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→Derivation of gravitational field outside of a solid sphere: fixing some rather clumsy TeX usage |
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There are three steps to proving Newton's shell theorem. First, the equation for a gravitational field due to a ring of mass will be derived. Arranging an infinite number of infinitely thin rings to make a disc, this equation involving a ring will be used to find the gravitational field due to a disk. Finally, arranging an infinite number of infinitely thin discs to make a sphere, this equation involving a disc will be used to find the gravitational field due to a sphere.
The gravitational field <math>E</math> at a position called <math>P</math> at <math>(x,y)=(-p,0)</math> on the ''x''-axis due to a point of mass <math>M</math> at the origin is<blockquote><math>E_\text{point}=\frac{GM}{p^2}</math> [[File:Point2.png|frameless|300x300px]]</blockquote>Suppose that this mass is moved upwards along the y-axis to point <math>(0,R)</math>. The distance between <math>P</math> and the point mass is now longer than before; It becomes the hypotenuse of the right triangle with legs <math>p</math> and <math>R</math> which is <math display="inline">\sqrt{p^2+R^2}</math>. Hence, the gravitational field of the elevated point is:<blockquote><math>E_\text{elevated point}=\frac{GM}{p^2+R^2}</math> [[File:Pointy2.png|frameless|270x270px]]</blockquote>The magnitude of the gravitational field that would pull a particle at point <math>P</math> in the ''x''-direction is the gravitational field multiplied by <math>\cos(\theta)</math> where <math>\theta</math> is the angle adjacent to the ''x''-axis. In this case, <math>\cos(\theta)=\frac{p}{\sqrt{p^2+R^2}}</math>. Hence, the magnitude of the gravitational field in the ''x''-direction, <math>E_x</math> is:<blockquote><math>E_x=\frac{GM\cos{\theta}}{p^2+R^2}</math></blockquote>Substituting in <math>\cos(\theta)</math> gives<blockquote><math>E_x=\frac{GMp}{(p^2+R^2)^{3/2}}</math></blockquote>Suppose that this mass is evenly distributed in a ring centered at the origin and facing point <math>P</math> with the same radius <math>R</math>. Because all of the mass is located at the same angle with respect to the ''x''-axis, and the distance between the points on the ring is the same distance as before, the gravitational field in the ''x''-direction at point <math>P</math> due to the ring is the same as a point mass located at a point <math>R</math> units above the y-axis:<blockquote><math>E_\text{ring}=\frac{GMp}{(p^2+R^2)^{3/2}</math> [[File:Wider ring2.png|frameless|280x280px]]</blockquote>To find the gravitational field at point <math>P</math> due to a disc, an infinite number of infinitely thin rings facing <math>P</math>, each with a radius <math>y</math>, width of <math>dy</math>, and mass of <math>dM</math> may be placed inside one another to form a disc. The mass of any one of the rings <math>dM</math> is the mass of the disc multiplied by the ratio of the area of the ring <math>2\pi y\,dy</math> to the total area of the disc <math>\pi R^2</math>. So, <math>dM=\frac{M\cdot 2y\,dy}{R^2}</math>. Hence, a small change in the gravitational field, <math>E</math> is:<blockquote><math>dE=\frac{Gp\ dM}{(p^2+y^2)^{3/2}</math> [[File:Wider ring with inside ring2.png|frameless|350x350px]]</blockquote>Substituting in <math>dM</math> and integrating both sides gives the gravitational field of the disk:<blockquote><math>E=\int \frac{GMp\ \cdot \frac{2y\ dy}{R^2}}{(p^2+y^2)^{3/2}</math></blockquote>Adding up the contribution to the gravitational field from each of these rings will yield the expression for the gravitational field due to a disc. This is equivalent to integrating this above expression from <math>y=0</math> to <math>y=R</math> , resulting in:<blockquote><math>E_\text{disc}=\frac{2GM}{R^2} \left( 1-\frac{p}{\sqrt{p^2+R^2}}\right)</math></blockquote>To find the gravitational field at point <math>P</math> due to a sphere centered at the origin, an infinite amount of infinitely thin discs facing <math>P</math>, each with a radius <math>R</math>, width of <math>dx</math>, and mass of <math>dM</math> may be placed together.
These discs' radii <math>R</math> follow the height of the cross section of a sphere (with constant radius "<math>a</math>") which is an equation of a semi-circle: <math>R=\sqrt{a^2-x^2}</math>. <math>x</math> varies from <math>-a</math> to <math>a</math>.
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