What is the closest one could get to a black hole and still stand a realistic chance of escape?

Question

This question is inspired by the one asked here:

Is there any stable orbit around a black hole?

and I note that (though I've heard of these before) there are, apparently, 3 important fundamental "zones of inescapability" with regard to a black hole that can be identified on the basis of physical principles and which can be defined by the Schwarzschild $r$ coordinate and its critical radius $r_S$: one of these is, of course, the event horizon, at $1\ r_S$, from which no escape at all is possible. The second one is the "photon sphere", here referred to as the zone at the "innermost bound circular orbit", at $1.5\ r_S$, at which a photon (particle of light) traveling at right angles (i.e. constant $r$-coordinate) can complete one circuit around the hole and within which only impossibly powerful rockets could get you out. Orbits above this point but below the "innermost stable circular orbit", out to $3\ r_S$, are possible but require a continuous application of (likely ungodly large) engine propulsion. Above this, more-or-less normal orbits are possible, though they precess more and more rapidly as the general-relativistic effect that is incipient with the orbits of the planets becomes fully mature in these environs.

Now, the trick, of course, is the qualifiers regarding propulsion: "impossibly powerful", "ungodly large" which basically are likely words for "it can't be done with realistic technology". Which, then, of course, is what begs this question - what can be done with such?

For my "realistic" scenario, a generous interpretation will be used which is "a matter-antimatter fueled spacecraft with a wet mass 100x its dry mass". The goal is for the craft to execute one of two manoeuvres:

1. lower itself into an orbit around the hole, orbit a few times, and then take off again (so it must have fuel for both trips) and return to where it launched from, and
2. establish a "slingshot" trajectory around the hole that takes it as close as possible before shooting it back out into space (this may be easier?).

I consider this "reasonable" because any travel to a black hole will already be an extreme problem in interstellar travel, much less the kinds of targets that will already be in use for this kind of proposal which will almost surely have to be supermassive black holes to be interesting and thus requiring travel at least to the center of the Galaxy (~200 000 Pm) if not intergalactically (distances of many Zm), with tera- or petaseconds of travel time (thousands to millions of years) at least, or something like wormholes if they are possible. Hence I presume the technology to make such a craft will be readily available to any civilization that is capable of pulling this feat off to begin with.

The reason we need supermassive black holes is tidal forces and ideally we want to get as close to the horizon as possible. Possible targets may be Sagittarius A* and M87* (the black hole that the picture was taken of), both supermassive black holes, the latter much more so still, at the centers of galaxies. However, M87* presents the additional challenge of an extremely active matter environment around it, which may foreclose that possibility even further, though even Sgr A* might be too much. Hence, failing that, we can just consider a hypothetical black hole of mass equal to either, that is suitably "quiet", if nothing else is possible. (or perhaps imagine that with M87*, it has entered a quiescent period by the time the ships arrive as compared to the time when it sent the images we saw)

What would be the closest approach possible, in both scenarios, while still having enough fuel to get away (or not being destroyed by other effects)? We should also assume that the craft starts out at great distance where gravity is hopefully unremarkable, but do not budget for more than 250 Ms shipboard total mission time (a bit less than 2893 days of 86.4 ks each), measured from time of release from mission departure station to time of return.

Answer 1

The simplest case is for a rocket that does not fire its engine when it approaches the black hole from afar, and hence follows a geodesic trajectory. In this case the closest you can get is $3r_s$, but you can make an arbitrary number of turns around the black hole.

The reason for this is that the counterpart of a hyperbolic orbit corresponds to moving in the effective potential $V(r)=-GMm/r+L^2/2\mu r^2 - G(M+m)L^2/c^2\mu r^3$ (this is the Wikipedia version of the formula; different authors use different measures). For a given energy and angular momentum this means that either you can come in from infinity and fall into the black hole (for low energy and angular momentum) or the potential has a maximum greater than your energy and you will turn back. The closest you can get the maximum to go to the hole is $r=3r_s$ (as your angular momentum increases).

