An introduction to the mathematical theory of attraction ...

Examples. 63
It is here supposed that the material surface of the sphere is equally expanded in all directions so that the tension is the same along all great circles.
The normal pressure at a point P on the sphere is equilibrated by the tensions
along all the great circles passing through P. Consider now a small circle
having its pole at P, the arc ds of a great circle from P to its circumference
being infinitely small. If d5r be the angle between two tangents to a great
circle passing through P at the points where it meets the small circle, we have
adr = 2ds. Again the resultant of the two tensions acting at P along the same
great circle is 2Tds dO sin ~dr. Hence the resultant of all the tensions passing
through P is
r Tdr ds that is 2r - ds2,
a
and this must equilibrate the total force acting normally to the spherical area
enclosed by the small circle having ds for radius. Hence if F be the magnitude
of this force per unit of area,
birds2 = 2r - ds2.
a
Substituting for F from Art. 35, and remembering that o = 4a- we obtain
the equation required.
10. Prove that at any point of the surface of an ellipsoidal mass of homogeneous liquid, rotating in relative equilibrium round a permanent axis, the
force acting on a fluid particle is proportional and parallel to the corresponding
semi-diameter of the reciprocal ellipsoid.
The components of the force are
-(A- w2), -(B - 2)y, and - Cz,
whence, by Ex. 6, Art. 24, the above result is obvious.
11. Show that, for any distribution of mass, every line of force which does
not encounter mass or pass through a point of equilibrium has an asymptote
passing through the centre of mass.
From Art. 31 it appears that the line of force has a point at infinity; and, as
the resultant force at this point passes through the centre of mass, the truth of
the theorem is obvious.
12. Find the equation of a line of force due to masses m', n", m"', &c.,
situated at points on the same straight line.
By Art 34 we have
1 = 2r (m (1 - cos 0') + m" (1-cos 0") + &c.;
whence m' cos 0' - m" cos 0" + &c. = M-
where M is the sum of the masses. If I be constant, this equation represents a
line of force.
It can readily be obtained from the consideration that the component of the
resultant force perpendicular to a line of force at any point on this line is zero;
hence
nm' r'dO' m" r"dO"
-2 ds &c. = 0;
r'2 ds r"~ ds