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If I understand correctly, the radius of the M87 event horizon is approximately 19 billion km. I used an escape velocity calculator as a sanity check, entering 1.28E40 kg and 1.9E10 km and the calculator yielded an escape velocity of 299,879 km/sec.

I used the same tool to calculate the escape velocity for the ISCO, which is at 3R, which is a distance of 5.7E10 km from the center. The escape velocity from ISCO is 173,135 km/sec. which means it can send us pictures of what it sees, and we can receive them if we adjust for the expected redshift.

Suppose we drop an internally powered spherical lightbulb into the black hole. As it reaches a radius of 1.9E10 km, the gravitational gradient between our satellite at ISCO and the lightbulb will only be (approximately) c-173,135km/sec = approx 127,000 km/sec. Our lightbulb will only disappear from view when the relative difference in the PE of our positions in the gravitational field creates an escape velocity equal to C. And that happens only when the lightbulb is far inside the event horizon as it appears to distant observers. For example, at a radius of 1.2E10 km, the escape velocity is 377,000 km/sec, a meaningless number to a distant observer. But to our satellite at ISCO, the escape velocity difference is only 377,000 km/sec - 173,000 km/sec = 204,000 km/sec. This means we would notice some redshift and some time dilation but it seems as if the lightbulb would be quite visible. Now - what if our lightbulb was a mini sat that was equipped with lights, cameras, and transceivers. And what if we dropped a sequence of them so they could take pictures of what is below as they fell and transmit those images to the mini sat directly above them. As long as the difference in the relative escape velocities remained well below c, redshift and time dilation will be modest, etc. Given that coil d we not daisy chain our way down to the point where spaghettification began destroying our equipment?

Deschele Schilder
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Mark Moessinger
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    It's not valid to just subtract escape velocities this way. There's no timelike or null geodesic connecting the black hole interior to any point on the black hole ISCO (or any exterior point, for that matter), which means that information from the inside cannot reach any point on the outside, no matter what scheme you use. – Zo the Relativist Oct 01 '20 at 20:34
  • C - 173,135KM/sec = approx 127KM/sec Aren't you off by three orders of magnitude? – G. Smith Oct 01 '20 at 20:45
  • @G.Smith I already edited. – Deschele Schilder Oct 01 '20 at 20:55
  • the gravitational gradient between our satellite at ISCO and the lightbulb will only be (approximately) c-173,135km/sec = approx 127,000 km/sec What do you mean by this? Isn't this the mean gradient of the escape velocity? – Deschele Schilder Oct 01 '20 at 20:57
  • @Mark Moessinger Good question. I posed a similar question: https://physics.stackexchange.com/questions/575545/does-the-apparent-radius-of-the-event-horizon-of-a-black-hole-depend-on-the-dist I ended up answering it myself - probably incorrectly. I'm finding it really difficult to get any intuitive grasp of these concepts. – Roger Wood Oct 01 '20 at 23:13

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No signal from any spacetime point inside the black hole can reach any spacetime point outside it. You can't evade that fact by relaying signals between devices. That's just another form of signaling.

There are situations in which speeds of the form $c-v$ can be meaningful (there are some examples in Einstein's original 1905 paper), but this isn't one of them.

The fact that when you plug a black hole mass and radius into a Newtonian escape-velocity calculator you get $c$ and not $c/2$ or $2c$ or something is as far as I know just coincidence. Escape velocity is only meaningful for objects moving only under the force of gravity; an object propelled in some other way (e.g. by a rocket or ground-based laser or space elevator) can escape a planet at a maximum speed arbitrarily lower than the escape velocity at the surface. This is not the case for black holes which can't be escaped by any means at any speed.

benrg
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  • From reading a wiki article, I believe it is fair to say that the event horizon is defined by the radius at which the escape velocity equals the speed of light. For simplicity I would like to assume a non rotating black hole. And according to wiki I absolutely can compare the gravitational gradient between two points using the velocity it would take at X(1), to reach X(2). This is opposite in sign to the velocity I would get dropping a ball at X(2) and recording its instantaneous velocity as it falls through X(1). – Mark Moessinger Oct 02 '20 at 16:57