Abstract: To get motion that is uniform (velocity is not changing in modulus) but circular (velocity is always changing in direction), you need a force that is applied ever-perpendicularly (always giving a new direction, never giving more of the same direction) and you can only do that if the acceleration comes with a certain modulus, which in turn is proportional to (i) directly, the velocity that the object comes with and (ii) inversely, the distance, i.e. the radius. That is logical: if the object comes faster, you need more force to give it a new direction; if it is farther away, attaining circularity is less of a change of direction. In practice, you can achieve this by adjusting the force to the velocity (tension force) or by adjusting the velocity to the force (gravity force). To get to the specific formula (v^2/r), we take the approach of a sometimes given but less popular route: consider first velocity change. If you paint the evolution of the velocity vector after one round, with always same modulus but ever-changing direction, you get that velocity has changed 2πv. But to get acceleration, we should divide by the time, which is the path (2πr) divided by v. Thus we get 2πv/2πr/v = v^2/r.
We start with an object that is moving by inertia and hence with constant velocity v, i.e. in a straight line (no change in direction) and with always the same speed v (no change in modulus).
After applying acceleration generated by whatever cause, we stipulate that the object:
- takes a "circular" trajectory, which means that it will keep the same distance away (the radius) from whatever reference it is orbiting around
- and keeps "uniform" circular motion, which means that the speed v of the object remains the same.
But, if you think of it, our acceleration source must take care of the two things (circularity and uniformity) together, it cannot produce one without the other: if something is causing circular motion, it cannot be at the same time changing the speed of the object. It may happen that the source causes a curve where both the direction and the speed change, but that will not be a circle: it may be an open curve (a parabola for example) or even a closed orbit if you wish (an ellipsis), but not a perfect circle. Why so? Think that any acceleration vector can be of two types: being perpendicular or not. If it is perpendicular, it means that it has no component at all in the direction of the body. (That is what perpendicularity means, total "unrelation": if you project one perpendicular vector onto another, the shadow is none, that is to say, one vector has no component at all in the direction of the other). Therefore, it will have no effect at all in such former direction, it will not add or subtract anything to the modulus thereof. Keep doing it at each ideal instant and you will get that the object is always acquiring a new direction, but never more of the previous one. That is a circular trajectory, which we can thus define as something caused by an "ever-perpendicular" source having, therefore, an effect only in terms of direction and not modulus.
(Before you start disbelieving, note that I am talking about a single source of acceleration. If you factor in several sources, it may be, like in some example below, that you get non-uniform circular motion.)
Now, have we thus finished? Is it then a sufficient condition that the acceleration vector is constantly perpendicular to the inertial trajectory of the object? Well, yes... but you will not get that "constancy in perpendicularity" of the acceleration vector unless the latter also comes with a certain modulus. To understand why, it is convenient to jump, provisionally, from kinematics to dynamics and consider two idealized situations where you can attain such a result.
One possibility is with tension force. Think of this example: an asteroid that is flying by and you attach it to a planet with a very long string, so long that the planet, despite being very massive, is not exerting any significant gravity acceleration, all the job being done by contact force as transmitted by the string. Tension force is similar to normal force in that they are constraint forces: they will gather as much force as necessary to keep the body from penetrating the floor (in the case of normal force) or escaping away (in the case of tension). So the possibility that the asteroid flies away, as long as the string does not break, is not an issue. We can also trust that tension will always act perpendicularly because the string is attached to a pivot and hence rotating with the asteroid, so it will always act by giving it precisely the direction that it does not have. Thus we have attained our objective (circular trajectory), but only because we have somehow cheated: we have chosen a display where the acceleration is automatically adjusted to whatever is needed, not only in terms of direction (thanks to the pivot) but also in terms of modulus (thanks to the string acting as a constraint force).
(Once inside the circle, the asteroid, if endowed with an engine, could accelerate and achieve non-uniform circular motion, because the tension force will automatically adjust to do what is required at each new instant so as to avoid that the satellite flies away, but that non-uniformity is the work of the engine, not of the string.)
