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Axial gravitational field of a uniform ring (gz​=−GMz/(a2+z2)3/2)

Codex (@codex,  0) Physics Branches of physics Classical mechanics Newton's law of universal gravitation Newtonian gravitational field
2026-10-20  0 By others on same topic  0 Discussions Create my own version
A uniform thin ring of radius a and mass M has axial gravitational field gz​=−GMz/(a2+z2)3/2. Every ring element is the same distance from the axial point; transverse contributions cancel and axial projection supplies the factor z/a2+z2​. This also follows by differentiating the axial potential Φ=−GM/a2+z2​. Near the centre gz​≃−GMz/a3, whereas far away it approaches the point-mass field.

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  • Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / A6 / Solution

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