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If something like a spaceship or a moon was falling towards a black hole and happened to cross the event horizon what would that look like to outside observers? On one hand, photons can not escape at that point, so the object should no longer be visible. On the other hand, the object's time will seem to slow down from the point of view of outside observers.

So, will the people watching it see the object pop out of existence at some point, or will it look it just hangs there forever or will some other effects occur? And if it does disappear from view how long will it take (or what would be the formula, since actual time will probably depend on black hole size and distance from the observers)?

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    $\begingroup$ Any number of higher-quality educational materials or lectures pertaining to black holes have discussed this. My advice is that you look at some of them. $\endgroup$ – can-ned_food Jul 25 '17 at 23:34
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Hard science: from outsiders perspective, an object would not appear to be crossing the event horizon. It would appear to be moving slower and slower the closer it gets to the event horizon, and ultimately fully stop exactly at the event horizon, since for outside observer, time appears to be stopped at the event horizon.

Secondly, as the object seem to move slower and slower, it'll become more and more red, as the frequency of any radiation, including light, becomes more and more redshifted - eventually all frequencies becoming 0 Hz at event horizon.

Lastly, the intensity of any radiation becomes fainter and fainter, since from free-falling object's perspective, it still emits x photons per second, but for observer those seconds are becoming longer and longer, thus less photons per observers' second

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The color would start shifting (blue shift I believe) and eventually darkening and disappearing. At around the crossing, the object in question would cease moving and appear to be still (it will well be past the cross, however, the time dilation would make it appear like it has yet to do so. This depends on how far out it is. In all likely hood, if you are close enough to see something like a spaceship or person fall into the blackhole, you're probably too close to stop yourself from doing so as well.

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    $\begingroup$ It would actually start redshifting. $\endgroup$ – AngelPray Jul 25 '17 at 18:43
  • $\begingroup$ Thanks... knew it was one, but couldn't remember. Anywho, here's a good go to for Black Holes for Dummy Writers... don't click on other links... it's like Wiki walking with drugs if you do. tvtropes.org/pmwiki/pmwiki.php/UsefulNotes/BlackHoles $\endgroup$ – hszmv Jul 25 '17 at 18:45
  • $\begingroup$ If you believe information in an answer of yours is incorrect, it is probably good manners to change it … :-) $\endgroup$ – can-ned_food Jul 30 '17 at 15:08
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I like more the Schwarzschild radius/sphere terms. "Event horizon" term sounds senseless to me. Why horizon? What event?

You will never see anything crossing S.sphere from outside. Any object that is crossing the S.sphere from its own point of view will look for you as object, closing to the sphere, more and more slowly. From the outside, it will reach the sphere at the moment +infinitum.

Simultaneously the object will look more and more short - in the direction to the center of the black hole. Also it will radiate energy - in more and more short waves from its point of view, but in more and more long waves - for the outside observer.

If the object returns before reaching the S.sphere, it will do it in future - the more distant future if the object was closer to the sphere.

Yet another effect is that if you'll look from the side, you will see the object turning its back to you - does not depend, from what direction you are looking. When the object is near the sphere, every outside observer will see only the back end of the object.

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