Page:EB1911 - Volume 08.djvu/831

lengths of the ten measured bases is 29,863 toises, so that the average length of a base line is 19,100 ft. The azimuths were observed at fourteen stations. In high latitudes the determination of the meridian is a matter of great difficulty; nevertheless the azimuths at all the northern stations were successfully determined,—the probable error of the result at Fuglenaes being ± 0″·53.

Before proceeding with the modern developments of geodetic measurements and their application to the figure of the earth, we must discuss the “mechanical theory,” which is indispensable for a full understanding of the subject.

Mechanical Theory.

Newton, by applying his theory of gravitation, combined with the so-called centrifugal force, to the earth, and assuming that an oblate ellipsoid of rotation is a form of equilibrium for a homogeneous fluid rotating with uniform angular velocity, obtained the ratio of the axes 229 : 230, and the law of variation of gravity on the surface. A few years later Huygens published an investigation of the figure of the earth, supposing the attraction of every particle to be towards the centre of the earth, obtaining as a result that the proportion of the axes should be 578 : 579. In 1740 Colin Maclaurin, in his De causa physica fluxus et refluxus maris, demonstrated that the oblate ellipsoid of revolution is a figure which satisfies the conditions of equilibrium in the case of a revolving homogeneous fluid mass, whose particles attract one another according to the law of the inverse square of the distance; he gave the equation connecting the ellipticity with the proportion of the centrifugal force at the equator to gravity, and determined the attraction on a particle situated anywhere on the surface of such a body. In 1743 Clairault published his Théorie de la figure de la terre, which contains a remarkable theorem (“Clairault’s Theorem”), establishing a relation between the ellipticity of the earth and the variation of gravity from the equator to the poles. Assuming that the earth is composed of concentric ellipsoidal strata having a common axis of rotation, each stratum homogeneous in itself, but the ellipticities and densities of the successive strata varying according to any law, and that the superficial stratum has the same form as if it were fluid, he proved that

${\frac {g'-g}{g}}+e={\frac {5}{2}}m,$

where $g,g'\!$ are the amounts of gravity at the equator and at the pole respectively, $e\!$ the ellipticity of the meridian (or “flattening”), and $m$ the ratio of the centrifugal force at the equator to $g$. He also proved that the increase of gravity in proceeding from the equator to the poles is as the square of the sine of the latitude. This, taken with the former theorem, gives the means of determining the earth’s ellipticity from observation of the relative force of gravity at any two places. P. S. Laplace, who devoted much attention to the subject, remarks on Clairault’s work that “the importance of all his results and the elegance with which they are presented place this work amongst the most beautiful of mathematical productions” (Isaac Todhunter’s History of the Mathematical Theories of Attraction and the Figure of the Earth, vol. i. p. 229).

The problem of the figure of the earth treated as a question of mechanics or hydrostatics is one of great difficulty, and it would be quite impracticable but for the circumstance that the surface differs but little from a sphere. In order to express the forces at any point of the body arising from the attraction of its particles, the form of the surface is required, but this form is the very one which it is the object of the investigation to discover; hence the complexity of the subject, and even with all the present resources of mathematicians only a partial and imperfect solution can be obtained.

We may here briefly indicate the line of reasoning by which some of the most important results may be obtained. If $\mathrm {X} ,\mathrm {Y} ,\mathrm {Z} \!$ be the components parallel to three rectangular axes of the forces acting on a particle of a fluid mass at the point $x,y,z\!$, then, $p\!$ being the pressure there, and $\rho \!$ the density,

$dp=\rho (\mathrm {X} dx+\mathrm {Y} dy+\mathrm {Z} dz);$

and for equilibrium the necessary conditions are, that $\rho (\mathrm {X} dx+\mathrm {Y} dy+\mathrm {Z} dz)\!$ be a complete differential, and at the free surface $\mathrm {X} dx+\mathrm {Y} dy+\mathrm {Z} dz=0$. This equation implies that the resultant of the forces is normal to the surface at every point, and in a homogeneous fluid it is obviously the differential equation of all surfaces of equal pressure. If the fluid be heterogeneous then it is to be remarked that for forces of attraction according to the ordinary law of gravitation, if $\mathrm {X} ,\mathrm {Y} ,\mathrm {Z} \!$ be the components of the attraction of a mass whose potential is $\mathrm {V} \!$, then

$\mathrm {X} dx+\mathrm {Y} dy+\mathrm {Z} dz={\frac {d\mathrm {V} }{dx}}dx+{\frac {d\mathrm {V} }{dy}}dy+{\frac {d\mathrm {V} }{dz}}dz,$

which is a complete differential. And in the case of a fluid rotating with uniform velocity, in which the so-called centrifugal force enters as a force acting on each particle proportional to its distance from the axis of rotation, the corresponding part of $\mathrm {X} dx+\mathrm {Y} dy+\mathrm {Z} dz$ is obviously a complete differential. Therefore for the forces with which we are now concerned $\mathrm {X} dx+\mathrm {Y} dy+\mathrm {Z} dz=d\mathrm {U} \!$, where $\mathrm {U}$ is some function of $x,y,z\!$, and it is necessary for equilibrium that $dp=\rho d\mathrm {U} \!$ be a complete differential; that is, $\rho \!$ must be a function of $\mathrm {U} \!$ or a function of $p\!$, and so also $p\!$ a function of $\mathrm {U}$. So that $d\mathrm {U} =0\!$ is the differential equation of surfaces of equal pressure and density.

