Shell theorem
Simplified gravity models for objects inside or outside a spherically symmetrical body
In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetric body. This theorem has particular application to astronomy.
Nº Q2222575 ★★
Uncommon · Knowledge
Shell theorem
Simplified gravity models for objects inside or outside a spherically symmetrical body
In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetric body. This theorem has particular application to astronomy.
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From Wikipedia
In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetric body. This theorem has particular application to astronomy. Isaac Newton proved the shell theorem and stated that: A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center. 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 r {\displaystyle r} from the center of the sphere. Then you can ignore all of the shells of greater radius, according to the shell theorem (2). But the point can be considered to be external to the remaining sphere of radius r, and according to (1) all of the mass of this sphere can be considered to be concentrated at its centre. The remaining mass m {\displaystyle m} is proportional to r 3 {\displaystyle r^{3}} (because it is based on volume). The gravitational force exerted on a body at radius r will be proportional to m / r 2 {\displaystyle m/r^{2}} (the inverse square law), so the overall gravitational effect is proportional to r 3 / r 2 = r {\displaystyle r^{3}/r^{2}=r} , so is linear in r {\displaystyle r} . 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...
Text: Wikipédia, CC BY-SA 4.0. · Image: Jim Wisniewski (CC BY-SA 2.5) ·
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