# Why does Goldstein's derivation of orbits from Newton's law not yield the orbit consisting of symmetric oscillation along a line thru center of Sun?

Why does Goldstein's derivation of Keplerian orbits from Newton's law of gravity not yield the orbit consisting of symmetric oscillation along a line passing thru the center of the Sun?

The fact that an orbiting body cannot penetrate the sun is irrelevant here. By Newton's theory, the oscillating body should accelerate to infinite velocity as it approaches the center, but then immediately decelerate symmetrically on the other side. The period will obviously be finite. This orbit is a maximally eccentric ellipse, and its foci are at the ends, but the sun is at the center!

• So the sun would not be at a focus? – BowlOfRed Nov 10 '15 at 5:49
• Goldstein talks about stable orbits not passing through the centre of force when he derives keplerian orbits. When he draws the graph of potential vs r, the bottom most point of the valley is the keplerian orbit. But depending upon the energy of the body and initial position, other orbits are possible. One of which is this. – sarat.kant Nov 11 '15 at 17:36

Inside the Sun, the force is no longer an inverse-square law. It actually grows linearly with $r$.
The second reason is that Goldstein (as well as any other classical mechanics book) is interested in orbits with a non vanishing angular momentum with respect to the center of the Sun. An oscillation along a line passing through this point has zero angular momentum. We cannot think of this orbit as the limiting case of the ellipse. The latter has eccentricity $$\epsilon=\sqrt{1+\frac{2EL^2}{mK^2}},$$ where $L$ is the angular momentum. By doing $L\rightarrow 0$ one gets $\epsilon\rightarrow 1$ which is a parabola (which does not correspond to an oscillation). This is inconsistent because in obtaining the eccentricity it is already assumed that $L\neq 0$.