We may now show that a homogeneous fluid mass in the form of an oblate ellipsoid of revolution having a uniform velocity of rotation can be in equilibrium. It may be proved that the attraction of the ellipsoid $x^{2}+y^{2}+z^{2}(1+\epsilon ^{2})=c^{2}(1+\epsilon ^{2})\!$; upon a particle $\mathrm {P} \!$ of its mass at $x,y,z\!$ has for components

$\mathrm {X} =-\mathrm {A} x,\mathrm {Y} =-\mathrm {A} y,\mathrm {Z} =-\mathrm {C} z,$

where

$\mathrm {A} =2\pi k^{2}\rho \left({\frac {1+\epsilon ^{2}}{\epsilon ^{3}}}\tan ^{-1}\epsilon -{\frac {1}{\epsilon ^{2}}}\right),$

$\mathrm {C} =4\pi k^{2}\rho \left({\frac {1+\epsilon ^{2}}{\epsilon ^{2}}}-{\frac {1+\epsilon ^{2}}{\epsilon ^{3}}}\tan ^{-1}\epsilon \right),$

and $k^{2}\!$ the constant of attraction. Besides the attraction of the mass of the ellipsoid, the centrifugal force at $\mathrm {P} \!$ has for components $+x\omega ^{2},+y\omega ^{2},0$; then the condition of fluid equilibrium is

$(\mathrm {A} -\omega ^{2})xdx+(\mathrm {A} -\omega ^{2})ydy+\mathrm {C} zdz=0,$

which by integration gives

$(\mathrm {A} -\omega ^{2})(x^{2}+y^{2})+\mathrm {C} z^{2}={\text{constant}}.$

This is the equation of an ellipsoid of rotation, and therefore the equilibrium is possible. The equation coincides with that of the surface of the fluid mass if we make

$\mathrm {A} -\omega ^{2}=\mathrm {C} /(1+\epsilon ^{2}),$

which gives

${\frac {\omega ^{2}}{2\pi k^{2}\rho }}={\frac {3+\epsilon ^{2}}{\epsilon ^{3}}}\tan ^{-1}\epsilon -{\frac {3}{\epsilon ^{2}}}.$

In the case of the earth, which is nearly spherical, we obtain by expanding the expression for $\omega ^{2}\!$ in powers of $\epsilon ^{2}\!$, rejecting the higher powers, and remarking that the ellipticity $e={\tfrac {1}{2}}\epsilon ^{2}\!$,

$\omega ^{2}/2\pi k^{2}\rho =4\epsilon ^{2}/15=8e/15.$

Now if $m$ be the ratio of the centrifugal force to the intensity of gravity at the equator, and $a=c(1+e)$, then

$m=a\omega ^{2}/{\tfrac {4}{3}}\pi k^{2}\rho a,\therefore \omega ^{2}/2\pi k^{2}\rho ={\tfrac {2}{3}}m.$

In the case of the earth it is a matter of observation that $m=1/289$, hence the ellipticity

$e=5m/4=1/231,$

so that the ratio of the axes on the supposition of a homogeneous fluid earth is 230 : 231, as stated by Newton.

Now, to come to the case of a heterogeneous fluid, we shall assume that its surfaces of equal density are spheroids, concentric and having a common axis of rotation, and that the ellipticity of these surfaces varies from the centre to the outer surface, the density also varying. In other words, the body is composed of homogeneous spheroidal shells of variable density and ellipticity. On this supposition we shall express the attraction of the mass upon a particle in its interior, and then, taking into account the centrifugal force, form the equation expressing the condition of fluid equilibrium. The attraction of the homogeneous spheroid $x^{2}+y^{2}+z^{2}(1+2e)=c^{2}(1+2e)$, where $e$ is the ellipticity (of which the square is neglected), on an internal particle, whose co-ordinates are $x=f,y=0,z=h$, has for its $x$ and $z$ components

$\mathrm {X} '=-{\tfrac {4}{3}}\pi k^{2}\rho f(1-{\tfrac {2}{5}}e),\;\mathrm {Z} '=-{\tfrac {4}{3}}\pi k^{2}\rho h(1+{\tfrac {4}{5}}e),$

the $\mathrm {Y} \!$ component being of course zero. Hence we infer that the attraction of a shell whose inner surface has an ellipticity $e\!$, and its outer surface an ellipticity $e+de$, the density being $\rho \!$, is expressed by

$d\mathrm {X} '={\tfrac {4}{3}}\cdot {\tfrac {2}{5}}\pi k^{2}\rho fde,\;d\mathrm {Z} '=-{\tfrac {4}{3}}\cdot {\tfrac {4}{5}}\pi k^{2}\rho hde.$

To apply this to our heterogeneous spheroid; if we put $c_{1}\!$ for the semiaxis of that surface of equal density on which is situated the attracted point $\mathrm {P} \!$, and $c_{0}\!$ for the semiaxis of the outer surface, the attraction of that portion of the body which is exterior to $\mathrm {P} \!$, namely, of all the shells which enclose $\mathrm {P} \!$, has for components

$\mathrm {X} _{0}={\tfrac {8}{15}}\pi k^{2}f\int _{c1}^{c0}\rho {\frac {de}{dc}}dc,\;\mathrm {Z} _{0}={\tfrac {16}{15}}\pi k^{2}h\int _{c1}^{c0}\rho {\frac {de}{dc}}dc,$