Lemma I.
Invenire gravitatem corporis longinqui ad circumferentiam circuli ex particulis materiæ in duplicatâ ratione distantiarum inversè attrahentibus constantem.
ESto NIK (Vid. Tab. [xxxiii.] Fig. 1.) circumferentia circuli, in cujus puncta omnia gravitet corpus longinquum S locatum extra planum circuli. In hoc planum agatur linea perpendicularis SH, et per circuli centrum X ducatur recta HXK secans circulum in I et K, et SR parallela ad HX: producatur autem SH ad distantiam datam SD, et agantur rectæ DC, XC, ipsis HX, SD, parallelæ. Tum ductâ chordâ quavis MN ad diametrum IK normali eamque secante in L, ex punctis M, N, demittantur in SR perpendiculares MR, NR, concurrentes in R; junctisque SM, SN, erit SM = SN, MR = NR, SR = HL. Dicantur jam SD, k; HX sive DC, h; XL, x; CX, z; XI, r; eritque HL = h - x, et SH = k - z. Est autem SM ad SH ut attractio 1 ⁄ (SM)² corporis S versus particulam M in directione SM ad ejusdem corporis attractionem in directione SH, quæ proinde erit SH ⁄ (SM)³: sed est SR = HL, et (SM)² = (SR)² + (MR)² = (SR)² + (SH)² + (ML)²; unde sit SH ⁄ (SM)³ = SH ⁄ ((HL)² + (SH)² + (ML)²)³⁄ ², et ductâ mn parallelâ ad MN, vis qua corpus S attrahitur ad arcus quàm minimos Mm, Nn, exponitur per SH × 2Mm ⁄ (SM)³ = SH × 2Mm × ((HL)² + (SH)² + (ML)²) ⁻³⁄ ². Est autem (HL)² + (SH)² + (ML)² = kk - 2kz + zz + hh - 2hx + rr, hincque ponendo kk + hh = ll, ((HL)² + (SH)² = (ML)²)⁻³⁄² = 1 ⁄ l³ + 3kz ⁄ l⁵ + 3hx ⁄ l⁵ - 3rr ⁄ 2l⁵ - 3zz ⁄ 2l⁵ + 15kkzz ⁄ 2l⁷ + 15khzx ⁄ 2l⁷ + 15hhxx ⁄ 2l⁷, neglectis terminis ulterioribus ob longinquitatem quam supponimus corporis S. Quarè, si scribatur d pro circumferentiâ IMKN, gravitas corporis S ad totam illam circumferentiam secundum SH, sive fluens fluxionis SH × 2Mm × ((HL)² + (SH)² + (ML)²) ⁻³⁄ ² evadit (k - z) × d in 1 ⁄ l³ + 3kz ⁄ l⁵ - 3rr ⁄ 2l⁵ - 3zz ⁄ 2l⁵ + 15kkzz ⁄ 2l⁷ + 15hhrr ⁄ 4l⁷. Simili modo obtinebitur gravitas ejusdem corporis S secundum SR. Q. E. I.