User:Kagurala/physics1st

This article writes the law that the gravitational force created by a mass point and an even mass sphere is equivalent to the gravitation created by the mass point and another mass point whose mass is the sphere's and location is on the ball's centre.

A Ring

We start with a ring whose thickness is infinitesimal and a mass point with gravition between them.
Assume a ring whose mass is M, and a mass point m which locates on the line perpendicular to the ring's plane and go through ring's centre. We call this line Axis may well. The distance between the mass point and ring's centre is l. The radius of ring is R.
Pick an infinitesimal length of ring dx. This small part gives a gravitation to the mass point:dF
$dF={\frac {GmdM}{R^{2}+l^{2}}}$
We make $\tan \theta ={\frac {R}{l}}$

It's easy to know that the component of dF on the plane perpendicular to the Axis is offset by the gravitation created by the symmetrical infinitesimal part on the ring. So,

$\int dF=\textstyle \int \limits _{0}^{M}\displaystyle {\frac {Gm\cos \theta }{R^{2}+l^{2}}}dM={\frac {GmM\cos \theta }{R^{2}+l^{2}}}$

$\because \cos \theta ={\frac {l}{\sqrt {R^{2}+l^{2}}}}$ ,

$\therefore \int dF={\frac {GmMl}{(R^{2}+l^{2})^{\frac {3}{2}}}}=F$

A Circle

Now we make the ring above stuffed evenly to be a circle, remaining its initial location. The F computed above acts as another infinitesimal force dF here. And the M in the latest formula is replaced by dM . So, the circle's mass is M ,and the radius is R. In this situation, we regard the infinitesimal ring on the circle is an elementary part. Every part gives the mass point a infinitesimal force dF.

$dF={\frac {Gml}{(r^{2}+l^{2})^{\frac {3}{2}}}}dM$

We can suppose the surface density is σ, and the infinitesimal ring's outer radius is r₁ whlie the inner radius is r₂.

so, $dM=(r_{1}^{2}-r_{2}^{2})\pi \sigma =2r\pi \sigma dr$

$\int dF=\int _{0}^{R}{\frac {2Gmlr\pi \sigma }{(r^{2}+l^{2})^{\frac {3}{2}}}}dr=-{\frac {2Gml\pi \sigma }{\sqrt {R^{2}+l^{2}}}}-(-{\frac {2Gml\pi \sigma }{\sqrt {0+l^{2}}}})=2Gm\pi \sigma -{\frac {2Gml\pi \sigma }{\sqrt {r^{2}+l^{2}}}}$

$\because \pi R^{2}\sigma =M$ ,$\therefore F={\frac {2GmM}{R^{2}}}(1-{\frac {l}{\sqrt {R^{2}+l^{2}}}})$.

A Sphere

In the last section, we deduce the gravitation that a even mass factate circle gives to a mass point can be equvalent to the force F $F={\frac {2GmM}{R^{2}}}(1-{\frac {l}{\sqrt {R^{2}+l^{2}}}})$ created from the centre of circle acting at the mass point. Now, we make infinite pieces of this kind of cicles whice have variable radius constitute a sphere. Every infinitesimal circle gives a gravitation dF, and we pick out one circle whose thickness is dl and radius is r. We can establish an coordinate axis called "l" which is coincide with the Axis come up before. Its origin is the mass point. We assume the distance between mass point and the sphere's centre is d+R. So,$dF={\frac {2Gm}{r^{2}}}(1-{\frac {l}{\sqrt {r^{2}+l^{2}}}})dM$,

$r^{2}=R^{2}-[R-(l-d)]^{2}=2dl+2Rl-2Rd-l^{2}-d^{2}$,

$dM=r^{2}\pi \rho dl$

$dF=2Gm\pi \rho (1-{\frac {l}{\sqrt {2dl+2Rl-2Rd-d^{2}}}})dl$

$\int dF=\int _{d}^{d+2R}2Gm\pi \rho (1-{\frac {l}{\sqrt {2dl+2Rl-2Rd-d^{2}}}})2dl={\frac {4GmR^{3}\pi \rho }{3(R+d)^{2}}}$

$\because$ ${\frac {4}{3}}\pi R^{3}\rho =M$,

$\therefore \int dF={\frac {GMm}{(R+d)^{2}}}$

Link

user:kagurala/physics