With these values we have pr' sin O'd r' d' d w' [r2+r12—2 rr'(cos @ cos 0'+sin 0 sin 0' cos(w—w')]} .(11) This equation is very much simplified if we place the origin of coördinates at the attracted point, for then r = 0, and equation (11) becomes V = SSS p r' sin 0' d r' d 0' d w' . . . . . . . . . . . . . = ..(12) From this equation we see that the value of Vis continuous, or does not become infinite as appears possible from equation (11). 6. Thus far we have considered Va continuous function, and made up of the sum of all the particles of the attracting mass, divided by their respective distances from the attracted point. There is another case in which we may consider the attracting particles (or masses) as independent and finite in number. Thus, if we wish to find the potential of the masses m', m", &c., for their attractive influence on the mass m, the distances being r', '', &c., we should have When we wish to find the potential for the disturbing force exerted by m', m", &c., on the motion of m, the form of the expression becomes changed, and it is usually called R, the Disturbing Function. This form of Vis employed in the Lunar and the Planetary Theory. 7. We shall now show how to find V for a sphere of uniform density. If we use equation (12) we must integrate with respect to w' between the limits 0 and 2 ; r', between the limits r' and r', for an external point; and between the limits r' and 9 for an internal point. If we put = cos 0' we shall have 1 2 V = π p S (r', 2 — r'22)d μ.... (14) and V=pf (r' ̧2 d μ............. (15) The equation of the generating circle of the sphere is c .... .(16) Put the distance between the attracted point and the center of the sphere, and take this line for the axis of x. We then have = 4 c μ v a2 — c2 + c2 μ2 r'22 = 2 c2 μ2 + a2 — c2 + 2 c μ √ a2 — c2 + 0 μ2. The limits of for an external point are +1, and c2 μ2 = c2 — a3; μ μ = and for an internal point the limits of u are +1 and 1. The equation -1 We can find the potential of a homogeneous spheroid of revolution, for a point situated in the axis of revolution, by a similar process, though the work is more complicated. Since the values of X, Y, Z can be calculated directly, we can find V for the spheroid by equation (5), the value of C being made the value of V for the center of the spheroid. SOLUTION OF TWO SIMILAR INDETERMINATE PROBLEMS. BY GEORGE R. PERKINS, LL. D., UTICA, N. Y. 1. Find three square numbers in arithmetical progression, such that if from each its root be subtracted, the remainders shall be squares. 2. Find three square numbers in arithmetical progression, such that if to each its root be added, the sums shall be squares. We will assume for the three square numbers as follows: 2 2 { } ( x + x ̄1) ± } } 2; { } (y+y ̄1)±} } 2; { } (~+~~1) ± } } 2. The upper sign corresponds with the first problem, and the lower sign with the second problem. Since the following condition is true for all values of x, we see that the square numbers as above assumed, will satisfy the con ditions of the problems as to the square of the remainders in the first, and the square of the sums in the second. It only remains therefore to determine x, y and z so that these assumed numbers shall be in arithmetical progression. That is, we must have { Hy+y ̃1)±} }3— { }(x+x ̄1)±} }2= { }(~+2−1)± ± }2 — { }(y+y−1)±} }2 ... (2) Since the difference of the squares of two quantities is equal to the product of their sum into their difference, we must have { \(y+y ̄1)+1(x+x ̄1)±1 }× { 1(y+y ̄1)−1(x+x-1) } { 1(2+2−1)+1(y+y ̃1)±1 }×{ †(2+2 ̄1)—1(y+y-1) } Condition (3) will be satisfied by the following assumed values: ‡(y + y−') + ‡(x + x ̄1) ± 1 = m....... ..(3) ..(4) Putting the values of (y + y1) ± 1, given by (8) and (11), equal, we find + Pn Ρ P+1 .(10) pn ..(11) Р We must now determinem so as to make these values of x, y and z rational, that is, we must have If we assume m2 + a m = (m + k), we shall find = _l — 4 cl3 + 2(2a c + 2 b c — a b)ľ2 — 4 a b c l + a2 b3 ̧ 4(l—b)2 14 cl2 + 2(2 a c + 2 be a b)2 - 4 a b c l + a2 b2 = .. (21) Assuming as follows: 1 - 4 cls + 2(2 a c + 2 b c — a b) l2 — 4 a b c l + a2 b2 = { 12 — 2 cl + 2 c(a + b − c) — a b ¦2, we find 2(a b + ac + b c) — (a2 + b2 + c2) 4( a + b + c) And substituting this value of k in (19) we find { 2(a b + a c + b c) — (a2 + b2 + c2 } 2 2 (22) (23) ..... (24) ± 8( a + b + c). (a − b + c). (a + b —c) ' m = -- 1)42 (2 p2 - · C + 2p+1). (2 p2+4p + 1) BUD· · (31) m b ± 8(2 p2 — 1) . — (2 p2 + 2 p + 1). (2 p2 + 4 p + 1) B C D· · (32) These results are general and hold good, in the case of the first problem, 3, we find A = 864571; B = 177; C = 673; D = 877; |