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Re: [Phys-l] Rocket Science



Regarding John Denker's question:

David Bowman wrote:

a particular definite integral ...
I ==Int{w_<, w_> | dw/sqrt(D(w)}

D(w) == sgn(E) - w^2 - V(L/(w*sqrt(2*m*|E|)))/|E|


I was wondering about the physical motivation and interpretation of
this formula.

From the way it's used, I suspect it must be something like the
integral of d(theta)/d(r) (in scaled units). Or is there something
else involved?

John, your guess is correct and that is the key to it all.

The way the derivation works is as follows:

We start with the conservation of total energy E by writing

E = m*v^2/2 + V(r) .

Next we write v^2 in polar coordinates as

v^2 = (dr/dt)^2 + r^2*(d[phi]/dt)^2 .

Next we use the chain rule, dr/dt = (d[phi]/dt)*(dr/d[phi])
and substitute in for the radial velocity in the above equation for
v^2 and use the distributive law. This gives

v^2 = ((dr/d[phi])^2 + r^2)*(d[phi]/dt)^2 .

Next we recall the conservation of angular momentum and write the
equation for the orbital angular momentum as

L = m*r^2*(d[phi/dt)) .

We solve this equation for d[phi]/dt in terms of the constants
L and m and the variable r giving

d[phi]/dt = L/(m*r^2) .

We next substitute this into the most recent equation for v^2
giving

v^2 = ((dr/d[phi])^2 + r^2)*L^2/(m*r^2)^2 .

Next we define u == 1/r and write du/d[phi] = -(dr/d[phi])/r^2 .

Replacing the dependence on the variable r with a dependence on the
variable u in the expression for v^2 gives

v^2 = ((du/d[phi])^2 + u^2)*(L/m)^2 .

Substitution this new expression for v^2 in to the equation for the
total energy E gives

E = ((du/d[phi])^2 + u^2)*L^2/(2*m) + V(1/u) .

Solving this equation for du/d[phi] in terms of the constants
E, L and m and the potential V(1/u) gives:

du/d[phi] = (+/-)sqrt(C(u))

where the function C(u) is defined as

C(u) = (2*m/L^2)*(E - V(1/u)) - u^2 .

Now the sign of the sqrt factor is determined by whether the particle
moves clockwise or counter-clockwise when it is moving closer to
the force center from farther out. But this depends on which side
of the orbital plane we view the motion from. For convenience we
can assume we are viewing the motion from the side that causes [phi]
to increase as u increases (i.e. r decreasing) so we can WOLOG take
the sign of the sqrt() as positive. Since du/d[phi] vanishes at any
apside that means that apsides correspond to roots of the C(u)
function. Since we only deal with positive radial distances that
means that we only need to consider positive u-values.

We next separate the differentials as

d[phi] = du/sqrt(C(u) .

If we integrate both sides of this equation from the apapsis (u_<) to
the periapsis (u_>) we get the expression for the definite integral
I, i.e.

I = Int{u_<, u_> | du/sqrt(C(u)} .

Since I represents a net angle change (in radians) for the orbiting
particle when it goes from apapsis to periapsis we see that its
value is dimensionless, but the integrand 1/sqrt(C(u) has the
dimension of length and the differential du has the dimension of
inverse length. We can write the integral for I in terms of a
dimensionless integrand multiplying a dimensionless differential
measure by changing the integration variable from u to w where

w = u*L/sqrt(2*m*|E|)

(assuming that L and E are both nonzero).

Doing this makes our expression for I become

I = Int{w_<, w_> | dw/sqrt(D(w)}

where

D(w) == sgn(E) - w^2 - V(L/(w*sqrt(2*m*|E|)))/|E| .

Because of the mirror plane symmetry I mentioned for the orbit about
a plane perpendicular to the orbital plane and containing an apside
and the force center (which is actually a consequence of time
reversal invariance, which for the orbit is a consequence of the
orbit equation depending on the square of du/d[phi]) this means that
the angle change in going from an apapsis to the next periapsis has
the same angle change as going from the periapsis to the next apapsis
in the orbit. Thus the total angle change in going through a full
radial apsidal cycle is 2*I. In order for an orbit to be a simple
closed figure (that doesn't cross itself) there must be a positive
integer number n of these radial cycles in one revolution of the
particle around the force center. This means that a simple closed
orbit requires:

2*[pi] = n*2*I

Solving for n gives n = [pi]/I where n is some positive integer when
we have a simple closed figure. If n is a rational number we have
a figure that crosses itself but eventually closes on itself after
some integer number of revolutions about the force center. Such an
orbit is one of those Spirograph (TM) orbits. If n is irrational
then the orbit never closes and never retraces itself.

David Bowman