Feynman's Lost Lecture (ft. 3Blue1Brown)

Because the acceleration vector is as much rotational as the radius vector. This tells us the tangent direction of each object in orbit; Whatever vector comes from our velocity graph touches that point theta degrees, that is the velocity vector of our orbiting object and therefore the direction of the tangency curve. In fact, let me start by drawing all of these velocity vectors as lines, because all we need to use is the information that you know about the slope of the orbital curve; the specific magnitude of each velocity will not be as important. Note that this is not the angle of the velocity vector at this time, but the angle θ is vertical.

No no no. The angle I'm referring to in the velocity diagram is approximately the center of the circle, which is almost certainly a little different from where the velocity vector is based. So the question is, what special curve satisfies the property of the tangent direction from the theta point in radians in the horizontal direction to this vector coming from a special eccentric point to the circular point in degrees theta on a circle from the vertical direction ? Well, here's the trick. First, rotate it in a full circle to set 90 degrees. Then take each individual speed direction and rotate them 90 degrees back the other way, so they are oriented like before, only each one is rooted in a different place.

AHA! We found the ellipse! But we still have some ideas and really understand how this emerging ellipse relates to astronomical orbits. Importantly, I don't just rotate your lines around any point, I rotate each one around its center, which means we can use the geometric proof we saw a few minutes ago. And it's those moments where you have that feeling and you frown and you think, "Wait, what happened to that evidence again?" One of the key points is when you have two lines, one from the center of the circle and one from the outside. -Central point, both are common points on the circumference of the perpendicular bisector to the eccentric line that will be tangent to the ellipse.

More importantly, the point of tangency is where it crosses the radial lines from the center. This means that our little ellipse has degrees theta horizontal, around the center of the circle, and has a tangent slope perpendicular to that eccentric line. And since the entire rotation is 90 degrees, that means it's parallel to the velocity vector that we need it to be. So this little emerging curve inside the velocity graph has the tangent property and we need our orbit! Therefore, the shape of the orbit must be an ellipse. QED. Well, pat yourself on the back, because all you need is ingenuity to follow this.

The first is this particular way of constructing an ellipse, the demonstration of which requires some geometric ingenuity. And then a very clever move to ask what shape the velocity vector will take when they move all their tails to the same place. And proving that it is a circle requires mixing the inverse square sum law with Kepler's second law, another clever move. But the intelligence doesn't end there! Showing how this velocity graph is rooted in terms of a vector at a point at the center of the circle implies that an elliptical orbit brings this neat trick of 90 degree rotation.

I like this. Watching Feynman do physics, even basic physics, is like watching Download Bobby Fischer. (Henry): Thanks again Grant, you should definitely check out his videos on

3blue1brown

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