= Solution
Far from its ends, approximate the body by an infinite uniform <cylinder>. By cylindrical <symmetry>, the <Newtonian gravitational field> is radial. Applying <Gauss's law for gravity> to a coaxial cylindrical surface of radius $s$ and length $\ell$ outside the material gives
$$
g(s)\,2\pi s\ell=4\pi G\rho\pi R^2\ell,\qquad g(s)=\frac{2\pi G\rho R^2}{s}.
$$
At the surface, $R=50\ \mathrm m$, so
$$
\boxed{g(R)\simeq2\pi G\rho R\simeq4.19\times10^{-5}\ \mathrm{m\,s^{-2}}.}
$$
The <gravitational acceleration> points inward toward the axis. Its small magnitude reflects both the small size and the modest <density>. The infinite-<cylinder> approximation uses the specified distance from both ends; a spherical approximation would discard the relevant geometry. No rotational contribution is included, since no rotation rate is supplied.
Back to article page
