If we sent someone on a path that passed as near as possible to a black hole without getting pulled in, how far into the future would they go due to time dilation? Let's assume the black hole is 5 solar masses (I am assuming the mass will affect the calculation). I would like to know from two considerations:

  • An observer that can survive anything (e.g. ideal case)

  • The forces that a human can tolerate (e.g. a human would not be able to get as close to a black hole as would be ideal)

  • 1
    $\begingroup$ You can come arbitrarily far into the the future. How far depends on the mass $M$ of the BH, the distance $r$ to the BH, and how long time $t$ you spend there. If you only give us $M$, you don't even need to go near a black hole; you can come $x$ years into the future by waiting $x$ years. $\endgroup$
    – pela
    Mar 12, 2016 at 19:18
  • $\begingroup$ Good point, although I am talking about the amount of time advanced due to time dilation :-) $\endgroup$
    – Jonathan
    Mar 12, 2016 at 20:50
  • 3
    $\begingroup$ For a 5 solar mass black hole there would be enormous tidal issues that would tear a space ship and/or person in the ship apart long before the time dilation got anywhere close to interesting. For a larger black hole, the question is asked and answered here: physics.stackexchange.com/questions/146105/… $\endgroup$
    – userLTK
    Mar 12, 2016 at 21:39
  • $\begingroup$ Jonathan, you can find the appropriate equation here. If you plug in your favorite numbers and find a time dilation factor of, say, $t=0.5$, this means that for every hour you spend at the BH, two hours will pass for an external observer, i.e. when you get back to civilization, you will experience it as having traveled one hour into the future. $\endgroup$
    – pela
    Mar 14, 2016 at 8:09
  • 1
    $\begingroup$ @Jonathan - Do you want a semi-practical answer that takes into account that your whole body can't get arbitrarily close to the event horizon since it has a finite thickness, along with the fact that the G-force you'd feel hovering above the horizon (your proper acceleration) would get too large to survive if the distance got small enough? Or just an ideal answer for a pointlike observer who can withstand any G-force? $\endgroup$
    – Hypnosifl
    Apr 23, 2016 at 0:50

2 Answers 2


The way that you have specified the question, the answer is as far as you like. You simply put your spaceship into any orbit around the black hole and wait.

A more sensible question is what is the largest time dilation factor that can be accomplished - i.e. that maximises your travel time into the future for a given amount of proper time experienced on the spaceship.

This in turn is governed by how close to the black hole you can come and still tolerate the tidal forces. If you don't put a limit on this (your first case), then the answer is again infinite; you can hover as close to the event horizon as you like, using an enormous amount of rocket fuel, and the time dilation (see below) can be arbitrarily large.

Your second case is more realistic. Roughly we can say that the tidal acceleration across a body of length $l$ is given by $2GMl/r^3$, where $M$ is the black hole mass and $r$ is the distance from the black hole. If we make this acceleration equal to say $1 g$, and your body length $l \sim 1$m, then for a $5M_{\odot}$ black hole $r \simeq 5000$ km (well outside the Schwarzschild radius of 15 km).

If you could "hover" at this radius, then the time dilation factor would be $$\frac{\tau}{\tau_0} = \left( 1 - \frac{2GM}{rc^2}\right)^{1/2},$$ where $\tau$ is the time interval on a clock on the spaceship and $\tau_0$ is the time interval well away from the black hole.

For $M=5M_{\odot}$ and $r = 5000$ km, this factor is 0.9985.

If the spaceship is in a circular orbit at this radius, the factor is $(1 - 3GM/rc^2)^{1/2} = 0.9978$.

If you ignore tidal forces ripping you and your ship apart then the smallest stable orbit you can accomplish is at $r=6GM/c^2$ - the so-called innermost stable circular orbit. Using the formula for the circular orbit above, then the time dilation factor becomes 0.816.

These factors are perhaps not as big as you might have imagined! If you want to improve on that then you must consider rapidly rotating Kerr black holes. The innermost prograde stable circular orbit (i.e. in the same direction as the black hole spin) can be much closer - approaching $r= GM/c^2$ and the time dilation factor calculated above can become arbitrarily small.

Of course the tidal forces are still there, so the way to get around this is to be in orbit around a much more massive black hole $>10^6$ solar masses, where it turns out the tidal forces at these radii might be tolerable to a human.

  • $\begingroup$ Even for the "realistic" case, I think it makes sense to consider a much larger black hole--after all, there seem to be plenty of supermassive black holes at the centers of many galaxies. Sagittarius A at the center of our own galaxy is about 4.3 million times the mass of the Sun, while the largest ones known are thought to have masses in excess of 10 billion times the mass of the Sun. And the tidal forces outside the horizon are negligible for very large black holes. $\endgroup$
    – Hypnosifl
    Apr 24, 2016 at 4:32
  • $\begingroup$ Also, this answer calculates an approximate answer for the time dilation at the innermost stable circular orbit (ISCO) for a rotating black hole...I noted in a comment that there are theoretical reasons to think the maximum value for the rotation rate $a = 1 - \epsilon$ (expressed as a fraction of the 'extremal' rotation rate where the black hole loses its event horizon and becomes a 'naked singularity') is about $a = 0.998$, or $\epsilon = 0.002$, giving a time dilation factor of 1 second experienced for every 10 seconds for distant clocks. $\endgroup$
    – Hypnosifl
    Apr 24, 2016 at 4:38
  • $\begingroup$ @Hypnosifl a 5 solar mass black hole is specified in the question. The tidal forces are essentially the same for even a maximally rotating black hole. $\endgroup$
    – ProfRob
    May 29, 2022 at 10:56

As far as we know, once beyond the event horizon, someone might experience all of time in the blink of an eye until the hawking radiation makes the black hole disappear and the person comes out again. More likely though you'd just get spaghettified..

In all seriousness though, there is no real limitation, although they aren't really transported to the future, the outside world just moves faster.

  • $\begingroup$ Does this mean time is essentially slowed down dramatically giving the illusion that time has moved faster in the outside world? $\endgroup$
    – NuWin
    Apr 21, 2016 at 23:36
  • $\begingroup$ Basically, yes, time does slow down $\endgroup$
    – Tanenthor
    Apr 22, 2016 at 1:32
  • $\begingroup$ And to whomever down voted my answer, could you please explain why? So I can correct the issue $\endgroup$
    – Tanenthor
    Apr 22, 2016 at 1:35
  • 1
    $\begingroup$ This isn't an answer to the question; the question specifically mentions going close to the event horizon without going past it, while the answer is about going beyond the event horizon. $\endgroup$
    – DylanSp
    Apr 22, 2016 at 12:31
  • 1
    $\begingroup$ @Rob Jeffries - In an eternal rotating black hole, you would see the entire history of the universe compressed to a finite time as you crossed the inner horizon, though as you say this doesn't happen at the outer event horizon (see the last paragraph on p. 153 of this book). In a more realistic rotating black hole that evaporates, presumably you only see the entire finite future history of the black hole, though I believe there is still an infinite blueshift of the signal so the speedup factor should still go to infinity. $\endgroup$
    – Hypnosifl
    Apr 23, 2016 at 0:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .