Pioneer Anomaly Anomalous No More.

Max, I'm going trim almost all your reply so we can
focus on the question of signed velocities. We can
come back to the other bits after that.

Max Keon wrote:
"George Dishman" wrote in message
oups.com...

.. Take Mercury's maximum
radial velocity relative to the Sun on its journey to and from
its perihelion, for example. v of course cannot be negative,
regardless of direction.
....
Now try your second equation which applies for inward
motion and hence v is negative. Let's take a similar toy
value of -0.01c:

What on earth is a negative velocity? A car moving toward me at
60km/h is not moving at -60km/h toward me. Nor can it be moving
at -60km/h away from me if it's going the other way. Wherever
it's going it is traveling at +60km/h.

That is the speed, not the velocity.

If I'm following along at
a constant distance behind the car, the car's velocity is zero
relative to me. If I increase my speed by 10km/h, is the car
traveling at -10km/h ?

Relative to you yes. The distance between you is
reducing so to get a negative distance change in
a positive time you must have a negative relative
velocity. Of course both cars are still moving forward
so the first might be moving at 80km/h while the
second is doing 90km/h and the relative velocity
is 80-90 = -10.

Of course it's not, it's moving toward
me at 10km/h. And it's moving away at 10km/h if I decrease my
speed at that rate.

If you decrease your speed to 70km/h then the same
calculation now gives you 80-70 = +10 as you expect.

The + - switch should be applied to direction, not velocity.

Velocity is a vector and includes the direction so if
we think in terms of a car then we have a 2D situation
and we can define the velocity as north-south and
east-west components. For a car moving at 80km/h
going east, the velocity would be (0, 80). If it was going
west, it would be (0, -80). North is (80,0) while west is
(-80, 0). The velocity of a car moving with a speed of
80km/h in a north-easterly direction is (56.6, 56.6).

Speed is a scalar and is the magnitude of the velocity
is if an object is moving at (vn, ve) then its speed is
sqrt(vn^2 + vs^2).

The location of a car can similarly be defined by
distance east and distance north of some reference
location. Velocity is the derivative of location so if
the distance east decreases, then the easterly
component of the velocity is negative.

What you are doing is invalid.

What I am doing is basic standard physics and
entirely correct. What you are doing only works if
you know the direction in advance, but near
perihelon and aphelion, you may not be able to
know the direction until you work th acceleration
to find out whether the object has changed from
outward to inward, so you don't know which
equation to apply until you have applied it. The
standard scientific method I am using always
works since the direction is encoded in the sign.

George