Shell theorem: Difference between revisions

Revision as of 08:18, 4 April 2022

Revision as of 00:45, 4 May 2022

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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 <math display="block">E_\text{point} = \frac{GM}{p^2}</math>

<blockquote>[[File:Point2.png|frameless|300x300px]]</blockquote>

:<math>E_\text{point}=\frac{GM}{p^2}</math>

Suppose that this mass is moved upwards along the ''y''-axis to point {{nowrap|<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 {{nowrap|<math display="inline">\sqrt{p^2 + R^2}</math>.}} Hence, the gravitational field of the elevated point is:

<blockquote>[[File:Point2.png|frameless|300x300px]]</blockquote>

:<math display="block">E_\text{elevated point} = \frac{GM}{p^2+R^2}</math>

Suppose that this mass is moved upwards along the ''y''-axis to point {{nowrap|<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 {{nowrap|<math display="inline">\sqrt{p^2+R^2}</math>.}} Hence, the gravitational field of the elevated point is:

<blockquote>[[File:Pointy2.png|frameless|270x270px]]</blockquote>

:<math>E_\text{elevated point}=\frac{GM}{p^2+R^2}</math>

<blockquote>[[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, {{nowrap|<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:

:<math display="block">E_x = \frac{GM\cos{\theta}}{p^2+R^2}</math>

Substituting in <math>\cos(\theta)</math> gives

:<math display="block">E_x = \frac{GMp}{\left(p^2+R^2\right)^{3/2}}</math>

Suppose that this mass is evenly distributed in a ring centered at the origin and facing point <math>P</math> with the same radius {{nowrap|<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:

:<math display="block">E_\text{ring} = \frac{GMp}{\left(p^2+R^2\right)^{3/2}}</math>

<blockquote>[[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 {{nowrap|<math>P</math>,}} each with a radius {{nowrap|<math>y</math>,}} width of {{nowrap|<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 {{nowrap|<math>\pi R^2</math>.}} So, {{nowrap|<math display="inline">dM=\frac{M\cdot 2y\,dy}{R^2}</math>.}} Hence, a small change in the gravitational field, <math>E</math> is:

:<math display="block">dE = \frac{Gp\,dM}{(p^2+y^2)^{3/2}}</math>

<blockquote>[[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:

:<math display="block">E = \int \frac{GMp\cdot \frac{2y\, dy}{R^2}}{(p^2+y^2)^{3/2}}</math>

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 {{nowrap|<math>y=R</math>,}} resulting in:

:<math display="block">E_\text{disc} = \frac{2GM}{R^2} \left( 1-\frac{p}{\sqrt{p^2+R^2}}\right)</math>

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 {{nowrap|<math>P</math>,}} each with a radius {{nowrap|<math>R</math>,}} width of {{nowrap|<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: {{nowrap|<math display="inline">R = \sqrt{a^2-x^2}</math>.}} <math>x</math> varies from <math>-a</math> to {{nowrap|<math>a</math>.}}

The mass of any of the discs <math>dM</math> is the mass of the sphere <math>M</math> multiplied by the ratio of the volume of an infinitely thin disc divided by the volume of a sphere (with constant radius {{nowrap|<math>a</math>).}} The volume of an infinitely thin disc is {{nowrap|<math>\pi R^2\, dx</math>,}} or {{nowrap|<math display="inline">\pi\left(a^2-x^2\right) dx</math>.}} So, {{nowrap|<math display="inline">dM = \frac{\pi M(a^2-x^2)\,dx}{\frac{4}{3}\pi a^3}</math>.}} Simplifying gives {{nowrap|<math display="inline">dM = \frac{3M(a^2-x^2)\,dx}{4a^3}</math>.}}

Each discs' position away from <math>P</math> will vary with its position within the 'sphere' made of the discs, so <math>p</math> must be replaced with {{nowrap|<math>p+x</math>.}}

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Replacing <math>M</math> with {{nowrap|<math>dM</math>,}} <math>R</math> with {{nowrap|<math>\sqrt{a^2-x^2}</math>,}} and <math>p</math> with <math>p+x</math> in the 'disc' equation yields:

:<math display="block">dE = \frac{\left( \frac{2G\left[3M\left(a^2-x^2\right)\right]}{4a^3} \right) }{\sqrt{a^2-x^2}^2}\cdot \left(1-\frac{p+x}{\sqrt{(p+x)^2+\sqrt{a^2-x^2}^2}}\right)\, dx</math>

Simplifying,

:<math display="block">\int dE = \int_{-a}^a \frac{3GM}{2a^3} \left(1 - \frac{p + x}{\sqrt{p^2 + a^2 + 2px}}\right)\, dx</math>

Integrating the gravitational field of each thin disc from <math>x=-a</math> to <math>x=+a</math> with respect to {{nowrap|<math>x</math>,}} and doing some careful algebra, yields Newton's shell theorem:

:<math display="block">E = \frac{GM}{p^2}</math>

where <math>p</math> is the distance between the center of the spherical mass and an arbitrary point {{nowrap|<math>P</math>.}} The gravitational field of a spherical mass may be calculated by treating all the mass as a point particle at the center of the sphere.