What can you say about the motion of an object with velocity vector perpendicular to position vector? Can you say anything about it at all?

I know that velocity is always perpendicular to the position vector for circular motion and at the endpoints of elliptical motion. Is there a general statement that can be made about the object's motion when the velocity is perpendicular to position?

Solution 1:

If you are constrained to move in a plane, then yes, circular motion is the only possibility. To see this, let $(x(t), y(t))$ be the coordinates at time $t$ on the path. The condition is $$x\frac{dx}{dt} + y\frac{dy}{dt} = 0$$ Integrate with respect to $t$: $$\int x\frac{dx}{dt}dt + \int y\frac{dy}{dt}dt = \int 0\,dt\\\frac {x^2}2 + \frac{y^2}2 = C$$ which is the equation of a circle.

So your path has to move around the circle. But that doesn't mean the path itself is a circle. It can speed up. It can slow down. It can stop and reverse direction, all without violating the perpendicularity condition (when it is stopped, the velocity is $0$, whose inner product with the position vector is still $0$ - this stretches the definition of "perpendicular" a bit, but generally it's allowed, as the physical objects being modelled occasionally do stop).

In three dimensions, a similar calculation shows you that the motion is constrained to some sphere. But you can move in all sorts of ways on a sphere that do not qualify as "circular".