= Solution
Choose the positive $z$ direction along the sphere's laboratory <velocity> $v\hat{\mathbf z}$. In its rest frame the far-field fluid velocity is $-v\hat{\mathbf z}$. Therefore the far-field potential is $-vr\cos\theta$. The impermeable sphere imposes $\partial_r\phi=0$ at $r=R$. The monopole coefficient vanishes because the sphere is not a source or sink, and the dipole coefficient follows from the wall condition:
$$
A_1=-v,\qquad B_0=0,\qquad B_1=-\frac{vR^3}{2}.
$$
Thus
$$
\boxed{\phi=-v\left(r+\frac{R^3}{2r^2}\right)\cos\theta+C,\qquad r\ge R.}
$$
The regular exterior <boundary value problem> has no forced higher multipoles; uniqueness fixes their coefficients to zero. Differentiating gives
$$
\boxed{u_r=-v\left(1-\frac{R^3}{r^3}\right)\cos\theta,\quad
u_\theta=v\left(1+\frac{R^3}{2r^3}\right)\sin\theta,\quad u_\varphi=0.}
$$
Here $u_\theta$ is the polar <velocity> component. An equivalent Cartesian form is
$$
\mathbf u=-v\hat{\mathbf z}+\frac{vR^3}{2r^3}\left[3\cos\theta\,\hat{\mathbf r}-\hat{\mathbf z}\right].
$$
Only no-penetration, not no-slip, is appropriate to this ideal inviscid solution. The sphere density $\rho_0$ does not enter the <velocity> field for an externally prescribed constant speed.<Fluid Mechanics by David Tong>[Section 2.3.2, Flow Around a Sphere, potential and impermeable-surface condition; the present far-field velocity has the opposite sign]{ref}
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