essentially L×p₀, in a direction perpendicular to p₀.

   Part B, due Friday March 31:

2. Text, problem 8.3 Assume the table is removed the instant m₁ is projected. Also describe what will happen differently if the table is not removed.
3. Text, problem 8.18
4. Text, problem 8.29. This problem is very popular, at all levels, but its answer is ill-defined. If we are giving up conservation laws (in this case, of mass), what are we keeping? It is better to specify how the sun's mass disappears: assume if flies out to infinity in a spherical shell, without colliding with the earth. The "sudden" change, as far as the earth is concerned, occurs when this shell crosses the earth's orbit, in a very short time. Now you know what quantity is conserved (it's a spherically symmetric shell).
5. Consider a small body (maybe an artificial satellite) in the field of two large bodies (like the earth and the moon). The large bodies are on circular orbits about their center of mass and unperturbed by the small body. There are several (5, actually) points, with respect to the large bodies, where the small body can be placed so that it will rotate rigidly together with the large bodies. (So that, in the rotating frame of the large bodies, the small body is at rest.) Let the masses of the large bodies be m and M, and their separation R. Find at least one of these equilibrium points for the small body.
   If the set-up seems too complicated, let m = M so each mass is at a distances R/2 from the CM. Three of the "equilibrium" points are on a line connecting the two large masses; these are unstable equilibria. The other two form a triangle with the large masses and can be stable. Prove that the triangle is equilateral. These are called Lagrange points, if you want to search in the internet :)

Please check that I have recorded all of your grades correctly. For privacy reasons the list below has no names, and is ordered by the grades on the midterm. This should enable you to find yourself. Let me know of any errors!

Hwk 1    Hwk 2      Hwk 3    Hwk 4        Hwk 5  1st exam    Hwk 6  Hwk 7a


                    5.5        4                   38
8.7      6          4.5        8          8.5      50                 2
9        8.5        2.5        9          5.5      53        1
9        5.5        7          5.5        5        54
9        7                     7.5        6        57
9        7.5        6.5        6.5        7.5      59        7        1.5
9.3      9          9          9         10        59                 2
9.5      9          7          9          5        60        8
8.5      7.5        5.5        8          8.5      61                 1.5
9.8      7.5                   9          5.5      62        9
8        6.5        3                     6        64        6
9.7      8          6          9.5        5.5      65
9.5      6          5.5        4          5.5      68        8.5
9.3      9          9                              70
9.5                                                71
10       7.5        6.5        9.5        7        73        9.5       2
9.2      5.5        7          7.5        8.5      73                  2
10       7.5       10          9         10        73       10         1.5
8.8      9          7.5        9.5        7.5      78        8         2
8.5      7          8          7.5        7        83        7.5       1.5
9        8          6.5                   6        83        8.5
10       9          6.5        9          6.5      85        9         1
10       8.5        8          8          6.5      88        7.5       1.5
9.5      9          7          9          7.5      89        9         1
8.5      6.5        9.5        8.5        9.5      94       10         2
10      10         10         10         10        96       10         2
10       7.5        9.5        9          9        97        9.5