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A satellite rotates around the earth; a ball on a string rotates around my hand. Each of these things experiences a net force equal to $ mv^2/r = F_{centripetal}$. Why does no object end up at the center of their circular paths?

An attempted answer: An object experiencing a centerwards force at every point of its orbit does move to its center -- but only momentarily. The combination of many momentary forces, all of them centerwards, stops any one vector from changing the object's orbit.

If this is true, I am not convinced. Would someone have a rigorous proof that a constant centerwards force produces a circular orbit, or, if my answer is mistaken, an answer?

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    $\begingroup$ The net force is not what you listed, the centripetal force is cancelled by the tension in the string. $\endgroup$ Mar 21 '15 at 19:00
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Why does no object end up at the center of their circular paths?

If the path is circular then, by definition, the particle maintains a constant distance from the center of the path. Perhaps you're asking why the particle has a circular path?

Would someone have a rigorous proof that a constant centerwards force produces a circular orbit

One needs something more for a circular orbit; one needs a force that is constant in magnitude and (always) perpendicular to the particle's velocity.

(The question of whether such an orbit is stable against perturbations is another question entirely.)

Perhaps the easiest way to think about this is to solve for the acceleration for a particle with uniform circular motion.

For example, in Cartesian coordinates, such circular motion is of the form

$$x(t) = R\cos(\omega t - \delta)$$

$$y(t) = R\sin(\omega t - \delta)$$

where $R,\;\omega$ and $\delta$ are constants Then, the acceleration components are

$$a_x(t) = -\omega^2 R\cos(\omega t - \delta)$$

$$a_y(t) = -\omega^2 R\sin(\omega t - \delta)$$

The magnitude of the acceleration is then

$$a = \omega^2 R$$

and it is clear that the acceleration is constant in magnitude.

The velocity components are

$$v_x = -\omega R \sin(\omega t - \delta)$$

$$v_y = \omega R \cos(\omega t - \delta)$$

and it's now easy to see that the acceleration is perpendicular to the velocity

$$\vec v \cdot \vec a = (\omega^3 R^2 - \omega^3 R^2)\sin(\omega t - \delta)\cos(\omega t - \delta)=0$$

We've shown that uniform circular motion implies a constant magnitude acceleration that is (always) perpendicular to the velocity.

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The object has to have an inertia that's pulling it away from the center. Otherwise it would end up at the center as you say. That's why orbiting objects have to constantly move. If the planets stopped orbiting the sun, there would be no inertia of their paths to keep them in place and they'd fall right into the sun. If the inertia isn't enough, then the orbiting object will spiral into the center of its orbit. If it's too much, it'll fly out of orbit. Orbits have to be perfectly balanced.

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