| u | du | = − | μ |
| dx | x2 |
(15)
whence
| u2 = | 2μ | + C. |
| x |
(16)
In the case of a particle falling directly towards the earth from rest at a very great distance we have C = 0 and, by Newton’s Law of Gravitation, μ/a2 = g, where a is the earth’s radius. The deviation of the earth’s figure from sphericity, and the variation of g with latitude, are here ignored. We find that the velocity with which the particle would arrive at the earth’s surface (x = a) is √(2ga). If we take as rough values a = 21 × 106 feet, g = 32 foot-second units, we get a velocity of 36,500 feet, or about seven miles, per second. If the particles start from rest at a finite distance c, we have in (16), C = − 2μ/c, and therefore
| dx | = u = − √ { | 2μ (c − x) | }, |
| dt | cx |
(17)
the minus sign indicating motion towards the origin. If we put x = c cos2 1⁄2φ, we find
| t = | c3/2 | (φ + sin φ), |
| √(8μ) |