= Solution
At the <terminal velocity>, buoyancy-corrected weight balances the <Stokes drag law>:
$$
\frac{4\pi a^3}{3}(\widetilde\rho-\rho)\mathbf g-6\pi\mu a\mathbf U=0.
$$
Hence
$$
\boxed{\mathbf U=\frac{2a^2(\widetilde\rho-\rho)}{9\mu}\mathbf g,\qquad
|\mathbf U|=\frac{2a^2|\widetilde\rho-\rho|}{9\mu}|\mathbf g|.}
$$
A denser sphere falls in the direction of $\mathbf g$; a lighter one rises. Equal densities give no sedimentation. The calculation assumes negligible fluid <Reynolds number>, an unbounded fluid, and negligible transient body <acceleration> after terminal settling. The fluid <mass density> enters through <buoyancy> and through the small-<Reynolds number> condition.
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