(8)
Since
| v2 = | h2 | = μ ( | 2 | ± | 1 | ), |
| p2 | r | a |
(9)
it appears that the orbit is an ellipse, parabola or hyperbola, according as v2 is less than, equal to, or greater than 2μ/r. Now it appears from (6) that 2μ/r is the square of the velocity which would be acquired by a particle falling from rest at infinity to the distance r. Hence the character of the orbit depends on whether the velocity at any point is less than, equal to, or greater than the velocity from infinity, as it is called. In an elliptic orbit the area πab is swept over in the time
| r = | πab | = | 2πa3/2 | , |
| 1⁄2h | √μ |
(10)
since h = μ1/2l1/2 = μ1/2ba−1/2 by (8).
The converse problem, to determine the law of force under which a given orbit can be described about a given pole, is solved by differentiating (5) with respect to r; thus
| P = | h2 dp | . |
| p3 dr |