ADVANCED DYNAMICS — PHY 4241/5227

HOME AND CLASS WORK – SET 5

(February 4, 2011)

Masses  Initial Positions  Initial Velocities
# i m_{i}  x_{i,0}^{1}  x_{i,0}^{2}  x_{i,0}^{3}  ẋ_{i,0}^{1}  ẋ_{i,0}^{2}  ẋ_{i,0}^{3}
1 1 0.651 0.585 -0.238 -0.755 -0.828 -0.865 -0.726
2 0.931 -0.096 0.000 0.357 -0.209 0.107 -0.660
2 1 1.510 0.460 -0.359 -0.234 -0.918 -0.941 -0.323
2 0.126 -0.066 -0.090 -0.809 0.789 0.788 0.620
3 1 1.328 -0.125 0.898 0.194 -0.452 0.172 0.125
2 1.999 -0.449 -0.085 -0.454 -0.976 -0.990 -0.968
4 1 0.180 0.204 -0.968 -0.753 -0.811 -0.632 0.784
2 1.560 -0.889 -0.979 0.854 -0.323 -0.774 -0.533

Table 1: Initial conditions for the Kepler problem (arbitrary units and set G = 1).

(15) Calculate the angular momentum vectors in the CM frame for the initial conditions given in the table above (you can download the data as text file Kepler.txt). Due February 7 before class 8 points.

(16) Let the interaction of two point particles be described by a potential which depends only on their distance:

L = \frac{m_{1}}{2} v_{1}^{2} + \frac{m_{2}}{2} v_{2}^{2} − U(r) , r = |r⃗| , r⃗ = r⃗_{1} − r⃗_{2} .

1. Is the energy of this system conserved (with reason) (1 point)?

2. Define the center of mass vector by R⃗ = (m_{1}r⃗_{1} + m_{2}r⃗_{2})/M, M = m_{1} + m_{2} and express r⃗_{1} and r⃗_{2} through R⃗ and r⃗ (2 points).

The center of mass (cm) frame is defined by R⃗(t) = 0. Show the following equalities in the cm system:

3. T_{cm} = m_{1} v_{1}^{2}/2 + m_{2} v_{2}^{2}/2 = μ v^{2}/2 with v⃗ = ˙ r⃗ and μ the reduced mass. Express μ through m_{1}, m_{2} and M (2 points).

4. L⃗ = r⃗_{1} × p⃗_{1} + r⃗_{2} × p⃗_{2} = μ r⃗ × v⃗ (2 points).

Due in class.

(17) Plot the effective potentials corresponding to the initial conditions of the table together with the CM energies (the Potential is U(r) = −G m_{1}m_{2}/r). Due February 11 before class 12 points.