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 Author Topic: No gravitational force inside a uniform shell of matter  (Read 8721 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: June 03, 2009, 10:59:53 pm » posted from:Taipei,T\'ai-pei,Taiwan

It is easier to understand that: a uniform sphere shell of matter exerts a gravitational force on a particle outside the shell as if all the shell's mass were concentrated at its center.

However, it might not be easy to see why : a uniform shell of matter exerts no gravitational force on a particle located inside it.

There are 2D view and 3D view in the applet.
The green surfaces (curves) represent mass distributed on a uniform shell.
The blue dot represent the particle inside the shell (with mass M).

The gravitation force is propotional to mass and r-2: $\vec{F}=-\frac{G M m}{r^2}\hat{r}$.
Let's consider mass distributed on those two green shell (m1 and m2).
The solid angles for both sirface related to particle are the same (solid angle=$\Omega$),
and the distance to both those two surface are r1 and r2.
Assume the surface density is $\sigma$
The surface area for two surface are $A_1=r_1^2 \Omega$ and $A_2=r_2^2 \Omega$
So $m_1=\sigma A_1 =\sigma r_1^2 \Omega$, and $m_2=\sigma A_2 =\sigma r_2^2 \Omega$

$F_1= \frac{GM\sigma r_1^2 \Omega}{r_1^2}=GM\sigma\Omega$, and $F_2= \frac{GM\sigma r_2^2 \Omega}{r_2^2}=GM\sigma\Omega$.
But those two force are in the oppositive direction so all the gravitational force cencel each other.
You can drag the particle with your mouse to watch how the surface area changed at the same time.

Another question for you: what happened to the rest of the mass on the surface (upper and lower parts)?
Does the gravitational force due to those mass also cancel each other? Why?

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