Half the Planet - Half as Hard to Escape?

Suppose the sphere's vertical axis is in the $z$ direction, $r$ is the horizontal perpendicular distance from the axis, $R$ is the sphere radius, and $M$ is the total sphere mass. The maximum $r$ for a particular $z$ is:

$\large{r_{max} = \sqrt{R^2 - z^2}}$

Suppose there is a small mass $m$ located at $(x,y,z) = (0,0,R)$ . The infinitesimal mass of a ring of radius $r$ and height $dz$ is:

$\large{dM = M \frac{2 \pi \, r \, dr \, dz}{\frac{4}{3} \pi R^3}}$

The distance of the ring from the small mass is:

$\large{D = \sqrt{r^2 + (R-z)^2}}$

The infinitesimal gravitational potential energy contribution is:

$\large{dU = \frac{G m \, dM}{D}}$

The total gravitational potential energy is (for the entire sphere):

$\large{U_{total} = \int_{z = -R}^{z = R} \int_{r = 0}^{r = r_{max}} dU}$

The gravitational potential energy for the half sphere $(z >= 0)$ is:

$\large{U_{half} = \int_{z = 0}^{z = R} \int_{r = 0}^{r = r_{max}} dU}$

The escape velocity for either configuration is:

$\large{U = \frac{1}{2} m \, v_e^2 \\ v_e = \sqrt{\frac{2 U}{m}}}$

When integrating over the entire sphere, this yields the following familiar answer:

$\large{v_e = \sqrt{\frac{2 G M}{R}}}$

Integrating over the half sphere $(z >= 0)$ results in an escape velocity equal to the original escape velocity multiplied by $\approx 0.815$ . Note that integrating over half the volume effectively results in half the aggregate planetary mass. But because it is the half of the planet closer to the object at the surface, it's gravity is more influential, and the escape velocity remains considerably more than half that of the original planet.