Physics - Classical Mechanics - Spherical Mass Distributions

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Introduction

Hey it's a me again @drifter1!

In this article we will continue with Physics, and more specifically the branch of "Classical Mechanics". Today's topic are Spherical Mass Distributions.

So, without further ado, let's dive straight into it!


Spherical Mass Distribution

Newton proved that, for all spherical mass distributions:

  • First Theorem:
    A body that is inside a spherical shell of matter experiences no net gravitational force from that shell.
  • Second Theorem:
    The gravitational force on a body that lies outside a spherical shell of matter is the same as it would be if all of the shell’s matter were concentrated into a point at its center.
Based on these two theorems, a spherical shell can be safely thought of as a point mass (all mass concentrated in its center).


Gravity outside a Spherical Shell

A spherical mass is built up of many infinitely thin spherical shells, which are nested inside each other. Let's consider a mass m is at distance r from the center of a spherical shell of mass M and radius R (R < r).

Based on the second theorem, the gravitational potential of the shell is given by:



because the shell can be though of as a point mass.

Proof

Let's proof this equation...

Based on the principle of superposition, the total gravitational attraction is equal to the vector sum of the forces of gravity that all particles of the spherical shells exert on mass m. But, it's simpler to calculate the gravitational potentials, as those are scalar and not vector quantities.

To begin, the shell is cut into rings, with each rings being at a distance l from m. Each ring has a width Rdθ and radius of Rsinθ. The surface area of the ring is:

Let's consider that the total mass M of the shell is evenly distributed over its surface. The surface area of the shell is 4πR2, and thus the mass of each ring is given by:

For infinitely thin rings, the total potential is given by the following integral:

From the Law of Cosines l2 can be replaced by:



Differentiating both sides results in:

Using it, the integral can be rewritten as:

The ring closest to m has a distance l of r - R, whilst the ring that is further has a distance of R + r. Now the indefinite integral can be turned into a definite integral, which gives:


Gravity within a Spherical Shell

Using the same math as before, inside the shell, the distance of each ring, l, extends from R - r to R + r, giving:


Simple Potential Examples

Hoop

A hoop can be thought of as a spherical shell.

Let's consider a hoop of radius r and mass M On the x-y plane the hoop can be simplified into a circle:



For an arbitrary point P (0, 0, l), the gravitational potential is given by:



where the denominator is the distance of the point P to the center of the hoop.

Disk

A disk of radius r in the x-y plane, can be described by the following equation:



For a mass density σ the total mass of the disk is πr2σ. Integrating the formula for the hoop, over a radius a from 0 to r (mass is now 2πaσda) results in:


RESOURCES:

References

  1. http://www2.yukawa.kyoto-u.ac.jp/~shinsuke.kawai/docs/TeachingMaterial/Gravitation.pdf
  2. http://astro.utoronto.ca/~bovy/AST1420/notes-2017/notebooks/02.-Spherical-Mass-Distributions.html
  3. https://www.sparknotes.com/physics/gravitation/potential/section3/

Images

  1. https://commons.wikimedia.org/wiki/File:Spherical_shell_moment_of_inertia.png

Mathematical equations used in this article, where made using quicklatex.


Previous articles of the series

Rectlinear motion

Plane motion

Newton's laws and Applications

Work and Energy

Momentum and Impulse

Angular Motion

Equilibrium and Elasticity

Gravity


Final words | Next up

And this is actually it for today's post!

Next time we will get into how Earth's Rotation affects Gravity..

See ya!

Keep on drifting!


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