:: M = 4* R3 x density = 9Rx7, which is therefore known. 3 F = 473. Generally when the force acts in parallel lines doy dir? y being parallel and I to the direction of the force. But in the Ellipse y? = (a – x®) a 2 But in the circle it will be likewise found that a F = Hence if the absolute force he changed from v = a* 10 w' = I an ellipse having the same diameter 2a as the circle, and any CENTRAL FORCES. axis-minor whatever may be changed ; and generally if it be required to ascertain the change in the absolute force requisite to make the body describe an ellipse of given axes 2a', 20', we have 8'4 re' :M:: a : a' a a and at the vertex and extremity of latus rectum is respectively a and sa 2 2 : 475. Since the body revolves in an hyperbola with the force in the centre, we know from Newton, Prop. X., or very readily from aolo (436) since p = that - a' + 6 VOC 476. 1, and p = 2a f = a - v(a - b) and a Q-(a'-6) atv (a— b) Now the mean angular velocity is that by which a body would uniformly describe a circle in the periodic time of the earth, and :: 27 ai ✓ V= с V = a 478. Generally, let R be the given distance or radiusvector, P the corresponding perpendicular upon the tangent, then since by the question and (436) cdp p3 dp cdp = Med? F = p3 and p = ca ....(a) + MR - Mega which is the equation to a conic section, and therefore the orbit has one apse at least. But the problem must be solved without considering the nature of the curve ; for since at an apse and :: 9* ca + PR). g f ::= ce I PR + 2F2 f and e will therefore have at least two real values, which indicate + + two apses. If the velocity of projection be given and equal to ß; then B = F (see 436). BP (6) B + 'R' – Mz which may be useful hereafter. 479. By 438. |