Convert velocity to spherical coordiates?

I think I authored a similar Thread here or at an other forum, but never got an answer I trusted.

There are well known equations for transforming a Cartesian point (xyz) to Spherical coordinates and vice versa. What are the equations for transforming velocity and/or acceleration vectors?

A while ago, I assumed that taking derivatives of the point conversion formulae would provide correct equations. Then an engineer claimed that a Cartesian point (x, y, z) is a position vector. He further stated that a vector is a vector is a vector. Hence, the equations for transforming a position vector should be correct for transforming velocity and acceleration vectors.

I considered the possibility that both methods were correct and would arrive at the same results. Some work with my MathCad7 software convinced me that this was not true. The two methods do not lead to the same results.

Does anybody have some thoughts on this? Ignoring the formulae for Longitude and Latitude, consider the following equation for the Spherical coordinate radius.

r = SquareRoot( x² + y² + z² )

Derivative analysis results in the following for transforming velocity.

Vr = ( x*Vx + y*Vy + z*Vz ) / r, where Vr, Vx, Vy, Vz are velocities.

Using the position vector formula on the velocity vector results in the following.

Vr = SquareRoot(Vx² + Vy² + Vz² )

Which of the above is correct for Vr? If you have an opinion, please explain.

My vote goes for the first equation, but I am not sure.

Guv,

http://www.csupomona.edu/~ajm/classes/phy321/delsph.pdf

But you've probably read that already?

Can't say I analysed it either :)

I would go for the second formula. The reason for this is that it does not depend on the current location. The first formula does, I think. I'm not sure, because I don't really understand the derivation. Why would you multiply the x-, y- and z-position of your current location by the velocity in the corresponding direction? If you're at the origin of the coordinate system, won't you move at all in that case?!

Position at time = 0: P₀ = (x₀, y₀, z₀)

Position at time = t: P_t = (x₀ + tVx, y₀ + tVy, z₀ + tVz)

Difference: P_{t - 0} = (tVx, tVy, tVz). Velocity is simply the square of the quadrates, which is formula 2 (for t = 1 time unit).

To my opinion the equation last mentioned is the right one.(Vr=SquaireRoot(Vx^2+Vy^2+Vz^2)
The equation first mentioned is ,I believe, the rate of change of r (r-dot),of course also a velocity,i.e. a velocity of the point (x,y,z)wrt the origin(0,0,0).

Here some additional information.

Here I have an example of the velocity and r-dot.
On my TI-83 calculator I have a program called ,, Planets,, , this program enables me to calculate the position,velocity,the rate of change of r (r-dot) etc for a given date. If I calculate the values for the velocity and r-dot for our planet Earth on 24 January 2003 I get
30.255 resp. 0.175 km/sec. Of course the last value is rather small wrt the velocity,it is also positive because the Earth is now moving away from the pericenter position,i.e the position at which the distance Sun-Earth is a minimum.

An excellent source for equations about velocity,acceleration,spherical coordinates etc is the ,,Eric Weisstein,,Website of Wolfram.
The Web link: http://scienceworld.wolfram.com/phys...Mechanics.html

I have to make a sincere apologize that i neglected to mention CMDP as the original arthor of this above posting ;)