Philosophy 167: Class 13 - Part 14 - De Motu Corporum in Gyrum: Problem 5, and Trajectories Under Inverse-Square and Uniform Gravity.
Smith, George E. (George Edwin), 1938-
We've done everything we need to for this case of inverse square except direct vertical fall. So that completes the accounts. We have to do direct vertical fall to have everything we need for inverse square. If you write the differential equation down for direct vertical fog, you find it a monster, okay?
But Newton had a very clever way of solving it non-symbolically. Start with the idea that you've got a very small velocity of projection. What are you gonna get? You're gonna get an ellipse. And if it's a small velocity of projection, it's gonna be a highly eccentric ellipse as shown here.
We know how to time anything in an ellipse. Kepler showed us how. Do the circumscribed circle, and the areas in the circle give you the corresponding areas of the ellipses. So you take the fraction of the total time, break it down on the circle, you can locate where the object is versus time all the way down.
And since you know period versus a cubed, you actually know you have enough that you can do calculations here. Next step, now let that velocity at the top go to zero, what's gonna happen to this? It's gonna move closer and closer to a straight line, and point b is gonna move closer and closer to s.
Okay, in the limit, you still can use this circle but the circle is now gonna, s and b are gonna coincide and the circle is going to be running through the center force. So this gives you a geometric solution for vertical fall versus time, under inverse square force.
It gives you more than that. In the notes, and I'll put some hand calculations up on trunk tomorrow so you can see this. You can compare uniform gravity vertical fall versus inverse square gravity fall, from any height you wish. Just set the problem up. Say, we know the distance of fall in the first second, call it 16 English feet.
That's roughly right. Okay now, let that be a fall from 64 feet or from 256 feet and let it go uniformly versus inverse square. Can we detect the difference over 256 feet between the two? The answer is, the ratio between the two differs in ten to the minus fifth.
It's way past detecting. In other words, he can say, here, if gravity, vertical gravity, terrestrial gravity, is an inverse square force to the center of the Earth, you couldn't tell the difference from any height we drop, between that and uniform gravity. Which is doubly striking because what he goes on to say in this passage, and this is the one place where gravity is not crossed out, he says under the assumption of hypothesis, he says, of inverse square gravity.
And oh by the way, gravity's a centripetal force. Under that hypothesis, problem four gives you the motion of a projectile in the absence of air resistance, and problem five gives you vertical fall in the absence of air resistance. Okay, so he's answered Hook's question about the trajectory, he's answered Galileo's question about the effect of curvature.
If it's in, he's done it twice. If it's constant it's those lobes that he did in the letter to Hook. If it's inverse square all the way down, it's an ellipse. But it's an ellipse of such high eccentricity, that it approximates a parabola at the end portion. Again you can do the calculations then compare the ellipse, given how far the center of the Earth is away, with the parabola.
And once again it will be undetectable, the difference between the two, for any reasonable size range for the ellipse. So what we have here are Galileo's two famous problems solve for inverse square forces. Fair enough? Look what we've done, we've tied all the things we've studied in this course into one very simple account of inverse square forces directed towards the center.
All of Kepler's laws, Galileo's laws, with the principle of inertia sitting in the background as a hypothesis.