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An automated Spamazon delivery starship suffers a catastrophic system failure and finds itself hurtling towards a black hole on a direct collision course. In particularly dramatic fashion, it manages to reignite its STL (0.99c) drives just in time to slowly escape from the black hole's gravitational influence. Any closer and it would be 'treading water' at full thrust, but it is just far enough to slowly increase the distance between it and the event horizon.

The possibility of dropping below a 4 star rating as a result of this delivery has terrified Corporate and they want to know -

How long is this going to take?

Should we wait for the ship to escape, or simply let it sink into the event horizon and just write it off with the insurance company?

(The when is the focus of the question and not the integrity of the starship. It is being assumed here that it will eventually escape intact in most cases. I'm also assuming that time dilation from the singularity will be playing a part here as well.)

EDIT:

Since some want hard numbers:

  • The black hole is 5 solar masses
  • The ship starts at 100% thrust at 0.99c, moving +1 m/s away relative to the singularity. (I'm not sure what a realistic ship mass looks like so that is up to interpretation)

FTL is only possible once out of the influence of the black hole's gravity well and to make it a hard cutoff, only after you are moving 0.99c relative to the black hole.

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    $\begingroup$ Simple. Which option costs less? Your story so you make up the numbers and circumstances to get the outcome you want. If you make a claim and the insurance company pays out, they own the ship when it finally does escape. $\endgroup$
    – DKNguyen
    Dec 30, 2020 at 20:54
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    $\begingroup$ I am curious to learn about the thought processes of somebody who asks for a numeric result without giving any numeric input data. (1) Mass of the black hole? (2) Mass of the ship? (3) Distance between the ship and the black hole when they start their engine? (4) Thrust of the engine? (Plus, the only number given in the question is not meaningful. A maximum speed tells nothing about acceleration. A Trabant has a top speed of 100 km/h; it takes 21 seconds to get to that speed. Downslope and pushed by the wind, of course.) $\endgroup$
    – AlexP
    Dec 30, 2020 at 20:58
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    $\begingroup$ So that's why my ice skates have been "out for delivery" for 250 years. $\endgroup$
    – user535733
    Dec 30, 2020 at 21:00
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    $\begingroup$ star ships don't have maximum speeds, only accelleration. $\endgroup$
    – ths
    Dec 30, 2020 at 21:05
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    $\begingroup$ You may be interested in en.wikipedia.org/wiki/Spaghettification. In particular "For example, for a black hole of 10 Sun masses, the above-mentioned [1m, 1kg, 10kN tensile strength rod] rod breaks at a distance of 320 km, well outside the Schwarzschild radius of 30 km. For a supermassive black hole of 10,000 Sun masses, it will break at a distance of 3200 km, well inside the Schwarzschild radius of 30,000 km." If you run through the numbers, at 320km from said black hole, that is a mere 95% of normal time. $\endgroup$
    – DKNguyen
    Dec 30, 2020 at 21:49

2 Answers 2

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The hard number that you are missing is the acceleration of the ship. In space it's not like you are swimming in water where you can only go at a certain speed, and a faster flow will suck you backwards. Instead, space allows you to accelerate without limit (provided you have enough reaction mass on board), getting faster and faster along the way. Your black hole adds another acceleration towards it, and in that sense you can hover over a black hole (provided your engines have enough thrust / you are far enough away, and you have reaction mass to waste).

However, the real point that you are missing is this: It would be insane to try to stop the ship and back out from the black hole. The only thing you need to do is ensure that you are not torn to pieces by the black hole (or fall into it). And that is easiest done by accelerating sideways so that you pass by the black hole in a safe distance. This takes much less $\Delta v$ than the brake-and-reverse maneuver, and it means that you do not loose the speed you picked up while falling in. This is the key: The speed that you picked up while falling towards the black hole will also carry you away from the black hole again. It's like riding a bike down a mountain without air or other friction to slow you down. You'll be moving insanely fast through the valley, and then ascent the next mountain to exactly the same height that you started from. All without even touching your pedals.

To give you some data points for determining the scales involved:

  • Stellar black holes have radii in the range of tens to hundreds of kilometers.

  • An object falling towards our sun from infinity reaches a speed of 617 kilometers at its surface, which is at a distance of 700000 kilometers (700 megameters, or 100 times the radius of the earth).

  • A human can only endure a few $g$ for a prolonged time which is $9.81\frac{m}{s}$. To stop a ship going at 600 kilometers per second at an acceleration of $10g$, you would need about 60000 seconds (17 hours), by which time you'd have hit the black hole.

