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!
$\endgroup$ 12 Answers
$\begingroup$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
$\endgroup$ 2 $\begingroup$From $0=2r\cdot r'=(r\cdot r)'$ we know that $r\cdot r$ is constant.
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