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Currently, the closest known black hole to Earth is between 1600-24000 light years away.

Even if we do send a drone travelling at the speed of light, it would reach the black hole long after we are all dead.

However, let's assume that NASA has some spare cash and decides to go forward with such an endeavour and send a miniature spaceship straight into the black hole's direction, so that future generations can study it more closely.

What will the natural processes be as the drone approaches the black hole or enters it (is this even possible)? Will it just be smashed like a pancake due to the high gravity as it approaches?

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closed as off-topic by horsh, gerrit, PearsonArtPhoto Sep 26 '13 at 16:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is about other space sciences (physics, weather, astronomy, etc), and does not directly pertain to space exploration as outlined in the help center." – horsh, gerrit, PearsonArtPhoto
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I don't have a full answer, relativistic effects are not my specialty. But I can say there's no pancaking - the opposite actually occurs: Spaghettification. (Yep, it's a real word: en.wikipedia.org/wiki/Spaghettification) $\endgroup$ – john3103 Sep 25 '13 at 15:22
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If by "enters it" you mean crossing the event horizon, then the easy answer is: we'll never know. No signal can reach us from the probe since photons can't escape. By definition the "material," if there is such a thing, at the center of a black hole is inside the event horizon. So the last signal you will get from the probe will be just before it crosses the event horizon. The time dilation noted in another answer here results in the signal being Doppler shifted to lower and lower frequencies, with the encoded data going to lower and lower data rates, and the intensity going down rapidly until you receive the last photon from the probe as it dims to nothing.

For supermassive black holes, spaghettification will not occur until after crossing the event horizon, so you can survive an approach to the event horizon.

Interestingly, what happens when you cross the event horizon is a subject of great debate. When considering the quantum mechanics and quantum information transfer at the event horizon of a black hole, it all becomes very complicated. You might make it across uneventfully (pun intended), or you might be completely destroyed right at the event horizon. See this excellent blog post for a discussion of the current debates.

The event horizon is defined as the distance from the center of the black hole at which the gravitational escape velocity equals the velocity of light. Therefore, given that the speed of light is a universal speed limit, nothing inside of the event horizon can ever escape. That distance for a simple, non-rotating black hole is called the Schwarzschild radius, equal to $2\mu\over c^2$. $\mu$ is the standard gravitational parameter for the mass of the black hole $M$, $\mu=GM$, and $c$ is the speed of light. If the Earth were somehow compressed into a black hole, its Schwarzschild radius would be about 1 cm.

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    $\begingroup$ Important to note that blackholes tend to spew some pretty serious radiation out of them, and our probe would have a difficult time surviving that. Not to mention other debris being sucked in being a collision hazard. $\endgroup$ – Matt Joyce Sep 26 '13 at 13:57
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Viewing far from the blackhole: As the probe falls in it will move slower and slower through time. It will appear redder, colder, and dimmer. As it approaches the event horizon their movement through time will halt, as they fade completely from view. Technically, you’ll never actually see someone fall into a blackhole, you’ll just see them get really close.

The Probe: First, torn apart and crushed, but let's skip that detail. Things farther from the blackhole move through time faster, so the rest of the universe will speed up from your probe's point of view. As a result the rest of the universe becomes bluer, hotter, and brighter. The blue shift of the incoming light turns it into gamma rays. So, right before the probe passes through the event horizon, it’ll get radiated with a universe’s lifetime worth of starlight and microwave background radiation. Whereas the event horizon itself is only special from an outside perspective. If you fall in you should pass right through it. However, what you see in the moment that you pass through the horizon is dependent on things we don’t know yet.

The video: very slow images until they stop and the data fades away.

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  • $\begingroup$ Mostly correct, except for the part about a universe's lifetime of radiation. That is a common misconception. See this explanation. $\endgroup$ – Mark Adler Sep 26 '13 at 5:24
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well a black hole has so much gravity it doesn't even allow light to escape. So first of all the camera wouldn't see anything. Second of all the camera, once inside it can't send any information back because even the electromagnetic wave it sends cannot escape. Third of all ,as soon as the camera enters the black hole it would be stripped apart. Such is the gravity of a black hole

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The probe would be pulled apart from the tidal forces of the black hole as it approached it. As explained in this Minute Physics video (specifically time code 1:39, but watch the whole video to understand tidal forces).

http://www.youtube.com/watch?v=gftT3wHJGtg

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Black hole has enormous density and gravity. So let's imagine we have a drone that can escape any gravitational forces and reach any place at any speed. As we reach the black hole we could see materials that compressed to the black hole, and that materials would become hot. We can see them, because as materials become hot enough they start emit light. But even light can not escape the gravity of black hole, so we will not see anything inside. But if you have a microscope on that drone may be you will see that the atomic structure is compressed, electrons are not turning around the atom nucleus but glued by gravitational force to nucleus surface.

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    $\begingroup$ This answer has some speculation beyond current accepted theory, and is probably the reason for the down votes. $\endgroup$ – James Jenkins Sep 26 '13 at 10:41

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