ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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Exterior potential sphere and zero ideal force — worked interpretation

Physical scope

Constructed parameter substitution into an inviscid impermeable-slip sphere solution, not a measured force or viscous-drag prediction.

Example

For the inviscid fixed-sphere model choose $a = 1 m$ , $U = (2, 0, 0) m s^{- 1}$ , $ρ_{0} = 1 k g m^{- 3}$ , and $p_{\infty} = 10^{5} P a$ . At polar angle $θ$ from the $x$ -axis, surface pressure is $p = 10^{5} + 2 (1 - \frac{9}{4} \sin^{2} θ) P a$ . It is $100002 P a$ at the front/back poles and $99997.5 P a$ at the equator. Antipodal algebra gives zero pressure-only resultant. This is not a no-slip viscous-sphere calculation.

Example references

Stewart, MIT 8.09, Fluid Mechanics, §6.4, printed pp. 120–123

Fitzpatrick, Fluid Mechanics, UT Austin, potential sphere

Facts & Assumptions

Given: $a = 1 m$ , $U = (2, 0, 0) m s^{- 1}$ , $ρ_{0} = 1 k g m^{- 3}$ , $p_{\infty} = 10^{5} P a$ and exact slip solution.

[F1]

Exterior potential sphere and zero ideal force — conditional physical prediction: surface velocity is tangent, pressure is $p_{\infty} + ρ_{0} (∣ U ∣^{2} - ∣ u ∣^{2}) / 2$ , and pressure-only resultant vanishes.

Proof

1.1givenF1algebra

At $r = a$ , speed is $3 ∣ U ∣ \sin θ / 2 = 3 \sin θ m s^{- 1}$ . Bernoulli gives $p = 10^{5} + \frac{1}{2} (4 - 9 \sin^{2} θ)$ Pa, yielding $100002$ Pa at $\sin θ = 0$ and $99997.5$ Pa at $\sin θ = 1$ .

2.1F1step 1.1algebra∎

Pressure depends on $\sin^{2} θ$ and is therefore unchanged at antipodal surface points, where normals are opposite. Their pressure-force contributions cancel in pairs; the zero result is only for inviscid pressure stress.

Depends on

Exterior potential sphere and zero ideal force — conditional physical prediction