A second possibility is that now there is no string. The planet must do its job only by means of gravity. But acceleration due to gravity is what it is, it does not adjust to the target. That is a problem. Imagine that we launch the satellite from the height of a mountain in parallel to the ground. At this initial stage, gravity is acting perpendicularly to the body's path. But such a body will start falling... Thankfully the horizon is also going down due to the curvature of the earth... Now, if the speed of the object is too high, it will fall at a lesser rate than the fall of the horizon, perpendicularity will be lost and the object will not follow a circular path; it may still be caught in an elliptical trajectory or it may even fly away. If the speed is too low, again perpendicularity will be lost and the body will be speed up towards the ground. But if it is the right speed, if we strike the adequate balance between the inertia of the object (as given by its speed) and the gravitational acceleration, then the body will fall at the same rate as the ground and we will get a circular trajectory. (That is why in nature gravitational orbits are usually not circular, but parabolic, elliptical and so on.)
We could look for other examples, but by now it should be clear that the applied force will have to match somehow the velocity of the object, either (i) because the force self-adjusts or (ii) because the velocity is the right one by chance or, more probably, because it is adjusted by very careful design. (Notice that we are not saying that the driving force that is responsible for the acceleration "has" a velocity v. The planet around which the satellite orbits does not have such velocity. This velocity is just the challenge that it faces and which determines what the force must be, to be up to the task.)
On another note, we can also guess that the distance to the object, which should become the radius "r" of the circle, also matters, but in this case in inverse proportion: the farther away from the center, the flatter that the circular trajectory will look like and so a change in direction at this softer rate will be less demanding.
Thus we have guessed that centripetal acceleration for circular motion must look like v/r...
To find out the exact formula, we can start looking, not for the time rate of the change in velocity (which is what acceleration is), but for the velocity change after a while, say for instance after a full revolution.
It helps us to know that the modulus of the initial velocity vector is always equal to the final velocity vector. Imagine that we paint those vectors as radii of a circle and we draw red lines between each initial and each final vector, which red lines correspond to the velocity increase vector for each lapse because indeed they correspond to the vectorial subtraction of final velocity from initial velocity. That is the picture drawn at the start of the post (sorry I cannot cite the source of the picture, I forgot where I took it from).
Because we have measured at intervals, the circle looks like a jagged wheel. But if we now imagine infinitesimal changes, we will compose a perfect circumference. The length of a circumference is 2π times its radius, which radius we have stipulated to be the original and never-changing modulus of the velocity. All this leads to a modulus (for the velocity increase) of 2πv.
Here there is a little logical jump to be made. Which is the "while", i.e. the temporal lapse that the said velocity increase corresponds to? The writings that use this method quickly assume that it is the time lapse required for the object to describe a full round. This is true but not so obvious. Let us not forget that the above picture does not emulate the spatial trajectory of the satellite: the v radii actually paint the constant speed of the orbiting object, which is always spatially tangential (perpendicular to the spatial radius), whereas the red vectors paint how this tangential velocity spatially changes direction (in parallel to the spatial radius). In other words, the v circumference is symmetrical to the spatial circumference. I am also assuming that this works, although I would like to see a more categorical confirmation.
Anyhow, this 2πv that we have landed upon is the accumulated velocity increase after a round. But we were looking for the acceleration, that is to say, a ratio of the change of velocity by time unit. Therefore we need to divide our velocity increase for the round by the time required by such round. This time is the space traversed by the object along such round (i.e. 2πr) divided by the constant velocity v. Thus we get 2πv/2πr/v = v/r/v = v^2/r.
Finally, just note that in real problems this weird "circular acceleration" may not be (in fact very often it is not) something happening autonomously: it is usually a component of another force, but it does help to know its value. And it may also happen that the speed of the body is changing (it is accelerating tangentially, like a bob in a pendulum), but it does help to know the acceleration that is needed at each instant (in view of the relevant v) to keep it in a circular track.
