Notebook for 262 session #2

Note: the main exercise is to complete the paper handout.
This notebook is here to give you the following incantation for drawing a moving spinning round thing.

draw[x_,y_,θ_]:= Show[{Graphics[Rectangle[{-1,-1/4},{5,0}]],Graphics[Line[{{-1,0},{-1,7}}]],Graphics[Circle[{x,y},1]],Graphics[Line[{{x,y},({x,y}+{Cos[θ],Sin[θ]})}]]}]

To use it, you need to define some x[t], y[t] and θ[t].  For example, here is something made up---this is not any physical
motion, just a ball which falls to the ground with a thud, rolls for a while, and then for no reason at all decides to change its speed.
Your challenge is to construct the corect x[t] etc. for Morin 8.21.

x[t_] = Piecewise[{{t,t<1},{2(t-1)+1,t<2},{3-(t-2),True}}]

Plot[x[t],{t,0,3}]

y[t_] = Piecewise[{{5(1-t^2)+1,t<1},{1,True}}]

θ[t_] = -2 t

Animate[draw[x[t],y[t],θ[t]],{t,0,3}]

Effective Potential

Supose we launch a small probe (mass m) with speed v0 toward  a large spherical planet (mass M, radius R) with gravitational potential -GMm/r.
Since M is much larger, the reduced mass μ simplifies to just m.
If the the probe is launched with “ impact parameter” b, the angular momentum about an origin at the CM (i.e. the center of the planet) is:

Li = (you fill this in)

As the probe falls in and just grazes the planet at radius R with some as yet  unknown speed v, the angular momentum is:

Lf = (you fill this in)

And since the gravitational force is “central”, i.e. radially directed, angular momentum is conserved:

eqnL = Li == Lf

Since such forces are also conservative, we also have conservation of total energy.
(Here we assume we are starting so far away that we can neglect the intial -GMm/r).

eqnE = (1/2) m v0^2 == (1/2) m v^2 - G M m / R

Your job: Solve these two equations for b (and v).

The effective potential approach does the same algebra in slightly different language.
Given the initial angular momentum, we construct the effective potential:

Veff[r_] = (Li)^2/(2 m r^2) - G M m /r

The total energy (as expressed as we start) is

E0 = (1/2) m v0^2

Here is a plot of Veff and the energy of our motion with all parameters set to one;

p = {m->1,M->1,G->1,b->1,v0->1}

Plot[{Veff[r],E0}/.p,{r,0,3}]

The distance of closest approach is the turning point where E0=Veff and r-dot is momentarily zero.
Demand that the turning point be R, i.e. Solve the equation Veff[R] = E0 to find b again.