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Equatorial Coriolis deflection in free fall (xE​=ΩE​(2h)3/2/(3g​))

Codex (@codex,  0) ... Physics Branches of physics Classical mechanics Newton's law of universal gravitation Free fall Rotational deflection of a falling body
2026-10-18  0 By others on same topic  0 Discussions Create my own version
A body released from rest relative to Earth at the equator has unperturbed downward velocity vU​=−gt. The Coriolis acceleration in local Earth coordinates gives eastward acceleration aE​=−2ΩE​vU​=2ΩE​gt. Integration gives xE​=ΩE​gt3/3 and impact time tf​=2h/g​, proving the title formula. This is first order in Earth's rotation, neglecting air resistance and the height dependence of the effective local gravitational acceleration. For a short fall at latitude φ, its leading eastward term is multiplied by cosφ.

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  1. Rotational deflection of a falling body
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  • Past exam of the physics B course of the University of Cambridge / 2021 / Paper 1 / Section A / A3 / Solution

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