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According to Newton, the gravitational force of two masses is F=G•m1•m2/D²,
where D is the distance. m1 is assumed to be 1 unit of mass (MU) and relatively small compared to the total mass.
If we take a small point mass mp of m2 at a distance d from m1, this partial force is ft= G•mp/d². However, the total force must be determined.
Let h be the perpendicular from mp to the straight line from m1 in the direction of the total force,
and a the length of the line from m1 to the base of the perpendicular.
The component of the partial force is f=ft •a/d = G•mp •a/d³ = G•mp•a/sqrt(a²+h²)³ according to Pythagoras and cosines.
The gravitational force on a hollow sphere (radius R) is to be determined. Let m1 be ld (in units of length) from its center.
(Mass density)•G = K. q is the projection of a point on the sphere onto the line. At the center: q=0.
If a is then (ld-q), the point fraction is: f=K•(ld-q)/sqrt((ld-q)²+h²)³
Since h is the same size at a cross-section circle (transverse direction) for the same q,
it is sufficient to integrate along a semicircle (longitudinal direction), with the modified mass density K•2 •PI •h,
i.e., the circumference of the cross-section circle times K. The arc s=R•phi is used as the path of integration.
With h=R•sin(phi) and q=R•cos(phi), the integrand is: (2•PI•K•R•sin(phi) •(ld -R•cos(phi))/sqrt ((ld-R•cos(phi))²+R²•sin(phi)²)³ )•R.
If we take the relative length L = ld/R, R cancels out, as the unit of length is contained in the mass density. We can further simplify with sin²+cos²=1 to:
Integral K•2•Pi•(L-cos(phi))•sin(phi)/sqrt( L(L-2•cos(phi))+1)³ von 0 bis Pi
This integral is solvable in an elementary way (substitution z = cos(phi)) and yields the solution:
4•PI•K/ L² for L greater than 1, 0 für L less than 1, discontinuous for L=1 (not solvable).
Within the hollow sphere, positive and negative forces cancel each other out – there is no ravitational force.
This makes it very easy to calculate the gravity of a star. Outside the surface, this results inm1•G•Gesamtmasse/D²,
inside, only the mass of the remaining core needs to be considered,
not including the masses of the spherical shells located above the path traveled. And if the density doesn't increase too much,
Gravitational force decreases steadily until it reaches the center. Thus, maximum gravitational force is at the surface.
Ludwig Resch
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