Orthogonal Position and Velocity Vectors Is it true that if the position and velocity vectors of  a moving particle are always perpendicular the path of the particle is on a sphere? If so how do I prove it?
Geometrically I believe it makes sense but I'm having a bit of trouble proving it algebraically.
What I have so far:
$$\bar{v}(t) \perp \bar{r}(t) \implies \bar{v}(t) \cdot \bar{r}(t)=0$$ 
Hence, we can say 
$$ x(t) x'(t)i + y(t)y'(t)j + z(t)z'(t)k = 0 $$
So we could create three functions that when you multiply the function by it's derivative would give $0$ (i.e all functions could be constants). However that wouldn't help us at all as the path of the particle would not lie on a sphere.  
So I'm not sure where to go from here or even if I'm on the right track, any help or tips would be appreciated!
 A: There are many way to prove this, but I will do so by proving that when the magnitude of the position vector is constant, its derivative must be tangent.  
Remember that $\vec r'(t)= v(t)$.  By definition, an object moving in circular motion must have a position vector ($\vec r(t)$) of constant magnitude:  $\Vert \vec r(t) \Vert $=c
Note that the position vector has a constant magnitude equal to the circle's radius.
Also, $\ \vec r(t) \cdot \vec r(t) = \Vert r(t) \Vert^2 \cdot \cos(0) = c^2 \ \ \ \ $ for all t.  This is the definition of the dot product.
If we take the derivative of this dot product, we get:
$\frac{d}{dt} \vec (r(t) \cdot \vec r(t)) = \frac{d}{dt} c^2 = 0$
By the chain rule, if we differentiate the dot product we also get:
$\frac{d}{dt} \vec (r(t) \cdot \vec r(t)) = [r(t) \cdot r'(t)] + [r'(t) \cdot r(t)] = 2 [\vec r'(t) \cdot \vec r(t)] = 0$  (the dot product is commutative)
We can simplify this to:
$\vec r'(t) \cdot r(t) = 0$, and by definition if the dot product of two vectors is 0, they are orthogonal.  QED
A: From $0=2r\cdot r'=(r\cdot r)'$ we know that $r\cdot r$ is constant.
