Escape velocity at an angle near a black hole What's the formula for the velocity required to escape at an angle (not normal) near a black hole?
I know for example that the required velocity not to impact the black hole at $90$ degrees is larger than the escape velocity for something going directly away once $R>2×Rs$. This is due to the fact that below 2Rs, the orbital velocity is greater the the escape velocity.
I know that in these case the $v_\infty$ will be quite large.
 A: It’s the same as every other mass and every other angle: $v_e = c \sqrt{r_s/r}$. I think you have become confused by inferring Euclidean consequences to geometries in non Euclidean curved space.
In short: it’s not escape velocity that varies with angle, it’s the length of the escape path, and whether or not the escape path is also a collision course.
$1.5 r_s $ is the minimum distance at which the black hole takes up exactly half of the sky, and therefore a path exactly tangent to the surface can describe a circular orbit, rather than a collision course. Because the circular  orbital velocity is close to c at that distance, the ellipse-describing spirals of collision courses and escape paths might be many orbits long before they finally collide or leave orbit forever. The closer the ellipse is to a perfect circle, the longer it orbits. This collection of nearly circular spirals near $1.5 r_s$ is the photon sphere.
At all distances outside the event horizon, light emitted at any angle pointed away from the black hole escapes (possibly after countless orbits, each getting slightly higher). At all distances, light pointed towards the black hole ends up at the center (possibly after countless orbits, each getting slightly lower). In the limit as distance from the horizon goes to infinity, this asymptotically approaches Euclidean geometry with the black hole taking up the solid angle defined by a cone with a base of radius $r_s$ and a height $r$. On the other hand, in the limit as the distance from the horizon goes to zero, the black hole occludes the entire sky.
