We shall find it convenient to express it in terms of the time of one revolution, say T. It is easily done, since plainly T = circumference/speed = 2πr/v; so the above expression for centrifugal force becomes 4π2mr/T2.
As to the fall of the body towards the centre every microscopic unit of time, it is easily reckoned. For by Euclid III. 36, and [Fig. 58], AP.AA' = AO2. Take A very near O, then OA = vt, and AA' = 2r; so AP = v2t2/2r = 2π2r t2/T2; or the fall per second is 2π2r/T2, r being its distance from the centre, and T its time of going once round.
In the case of the moon for instance, r is 60 earth radii; more exactly 60·2; and T is a lunar month, or more precisely 27 days, 7 hours, 43 minutes, and 11½ seconds. Hence the moon's deflection from the tangential or rectilinear path every minute comes out as very closely 16 feet (the true size of the earth being used).
Returning now to the case of a small body revolving round a big one, and assuming a force directly proportional to the mass of both bodies, and inversely proportional to the square of the distance between them: i.e. assuming the known force of gravity, it is
| V Mm |
| r2 |
where V is a constant, called the gravitation constant, to be determined by experiment.
If this is the centripetal force pulling a planet or satellite in, it must be equal to the centrifugal force of this latter, viz. (see above).
| 4π2mr |
| T2 |
Equate the two together, and at once we get