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# Is the Circular Velocity Hodograph in Kepler's Problem Circular?

Feynman starts his demonstration that the change in the velocity is constant, and then tacks on the phrase "at equal angles." His demonstration is this:

D v~ F Dt

where D stands for finite increments. The fact that the force decreases as 1/r^2 and the time increment as the square of r, means that Dv~1.

From this you can conclude that the velocity increments are all the same, but not that they are proportional through equal angles. In Feynman's words:

"the central core from which all will be deduced---that equal changes in velocity occur when the orbit is moving through equal angles." That requires Newton's second law

|dv/dt|=-GM/r^2.

If we eliminate r^2 by introducing the angular momentum, h, there results

|dv/dt|=R|d(theta)/dt|,

where R=GM/h, the constant velocity. Now this equation is transfer to finite increments

|Dv|=R|D(theta)| +R_0, (i)

where R_0 is a constant of integration, independent of time. To get proportionality we must set R_0=0. But is this consistent with the Kepler problem?

Introducing h into Newton's law gives

dv/d(theta)=-R(cos(theta),sin(theta),0)

Integrating this gives

v=R(-cos(theta),sin(theta),0) +c, (ii)

where c=(c_1,c_2,0) is a constant of integration. Does this "proves that the velocity vector v moves along a circle centered at c, around a circle with uniform velocity R from an origin that does not coincide with |c|." (Milnor 1983) So the question is what right do we have to set R_0 in (i) and retain a non-zero value in (ii)?

From (ii) the speed is

v=R(1+e^2+2 ecos(theta))^1/2, (iii)

where the eccentricity e=|c|/R. It is important to underline that no condition is imposed on e. They are all cardioids, no matter what e is.

Now, taking the finite increment of (iii) gives

Dv=(eR^2 sin(theta)/v) D(theta). (iv)

Comparing (iv) with (i) with R_0=0 (necessarily) we find

v=eR sin(theta). (v)

We have thus found two expressions for the speed, (iii) and (v) so they must be equal. The condition is

(theta)_max=cos^(-1)(-1/e). (vi)

Contrary to what is claimed, (iii) "the polar angle is limited to the interval |(theta)|<(theta)_max when e>1. That is, the hodograph coincides with the entire circle only when e<1. When |theta_max| is approached, the velocity becomes tangent to the hodograph, and the speed reaches a limiting value v_inf=R(e^2-1)^1/2....the planet is unbounded and moves asymptotically (as t->infinity) toward a point in velocity space.... v_inf." (Guillaumin et al, 2003). Rather, if both (i) [with R_0=0] and (v) are valid contemporaneously, then cos(theta)=-1/e, and v=(e^2-1)^1/2. This is whether we use (i) or (v).

Two points need to be emphasized:

1. the speed (iii) has absolutely no conditions on the value of the eccentricity. It is (v) that imposes the condition, and, instead of being valid for all (theta) it is valid only at the value given by (vi).

2. We have no right to maintain a finite constant of integration (ii) when we threw it out in (i). If we don't throw it out the proportionality in (i) no longer holds and consequently (v) no longer holds. Throwing out the constant of integration is tantamount to considering uniform circular motion, and there is no longer any necessity to discriminate the non-uniformity of point by the distance c is from the origin of the velocity circle.

Since we know that

r= h^2/GM/(1+e cos(theta))

is the orbit of an ellipse in configuration space, why shouldn't we consider (ii) as the corresponding velocity hodograph since the polar angles are the same? This removes any condition on the eccentricity. The statement that "if Kepler's second law holds, and the trajectory of the mass is a conic section then the hodograph is a circle" is inaccurate since the proof hinges on the inverse square law, apart from the fact that (i) is not the same as (iv).