This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1900 Excerpt: ...value at the Earth's surface at P, we at once see from the principle of relative accelerations that g is the vector difference of G and w?R cos A.. _ The direction of G is, by the theory of universal gravitation, PO. The direction of the apparent acceleration due to gravity is of course the vertical, defined as the ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1900 Excerpt: ...value at the Earth's surface at P, we at once see from the principle of relative accelerations that g is the vector difference of G and w?R cos A.. _ The direction of G is, by the theory of universal gravitation, PO. The direction of the apparent acceleration due to gravity is of course the vertical, defined as the direction taken by the plumb-line, or as the normal to the surface of a liquid at rest. The latter method is used for determining the vertical in Astronomy, and the latitude of the place (F) is defined as the inclination of the vertical so determined to the plane of the equator. If then, po, pq, qo represent the accelerations G, g, ui'R cos A, respectively, and we produce oq to t, angle poq--X, and angle pqt = l the latitude. Now, taking R equal approximately to 4000 miles, and neglecting the difference between the lengths of a sidereal and a mean solar day, so that "=24x60x60' we obtain 02 =-1117f.s.s. = f.s.s. nearly. If R = Earth's equatorial radius = 20,923,600 feet, and we take 9_ o equal to---radians per second, its true value, we obtain 86164 JR= '11126 f.s.s. Now g is equal to 32-2 approximately in the latitude of London, and varies from 32-088 at the equator to (probably) 32-253 at the poles; consequently the ratio---is very nearly equal to $ in all latitudes. It follows that the angle opq is very small, and drawing qr perpendicular to op we have G--g=ro = oq cos X=oPR cos2X, or g=G-o2R cos2 X, or =G-uPRcosH, writing I for X in the small term o2R cos2 X. Further, the circular measure of the angle I--X is approximately equal to rq uPR sin X cos X--, or, PI 9 for which we may write aPR sin I cos I Thus it appears that the effect of the Earth's rotation is to diminish the acceleration of a falling body near the Earth's surface in...
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