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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.

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    $\begingroup$ Do you mean rotating black hole? $\endgroup$ Dec 3 '21 at 2:02
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    $\begingroup$ No, just a Schwarzschild black hole $\endgroup$ Dec 3 '21 at 2:08
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    $\begingroup$ Where are you getting escape velocity varying with angle from? $\endgroup$
    – g s
    Dec 3 '21 at 3:34
  • $\begingroup$ Related (unanswered) question by the OP: physics.stackexchange.com/q/679556 $\endgroup$
    – PM 2Ring
    Dec 3 '21 at 7:07
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    $\begingroup$ @g-s At 1.5Rs, you have the photon sphere, here the orbital velocity is equal to the speed of light, but the escape velocity is lower. Clearly the velocity required to not enter the blackhole is dependent on angle when you are close enough. $\endgroup$ Dec 3 '21 at 7:47
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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.

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  • $\begingroup$ "A laser pointed out tangent to the surface from a low enough height (I get about 1cm) would hit the surface, even though orbital speed at earth radius is much less than c." This is definitely wrong. The rest of the answer isn't very helpful. $\endgroup$ Dec 4 '21 at 15:20
  • $\begingroup$ You're right, I checked my back-of-envelope estimate and realized that I accidentally made the Earth flat. I'll delete the offending section. I don't blame you for finding the rest of the answer suspect after a mistake that big. $\endgroup$
    – g s
    Dec 4 '21 at 17:10
  • $\begingroup$ 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 𝑟𝑠 r s and a height 𝑟 r The apparent size of a black hole at a distance si the size of the "black hole shadow" with had a raidus of approximately 2.6Rs. $\endgroup$ Dec 4 '21 at 17:13

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