Can we prove that a constant central force pointed towards a point produces circular motion? I was analysing the motion of a body when a single, radially inward force acts on a body, and I was trying to prove this is indeed circular motion, using geometry and looking at what happens at infinitesimal time scales. I ofcourse have been taught that this force is the reason for circular motion but let's forget that for a moment. When I think about the infinitesimal distances travelled in dt time difference, there is ofcourse a motion due to the initial velocity, along the tangential direction, and a small movement along the radially inward direction, due to the force. But when analysing in such small length scales, which operation happens first, or does both happen at the same instant? (The downward and sideways motion, due to the force and the velocity respectively). How to analyse the geometry of the case and come to the conclusion that this indeed will follow a circle equation?
 A: 
How to analyse the geometry of the case and come to the conclusion that this indeed will follow a circle equation?

You'd have to make a mistake, because in general this will not generate a circular trajectory.  Explicitly, let the particle start at the point $(0,R_0)$ with initial velocity $(v_0, 0)$, and let the force $\vec F = -F_0 \hat r$ pull the particle toward the origin.  The resulting motion will be circular if and only if
$$F_0 = \frac{mv_0^2}{R_0}$$
as per the standard centripetal force equation.  In general of course this need not be true; for a given force $F_0$, you may choose $v_0$ and $R_0$ completely arbitrarily and they need not satisfy this condition. If they don't, then the resulting motion will not be circular.
A: The thing with infinitesimal amounts is that they do not have time to make a difference.

*

*You start out with an initial velocity.

*Then you pull perpendicularly to that speed with a force.

*That perpendicular force causes a perpendicular acceleration

*which in turns creates a new, perpendicular velocity component.

With this new component added (via vector addition) to the initial velocity as another component, then you get a resultant velocity which now is slightly angled.
But this inwards velocity component is actually not happening! You, by definition, adjust your force immediately so that it always is perpendicular, and you thus also adjust the force angle to the new slightly angled velocity. Because you constantly at every single moment keep the force perpendicular, then no magnitude of velocity will ever be induced in this inwards directly.
A: for the general case you obtain those equations of motion. (the position vector to the mass is given with polar coordinates).
$$\ddot r-r\phi^2=\frac Fm\tag 1$$
and
$$r\,\ddot\phi+2\dot r\dot\phi=0\quad\Rightarrow \dot\phi=\frac{L}{r^2}\tag 2$$
thus only if
$$ F=-m\,r\phi^2\quad \Rightarrow\\
\ddot r=0\quad ,r(t)=r_0=\text{constant}$$
and from equation (2)
$$\dot\phi=\frac{L}{r_0^2}=\omega=\text{constant}$$
thus the force $~F=-m\,r_0\,\omega^2~$
A: As others have said, in general there will not just be circular motion, however from yourcomments I'd assume you'd want to prove that circular motion is A solution.
Given my radial force has the correct magnitude:
$$\vec{F}= -\frac{m|\vec{v}|^2}{r} \hat r $$
Applying newtons laws:
$$m\vec{a} = -\frac{m|\vec{v}|^2}{r} \hat r $$
Let's assume that there exists A solution in the form of a 2D circular trajectory.
$$\vec{x}(t) = r \cos(\omega t) \hat i  + r\sin(\omega t) \hat j$$
Differentiating
$$v(t) = -r\omega t \sin(\omega t)\hat i + r\omega \cos(\omega t) \hat j$$
$$|\vec{v}(t)|^2 = r^2\omega^2$$
Differentiating again
$$\vec{a}(t)= -r\omega^2  \cos(\omega t)\hat i - r\omega^2 \sin(\omega t) \hat j$$
$$\vec{a}(t) = -r\omega^2 \hat r$$
Substitute our trial solution into our differential equation.
$$-r\omega^2 \hat r = -r\omega^2 \hat r $$
$$1=1$$
Thus a circular path, is a specific solution to this differential equation
