Shell theorem: Difference between revisions

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# If the body is a spherically symmetric shell (i.e., a hollow ball), no net [[gravitational force]] is exerted by the shell on any object inside, regardless of the object's ___location within the shell.

A corollary is that inside a solid sphere of constant density, the gravitational force within the object varies linearly with distance from the center, becoming zero by symmetry at the center of [[mass]]. This can be seen as follows: take a point within such a sphere, at a distance <math>r</math> from the center of the sphere. Then you can ignore all of the shells of greater radius, according to the shell theorem (1). SoBut the point can be considered to be external to the remaining sphere of radius r, and according to (2) all of the mass of this sphere can be considered to be concentrated at its centre. The remaining mass <math>m</math> is proportional to <math>r^3</math> (because it is based on volume),. and theThe gravitational force exerted on ita isbody at radius r will be proportional to <math display="inline">\frac{m}{r^2}</math> (the [[inverse square law]]), so the overall gravitational effect is proportional to {{nowrap|<math display="inline">\frac{r^3}{r^2} =r</math>,}} so is linear in {{nowrap|<math>r</math>.}}

These results were important to Newton's analysis of planetary motion; they are not immediately obvious, but they can be proven with [[calculus]]. ([[Gauss's law for gravity]] offers an alternative way to state the theorem.)