and de and ? -Q.edu a er du :. dt = - Х 24 for the Integration of which see Lacroix's Differential and Integral Calculus, 4to edit., or Whewell's Dynamics, p. 15. F = 490. The tension is = the centripetal force necessary to retain the stone in the circle, or it is 472 g T2 gr being the length of the string. 4.r Ig :: 4:1 and :: by substituting and performing the arithmetical operations we shall get the time required. 491. Generally let it be required to find the velocity acquired through (n + 1) XR to the earth's centre; R being its radius. and v = 2gR' X 1 (n + 1)R Let e = nR, and we have 1 29R n+1 n.(n + 1) (6) n.(n + 1) the velocity acquiring in descending to the surface of the earth. Again, since within the Earth's surface Foc g, we have F:g::p : R хе (a') Lets = 0. Then and V' R = NgR .. (6) Hence V + V' = VgR x (1+ ✓nin+1 is the velocity required. vnosti) In the Problem n = l. :: V + V' =NgR x (1 +72) 492. Let a and 6 be the major and minor axes of the given ellipse. Then the greatest dist. = a + va'-b?=all+e) is known. Again, the Periodic Time in the Ellipse is (484) T = su and in the circle for the same force it is (considering it the limit of an ellipse) 2oat(1 + e) NE 493. and By (443) if v and v' denote the angular velocities of Po we have But p = dp .. podp 62 in the ellipse. e 2a ps 62 p2 62 PS 63 p3 x(2a –p) 1 = 26 vaje x (2ag – g) VOL. II. p3 oglo z Hence u' is least when 2ag pa = 2a 2p=0 p= a; or the angular velocity of p in the ellipse, the force being in the focus, is least at the extremity of the minor-axis. Rp 1 p= (aR - 1-1.p) and 0 = 1. Si+vn N (nR? – n-1p)-RJ17 2n 11-vn ✓(nR' – ņ– 1.3") +R7nS Hence whenę = 00 1 1 1+Nn 2 dn 1-yn √n 1-1 and the number of revolutions is 1 It n 1. 0 = 2 n N = .1.(1+2) *V 2 12.2.(72) + {**+ **)' + ** (in)* + &c.; NE Hence, if a few terms of this converging series be summed, and the sum be called S, we get 1 N 2 XS 22 which may be arithmetically computed to any degree of accuracy, by means of Logarithmic Tables. The whole number in the result will answer the problem. Let a, b be the semiaxes of the Ellipse; then since 495. (by 436) and v = 2 X 24 (r = p) dt = = V 3 Те a + Vai-b?, which is the distance from the farther apside, being put = r. Again, dę gde dr v (re- P + C} 2u 2 Let e = r; then and t = pe} 2 vers.--1 28 ? = 0. Then the time to the focus from the farther |