 Solution

At the initial instant the circulation around every closed curve is zero because the fluid is at rest. A smooth material-flow map carries closed curves bijectively between times. Conservation of circulation around every material curve therefore makes the line integral of the velocity zero around every closed curve at the current time, not just small contractible loops. Fix a reference point in a connected fluid region and defineThe vanishing closed-curve integrals make this independent of the path. Thus $\mathbf{u}=\nabla \varphi$: the fluid has a globally defined velocity potential. Its incompressible flow condition is $\nabla \cdot \mathbf{u}=0$, hence ${\nabla }^2\varphi =0$. This argument establishes the global potential without silently assuming the fluid region is simply connected.

Use the laboratory frame, with the sphere instantaneously centred at the origin and moving in the $\theta =0$ direction. Impermeability equates fluid and sphere normal velocities, while the fluid is at rest at infinity:Choose the additive constant so that $\varphi \to 0$ at infinity. Substitution of $\varphi =f(r)\cos \theta$ into the Laplace equation in spherical polar coordinates givesThe far-field condition gives $A=0$, and $f^{\prime }(a)=U$ gives $B=-U{a}^3/2$. ThereforeThe radial and polar velocity components areFor fluid density $\rho$, its kinetic energy isThus $T_f=\frac{1}{2}{m}_a{U}^2$ with added mass of a sphere $m_a=2\pi \rho a^3/3$, half the mass of displaced fluid.

For the initial acceleration after release, let $\dot{U}$ denote the upward acceleration, $\mathcal{V}=4\pi a^3/3$, and $m_b={\rho }_b\mathcal{V}$. At the instant $U=0$, the velocity-squared term in the Unsteady Bernoulli equation vanishes. The time derivative of the potential at the sphere is ${\varphi }_t(a,\theta )=-a\dot{U}\cos \theta /2$; centre-motion corrections also vanish at this instant. Relative to hydrostatic pressure, the additional pressure is consequentlyIts upward force on the sphere isAdding buoyancy and weight, Newton's second law gives $(m_b+m_a)\dot{U}=(\rho -{\rho }_b)\mathcal{V}g$. ThereforeThis is the acceleration of a freely falling sphere with added mass, with upward chosen positive. If ${\rho }_b>\rho$, the negative value represents a downward acceleration.