(adsbygoogle = window.adsbygoogle || []).push({}); The problem

Given Newton's law of gravitational attraction, that the force a body exerts on a particle in space is directed towards the body and has a magnitude proportional to the inverse square of the distance to the body, show that the force field is described by the 1-form

[tex]\frac{kx}{r^{3}}dx + \frac{ky}{r^{3}}dy + \frac{kz}{r^{3}}dz[/tex]

where k is a postive constant and r(x,y,z) is the distance from a point to the body.

The attempt

The 1-form for a constant force field is A dx + B dy + C dz, where A, B, and C are constant and represent the work required for a unit displacement in the relevant direction. So the force is in a direction (-A, -B, -C). Then the magnitude of a radial force is the total work required for unit displacement opposed to the force, which is just the unit vector in the direction -[tex]F[/tex]or [tex](\frac{A}{\sqrt{A^{2}+B^{2}+C^{2}}},\frac{B}{\sqrt{A^{2}+B^{2}+C^{2}}},\frac{C}{\sqrt{A^{2}+B^{2}+C^{2}}})[/tex]. Thus, [tex]F = \sqrt{A^{2}+B^{2}+C^{2}}[/tex].

The problem I'm having is showing that the general 1-form of a radial force is cx dx + cy dy + cz dz, where c > 0.

If I could show that, then I know by Newton and the first part of the solution that [tex]\sqrt{(cx)^{2}+(cy)^{2}+(cz)^{2}} = \frac{k}{r^{2}}[/tex], where k is some constant. Solving for c yields [tex]c = \frac{k}{r^{3}}[/tex] and the problem is done.

Can anyone help me understand why to say that the force at (x,y,z) is radially outward means that it is of the form mentioned above?

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# Homework Help: 1-form of a central force field?

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