A sphere of radius a moves with constant velocity $U$ in a fluid otherwise at rest. How far ahead of the sphere is there a disturbance of magnitude $1/20.U$? Show that the acceleration of a fluid particle at distance x ahead of the centre of the sphere is $$3U^2(\frac{a^3}{x^4}-\frac{a^6}{x^7})$$
I think I've got that the potential function is $$-U \frac{a^3}{2r^2}\cos{\theta}$$ but i'm not sure how to go further to answer the question.
Not sure that I agree with your potential function. $v_r$ must go to zero at $r=a$, which it does not with your potential function. Though I have to admit that someone does agree with you. I prefer the function you can find here, since it gives a correct $v_r$ at $r=a$:
$\Phi=U(r+\frac{a^3}{2r^2})\cos\theta$
Anyway, once you decide on your potential, $\Phi$, the first question involves just taking $v_r=\partial\Phi/\partial r$, setting $\theta=0$, and solving for where $v_r$ equals your given number.
For the second part, acceleration = $dv/dt = d (\partial\Phi/\partial r)/dt=dr/dt\cdot\partial^2\Phi/\partial r^2=$ $v\cdot\partial^2\Phi/\partial r^2=\partial\Phi/\partial r\cdot\partial^2\Phi/\partial r^2=1/2\cdot \partial_r((\partial_r\Phi)^2)$
