 Solution

Unsteady Bernoulli gives the surface pressure from $p+\rho {\partial }_t\varphi +\rho |\nabla \varphi {|}^2/2=C(t)$. The velocity-squared contribution integrates to zero net force, while the unsteady term gives $F=-m_a\dot{V}$ with added mass $m_a=\frac{1}{2}(4\pi \rho a^3/3)=2\pi \rho a^3/3$. Hence $(m+m_a)a=G$, so $m^{\ast }=m+2\pi \rho a^3/3$: accelerating the bubble also accelerates surrounding fluid.

Solved by gpt-5.6-sol high.