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Let the fluid be at rest at infinity, and write its velocity potential as $\varphi =f(r)\cos \theta$, the angular dependence selected by translation along the polar axis. The separated Laplace equation iswith radial solutions $r$ and $r^{-2}$. Decay at infinity eliminates the former. The no-penetration condition on the moving sphere is $u_r(a)=U\cos \theta$, giving $f(r)=-U{a}^3/(2r^2)$. Thus the potential flow around a translating sphere isFor density $\rho$, integrate its kinetic energy over the exterior. The angular factor satisfies $\int ({\cos }^2\theta +{\sin }^2\theta /4)d\Omega =2\pi$, and ${\int }_a^{\infty }{r}^{-4}dr=1/(3a^3)$. HenceThe added mass of a sphere is consequently $M_f/2$. For a heavy sphere, $M>M_f$, falling through distance $h$, the net gravitational work after buoyancy is $(M-M_f)gh$. Its own kinetic energy plus that of the fluid is $(M+M_f/2)U^2/2$. Conservation of energy from rest therefore givesThis is the unbounded inviscid-fluid model; walls or viscous dissipation would alter the flow and its energy.

Irrotational flow permits $\mathbf{u}=\nabla \varphi$, and incompressible flow imposes $\nabla \cdot \mathbf{u}=0$, hence $\Delta \varphi =0$. In the laboratory frame, the no-penetration condition at the translating sphere is ${\partial }_r\varphi (a,\theta )=U\cos \theta$.

For separation of variables, take the angular factor $g(\theta )=\cos \theta$, which satisfies $\frac{1}{\sin \theta }(\sin \theta g^{\prime }{)}^{\prime }=-2g$. The radial equation iswith solutions $r$ and $r^{-2}$. Rest at infinity excludes the $r$ term. The normal boundary condition fixes the remaining coefficient, giving the potential flow around a translating sphereThere is no azimuthal velocity. With fluid density ${\rho }_f$, the kinetic energy isThe angular integral is $2\pi$, and the radial integral is $1/(3a^3)$, so $K_f={\rho }_f\pi a^3{U}^2/3$. Since $M=(4\pi a^3/3){\rho }_s$ and $k={\rho }_f/{\rho }_s$, $K_f=kM{U}^2/4$. Thus the added mass of a sphere is $kM/2$.

During a fall through $h$, gravity does work $Mgh$ and buoyancy does work $-kMgh$. Starting from rest, conservation of energy givesConsequently $U=\sqrt{4(1-k)gh/(2+k)}$. The mass cancels: the reduced speed comes from both buoyancy and the fluid's added mass.

Take the sphere to move in the positive $z$-direction and use spherical coordinates centred on it. An axisymmetric harmonic potential that decays at infinity has the dipole form $A\cos \theta /r^2$. The no-penetration condition in the laboratory frame isIt fixes $A=-U{a}^3/2$, so the potential flow around a translating sphere isThe laboratory-frame velocity components are

The fluid kinetic energy iswhere $V=4\pi a^3/3$. Thus the added mass of a sphere is $\rho V/2$.

When the sphere falls, replacing fluid of density $\rho$ by material of density ${\rho }_s$ lowers the gravitational potential energy at rateThe total kinetic energy of sphere and fluid isConservation of total energy givesFor $U\ne 0$, the acceleration of a freely falling sphere with added mass isThe same formula holds at release by continuity.