However, the angular velocity behaves as $d\theta / d\tau = (L/\mu)/r^2$: you will swirl around a lot as you get close to the black hole. Since a carefully selected trajectory will have energy and angular momentum so that you turn around almost exactly at the potential maximum where the derivative $V'(r)\approx 0$ the proper time spent there will be long and you can get as many turns as you want. The orbital frequency as seen from the outside is 2200 Hz $(M_\odot/M)$ (!) (source).

Note that all of this is freefall: the only thing to worry about is tidal forces, but that is not much of an issue near a SMBH. The velocity needed is of course in principle zero at infinity if you come in along a parabolic orbit.

The second question is if one can "slingshot". I assume this refers to an Oberth manoeuvre rather than a gravitational slingshot, which is unpowered. Gravitational slingshots from black holes can give you any change in angle, but the velocity change is just like the classical case. Rotation does not affect this, except by making the angular shift weirder. Note that for prograde orbits around an extremal spinning black hole the ISCO approaches the horizon radius, so the above unpowered flybys can in principle get right up to the horizon.

For the powered manoeuvre I don't have a full answer; I would actually like to see it answered (see this question and my half-answer to it). The basic story looks like that for velocity-at-infinity increasing manoeuvres you should not get closer than $3r_s$. But for a sightseeing trip close to the horizon spending some of your fuel (and mass) ought to work.

What you need to do is to fire the engines powerfully enough to change the sign of your radial velocity and retain more total kinetic energy than the potential maximum. The potential maximum is due to angular momentum.

For the freefall radial trajectory with no angular momentum you will have $(dr/d\tau)^2=(E/mc^2)^2-c^2(1-r_s/r)$, so if you start with zero velocity at $r=\infty$ (hence $E=mc^2$) the inward radial velocity will be $dr/d\tau = c\sqrt{r_s/r}$. So you need $\Delta v =2c\sqrt{r_s/r}$ to escape again this way. We can rearrange the formula into $$r_{escape}= \frac{r_s}{\left(\Delta v / 2c\right)^2}$$

In your case $m_0/m_1=100$ and I assume $I_{sp}\approx 0.6c$, $\Delta v = c \tanh \left( \frac{I_{sp}}{c}\log\left(\frac{m_0}{m_1}\right)\right) = 0.9921c$, and $r_{escape}=4.0642 r_s $. That demonstrates that the purely radial approach is not a good choice; adding a bit of lateral thrust would allow a closer approach.

I don't have the full answer for the lateral case. If you come in along a flyby orbit that approaches the ISCO you can slow down a bit so you now are on an inspiralling trajectory. To escape you need to pay roughly twice the potential difference between your final radial distance orbit and the maxmimum of the new potential.

Again, rotating black holes allow you to use the Penrose process to gain a lot of angular momentum by dumping reaction mass in the right direction. However, the efficiency is proportional to the total momentum of the reaction mass and is rather negligible for non-relativistic rockets. Hence rockets not powered by antimatter will likely not benefit much from this possibility.

Answer 2

2. We can imagine that the Sun was a black hole with the same mass it has now. Then if you accelerate a rocket 30/km per second from the Earth in the direction opposite to the direction of the Earth around the Sun, then that rocket would have no angular momenta with respect to the Sun and fall right into it. This is because the Earth travels at 30 km/s in a rouhly spherical orbit. In order to establish a really close encounter, and ignoring tidal effects, you thus only have to accelerate the velocity of the rocket with less than 30 km/s.

Since the solar system travels Sagittarius A with a velocity of roughly 230 km/s you would have to accelerate the rocket with less then 230 km/s to get a really close encounter with Sagitarrius A. If this is possible, you have to ask a rocket engineer.

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1. In general relativty, the energy of a body in circular motion around a spherically symmetric gravitational field can be written as (I belive this is correct):

$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{3GM}{rc^2}}}\right)$$.

setting $E=mc^2$ at rest at infinity. In case of Sagitarrius A you could use this expression for, depending on how good your rocket engines are, how close to Sagittarius A you can accelerate/brake your self into a circular orbit and then get back up again.