Centripetal acceleration vector

[Andrew Mason]

What you did was to show that these assumptions are consistent with an inverse square law force at those two points, and at those two points only. Showing that elliptical orbits with the central object at a focus means an inverse square law force everywhere takes a bit more work.

But if one starts with Kepler's observation that ALL planets move in ellipses with the sun at a focus, then one has 6 pairs of points, all of which fit the 1/r^2 law. To get any number of pairs of points on a particular ellipse that are diametrically opposite, one could simply imagine a planet whose elliptical path has perigee and apogee at those points. Using Philip's method, one could show that the relationship between the accelerations of those bodies and the distance from the sun would be $a \propto 1/r^2$.

Of course he didn't. The mathematical machinery to deduce elliptical orbits from an inverse square law, or to deduce an inverse square law from Kepler's observations, did not exist in Kepler's time.

And Feynman's lecture (Feynman's Lost Lecture) shows just how remarkable that mathematical achievement that was for Newton.

AM

 

[A.T.]

One could call it an instantaneous centripetal component of acceleration.

One could. But acceleration is always an instantaneous quantity, so you can drop the "instantaneous" qualifier. And "component" is implied, unless you explicitly state that something is a net acceleration.

"Normal acceleration" is an alternative. But usually "normal" means any perpendicular vector, which is not unique in 3D.

 

[D H]

One could call it an instantaneous centripetal component of acceleration.

In which frame of reference? A curve in one frame of reference can look very different in another frame of reference. Consider the Earth's Moon. The path traced by the Earth's Moon looks very different from the perspectives of a geocentric versus a heliocentric point of view.

But since Newton called the acceleration toward the central point "centripetal" it is somewhat confusing to call it "centripetal" at all.

Exactly. Nowadays we call those things "central forces" rather than "centripetal forces". Looking all the way back to the original sources is a good subject for philosophy of science. It is not a good idea for learning or doing science. The historical documents result in a great deal of grief here at PhysicsForums.

One need look no further than the movie Galaxy Quest to see the kinds of troubles that can ensue due to blind faith in the historical documents.

 

[Philip Wood]

What you did was to show that these assumptions are consistent with an inverse square law force at those two points, and at those two points only.

Yes, that's right. But if we assume that the force from the Sun varies with distance from the Sun as a simple power law, then just these two points suffice to show that it's an inverse square law. That's all I've ever claimed. Surely that's not objectionable.

Showing that elliptical orbits with the central object at a focus means an inverse square law force everywhere takes a bit more work.

What does need a bit more work is justifying what I assumed - a simple power law. And a full treatment of elliptical orbits delivers this: the inverse square law emerges without having to assume a power law force.

Of course he didn't. The mathematical machinery to deduce elliptical orbits from an inverse square law, or to deduce an inverse square law from Kepler's observations, did not exist in Kepler's time.

Good. We're agreed here. I only raised this because I didn't - and still don't - understand your accusation of circularity (no pun intended) in my argument. You'll remember claiming that I was implicitly assuming an inverse square law.

 

[Andrew Mason]

In which frame of reference? A curve in one frame of reference can look very different in another frame of reference. Consider the Earth's Moon. The path traced by the Earth's Moon looks very different from the perspectives of a geocentric versus a heliocentric point of view.

The acceleration is the same in any inertial frame of reference. The path depends on the velocity of the origin of the frame you are measuring it in but that does not affect the determination of the centre of rotation.

AM

 

[jhae2.718]

$\renewcommand{\vec}[1]{\boldsymbol #1}$

From this last result, we have the centripetal acceleration given by $-r\dot{\theta}^2\hat{\vec{b}}_1$ (this result may look more familiar if we define $\omega \equiv \dot{\theta}$ and write $-r\dot{\theta}^2\hat{\vec{b}}_1 = - r\omega^2\hat{\vec{b}}_1 = -v^2/r\hat{\vec{b}}_1$). But we know that $\hat{\vec{b}}_1$ points outwards, so the centripetal acceleration must point toward the center.

Upon rereading my earlier post, I feel as if I should clarify that the $v$ in $-v^2/r\hat{\vec{b}}_1$ is not the magnitude of the inertial velocity vector, but the magnitude of the transverse velocity $r\dot{\theta} = r\omega$.