= Solution
For <contact loss on a smooth sphere>, let $\theta$ be the angle from the upward vertical. In the limit of an infinitesimal initial displacement, <conservation of energy> gives $v^2=2gR(1-\cos\theta)$. The inward radial equation is
$$
mg\cos\theta-N=mv^2/R.
$$
Contact is lost at $N=0$, so $\cos\theta=2(1-\cos\theta)$ and $\theta=\arccos(2/3)$. The distance travelled \b[along the sphere] is therefore
$$
\boxed{s=R\arccos(2/3)=0.841\,\mathrm m.}
$$
The vertical drop is instead $R/3$; it is not the requested arc length. Exact rest at the unstable top would persist in the ideal mathematical solution, so the infinitesimal-displacement limit in the question is essential.
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