Wrong. In polar coordinates the acceleration is of constant magnitude and direction at a given radius(radially inward.) You must confusing Cartesian with polar coordinates.

You mean you are intentionally using dishonest coordinates to pretend it is constant to cover up your earlier mistake?

Like I said, if it is constant, there are no higher derivatives.

The derivative of a constant is 0. That means no jerk and your entire claim falls to pieces.

But even with polar coordinates, the vector is changing direction.

With polar coordinates you have a magnitude and a direction, which will be expressed as a magnitude and direction rather than an x and y component (dealing with 2D).

When the orbiting object has a position of (r,0), the acceleration vector is (a,π). When the position is (r,π), the acceleration vector is (a,0).

Are you really going to claim that (a,π) and (a,0) are the same?

The frame it remains constant in is a rotating reference frame, in which the position remains constant and there are 2 forces, an inwards force pulling the object towards the centre, and an outwards centrifugal force pulling it to the edge. This keeps the position constant and all derivatives of it as 0. This means no jerk.

So what is it going to be?

Is the acceleration changing, or is it static and thus there is no jerk?

But, as demonstrated a constant inward acceleration is not able to support a circular or elliptical orbit.

WHERE?

You keep asserting this but you are yet to provide any justification of it.

Yes, the blog post you linked does show there are higher order derivatives due to this inwards acceleration.

It even shows it produces a circular orbit.

It in no way disproves it.

Instead it just jumps to the completely unsupported conclusion that because jerk is a thing it magically can't do it.

There is absolutely no basis for it at all.

There can be no "higher order rates of change of the acceleration due to the central force" if there is only a non-varying central force as Newtonian mechanics asserts. There can be no effects without causes.

So you admit your blog post is garbage?

Such irony! And Newton's second law isn't pertinent here, we are talking about DERIVATIVES of Newton's second law.

No, that is what is extremely important.

It indicates that if you have a force acting on an object then it

Yes, I am saying that. An object will move in the direction of the net force acting upon it if the net force does not vary with time.

Try actually addressing the question.

This object isn't just sitting there having a net force on it. It is already moving.

What happens?

We can easily see that the force applied will accelerate it towards the centre of the orbit, but the sideways motion will mean that it can't just go straight down and instead remains the same distance, circling.

Can you explain what you think should happen and why.

gravitation is postulated to merely provide an acceleration and no higher order rates of change of acceleration.

Again, it is the motion of the object which provides them.

If the object was stationary (and held stationary) then the acceleration from gravity would be constant.

But it isn't. It moves. As it moves, it can change the magnitude (if it moves radially) and the direction (if it moves tangentially). This is all that is required to provide these higher order terms.

You don't need magical higher order forces.

Using the wrong radius for your Cavendish experiment

Even when using the 4 m, you don't get anywhere near the number you provided. You would need roughly 15 m, which is a massive blunder.

So either you pulled a number out of thin air, or you just failed at a simple calculation.

The first was your fault

No, it was entirely your fault for making such a horribly flawed assumption. There was no reason at all for you to use roughly 15 m.

You could estimate it from the footage and masses provided like I did, or you could have asked.

I guarantee that my math skills are more than adequate to assess this matter.

Your math skills do seem to be quite poor.

You are claiming the acceleration/force is constant, but also changing by virtue of having derivatives.

As a nice simple question, what is the derivative of a constant?

More importantly, what about your physics skills which seem to be severely lacking, with you repeatedly claiming you need magical higher order forces?

This is a poor analogy. A rigid body undergoing torque (turbine, spinning shafts) is not the same as a body orbiting around another point due to a central force directed from that point.

Then how about a bunch of objects on a string, setup so they are equally spaced around a circle with an even number of them.

That way there is no net force on the shaft, just a torque to keep it spinning.

According to you, all the objects should magically fly to the centre.

We can even have it set up to be on a spring so you can't complain that the string gets loose and lets the balls fly back out, or you can attach a load cell to each string/object to confirm it is constantly being pulled towards the centre.

The spring version already exists in the form of a centrifugal clutch.

At low speed, the spring is providing a force greater than the centripetal force and the clutch pulls inwards. At a very specific speed it is balanced and would remain in place. At a higher speeds it goes outwards even though the only force acting on it is the force (not torque) from the springs pushing it inwards.

Now again, can you use your allegedly adequate math skills to explain exactly what will happen to a object experiencing a force directed towards the origin with a magnitude of |F|=mv

^{2}/r to produce an acceleration of |a|=v

^{2}/r.

Can you provide a formula (in either Cartesian or polar coordinates) to describe the path taken, preferably in parametric form where t is the time in seconds?

For someone as allegedly skilled as you, this should be no problem.