  • After a lateral $10g$ acceleration for 400 seconds, you would already be 1000 kilometers away from your original trajectory, more than enough to miss the black hole.

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  • $\begingroup$ I completely agree on your point that preserving your speed and making an 'orbit' of the black hole is the best option. The implication here though was that control was regained at the 11th hour, where the only remaining option was to thrust directly away from the black hole like you were lifting off a planet. Not sure how realistic that would be just by itself though. $\endgroup$
    – Razmode
    Dec 30, 2020 at 23:22
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    $\begingroup$ @Razmode If you can still brake-and-reverse, you can also still miss-and-go. You'll be able to do the miss-and-go far longer than the brake-and-reverse. As I've shown, being one sun radius away from a one sun mass black hole is already way beyond the brake-and-reverse maneuver, but still well within the realm of miss-and-go. $\endgroup$
    – cmaster - reinstate monica
    Dec 30, 2020 at 23:42
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Assuming you mean an Acceleration of 0.99C/sec and not a Speed of 0.99C which is meaningless for this problem...

As far as time dilation goes, it depends on who's perspective you are talking about. Do you mean 1 m/s relative to the ship or to the outside world. Time dilation affects how the ship experiences time; so, what we may see as taking a long time would be a very short time as far as the ship is concerned.

Either way, this problem defies real world logic because no matter which way you look at it. 1m/s puts you WAY too close to the point of no return to make since. Let's say you mean 1m/s to the outside world since that is actually the more realistic situation. The acceleration of gravity of a 5 solar mass black hole hits .99C at a radius of 1,495,417.9953 m, and 1m/s of gravity less than that is at a radius of 1,495,417.9977 m which is only a difference of 2.5 mm. Since your ship is probably more than 2.5mm in length, not only will part of your ship will be past the point of no return, but the shear on your ship would be unimaginable. If you are talking about 1m/s from the perspective of the ship which is experiencing time in slow motion, the you are talking about being within less than a nanometer of the point of no return... so let's just through that notion out of the window.

This is so beyond unrealistic, but let's say you somehow get your ship into that 2.5mm sweet spot and have something similar to Star Trek style inertial dampeners/integrity fields or what not so that you are magically not spaghettified so that you are actually moving away at 1m/s from an undilated perspective. Every meter you move away will vastly increase your acceleration such that by the time you get just 1km above the point of no return, your will already be moving away at about 0.13C. A second after that you will be at a radius of about 1,892,958 m moving at about 0.63C.

So in theory, your total escape into relatively normal space will only take a few seconds from the perspective of us normal paying customers, and even less time as far as your ship is concerned; so, no need to worry about your delivery being late... in practice though, your ship is actually going to get torn apart and your goods lost to the black hole if you try actually getting that close to it.

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  • $\begingroup$ I suppose in this case it would be the tail end that is 2.5mm away from the singularity, with the rest being farther, naturally. Ah, so it would not be long is the answer then. $\endgroup$
    – Razmode
    Dec 30, 2020 at 22:38
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    $\begingroup$ @Razmode Pretty much. Yes. That said, based on the context of your question , I assumed you meant acceleration of .99C and not a speed of .99C since speed is not a meaningful variable here. But, if your ship has a much lower acceleration with a max speed of .99C, then your ship will get "trapped" farther out, and take longer to escape, as the falloff gradient will be much shallower farther from the black hole, but in this case, relativistic physics will not really apply, you're just looking at a good ol' take-off from gravity well type problem. $\endgroup$
    – Nosajimiki
    Dec 30, 2020 at 22:45
  • $\begingroup$ I believe you are correct on acceleration. I will have to look into reference frames and such again it seems. A more realistic example would be a ship near the spaghettification boundary DKNguyen mentioned and a 'soft' event horizon would also be whatever the max acceleration a more reasonable ship could output, meaning it was doomed but not yet destroyed. However, I chose the event horizon itself because it had the distinct advantage of being the most dramatic :) $\endgroup$
    – Razmode
    Dec 30, 2020 at 23:03
  • $\begingroup$ @Razmode The "spaghettification boundary" is based on the structural integrity of your ship. The more rugged your ship the hard it will be for the black hole to stretch it out; so, just be aware that that will be a soft cap rather than a hard cap. $\endgroup$
    – Nosajimiki
    Dec 31, 2020 at 14:23

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