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SURFACE OF EQUAL ATTRACTION
Surface studied by the Marquis de SaintJacques in 1750, and by Gauss in 1830. 
Spherical equation: ,
pour .
Cartesian parametrization: . Volume: . Attraction of the homogeneous solid with density and mass M on a massive point with mass m located at O: . 
Given a massive point, the surface of equal attraction is the locus of the points contributing with the same given intensity to the (Newtonian) attraction force applied to the massive point, such force resulting from the various attraction forces of the whole surface.
Proof: If Oz is the direction of the resulting force, and the attracted massive point is located at O, then the component of the attraction of a point with spherical coordinates ( = longitude, = latitude) is proportional to , hence the equation of the surface.
It is the surface of revolution obtained by rotation of a half curve of the dipole (or curve of equal attraction) around its symmetry axis.
It can be proved that the homogeneous solid with mass
M such that the (intensity of the) attraction applied to a given
massive point is maximal among all the solids with mass M, or solid
of maximal attraction, is the solid encompassed by this surface.
Comparison between the attractions of this solid (Smax) and of a sphere with same volume and same density:  
Radius of the spherical ball with same volume as (Smax): .
Attraction of this ball applied to a point placed on its surface: . Ratio of the attractions: : (Smax) wins by a short head... 
Proof of the maximality of this solid, given by Bertrand
in 1864 (see this
book page 430, and see also this
book page 45).
Since the attraction of a body on a massive point is always,
according to previous statements, of finite intensity, we can look for
the shape of a homogeneous body with given mass, for which its attraction
on a given molecule would be the maximum possible.

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© Robert FERRÉOL 2017