Feynman's lost lecture The motion of the planets around the Sun

Page: 0 1 2 3 4 5 6

A planet orbitting the Sun The diagram shows a representation of the orbit of a planet around the Sun. In a certain time interval the planet would move from A to B, if no force were acting on it. In the next equal interval of time, again if no force were acting on it, the planet would move from B to c. The Sun's force is represented by an impulse at B resulting in a component of motion towards the sun BV. By the parallelogram law, the planet actually takes the path BC rather than Bc. The same procedure is repeated at each point, C, D, E...

Feynman, as Newton did, then goes on to show that the planet sweeps out equal areas in equal time, ie, that in the diagram below, the yellow area equals the blue area. Cabri shows the two areas. Try dragging some points. You will see that the two areas remain the same. But can this be proved?

The proof involves first demonstrating that, in the following diagram, triangles SAB and SBc are congruent. Since AB = Bc, they have equal bases, and they both have the same height given by SX. Therefore SAB and SBc are congruent. Move either of the green points B or V. It should be apparent that triangle SBC is simply a shear of triangle SBc and therefore has the same area. Thus triangles SAB and SBc have the same area.