The Earth is stopped dead on its orbit and begins to fall straight into the Sun. How long does the fall take? No integration is allowed — and the only number you may use is the length of the year.
Solution
About sixty-five days — the year divided by .
The straight fall is itself a Kepler orbit: the limit of ellipses stretched thinner and thinner around the Sun until the ellipse collapses onto a line. Falling from rest at the Earth’s distance means aphelion at and perihelion at the center, so the degenerate ellipse has semi-major axis . Now use the one fact everyone knows: the period of a bound Kepler orbit depends on its semi-major axis alone, . The three-halves power is mechanical similarity in the potential — Problema XI’s law — while the indifference to eccentricity is the extra degeneracy peculiar to , the same one that closes its orbits. Together they let a circle of radius be compared with a needle of semi-major axis : the needle’s full period is , and the fall is the aphelion-to-perihelion half of it, days.
The two natural worries dissolve on inspection. That the orbit runs into the point where the force law diverges costs nothing: the plunge is fastest exactly where the geometry is worst, and striking the Sun’s actual surface rather than its center trims only minutes off the answer — at the photosphere the Earth would already be moving at the solar escape speed, some . And nothing about the Earth itself entered: any object halted at — a planet, a pebble, a stalled comet — takes the same sixty-five days to arrive.
Astrophysicists carry this answer around as the free-fall time: stop anything at radius from a mass and it reaches the center in of the circular period at — the natural clock of gravitational collapse, from star formation to the emptying of a galactic nucleus.
Deeper in the notebook: 02. Geodesics, Orbits, the Effective Potential · the Classical Mechanics shelf, where the Kepler problem will live.