# Closest possible flyby to a nonrotating supermassive black hole

Imagine you think Laplace is the last word on black holes. That is, you are aware of the radius for which the escape velocity is $$c$$, but you think gravity is Newtonian. You set your spacecraft on a course meant to be a hyperbola that remains just outside the event horizon of a giant, nonrotating black hole. Tidal forces are not a problem, and your plan is simply to observe conditions near the boundary.

Unfortunately, gravity is not Newtonian, and once you get inside the photon sphere you will need to exert thrust to get back out. (I originally thought the ISCO was the key boundary here, but it's not.) The closer you get to the black hole, the more thrust you will need. My sense is that, even when tidal forces are too small to be a concern, there is a closest distance that any object made of matter can get to the event horizon and later return to infinity without being crushed by the acceleration it would have to undergo to escape the photon sphere.

What I would really like is a number, something like the closest you can get to the event horizon and then escape without any object larger than a meter in height being reduced to degenerate matter. But that is still too vague to answer. So instead, is there a formula for the peak thrust you need to return to infinity as a function of the minimum number of Schwarzschild radii from the center?

• Note that in a fly-by, with enough initial speed, you can get down to (just above) the photon sphere and escape without any thrust. ISCO isn't the limit here.
– Sten
Commented Oct 7, 2023 at 6:23
• Why would the object get crushed? (Rather than torn apart by tidal forces?) Commented Oct 7, 2023 at 20:27
• This is a supermassive black hole. They can be large enough so that tidal forces are negligible outside the EH. The point is that the area near the event horizon is still not survivable because your choices are either to fall in and get spaghettified later, or use rockets to avoid falling in but be crushed by the acceleration. You can't just sail safely by once you get (far enough) inside the photon sphere. Commented Oct 7, 2023 at 21:38

OP asked: is there a formula for the peak thrust you need to return to infinity as a function of the minimum number of Schwarzschild radii from the center?

Yes, you can find the working here. The amount of force required to radially espace a Schwarzschild black hole of mass $$M$$ is $$F=m a=\frac{G M m}{r^{2}} \frac{1}{\sqrt{1-\frac{2 G M}{c^{2} r}}}.$$

OP asked: What is the closest you can get to the event horizon and then escape without any object larger than a meter in height being reduced to degenerate matter.

Assuming that your object only extends in the radial direction with length $$\chi$$, the tidal force acting on this object can be calculated using the equation of geodesic deviation. In the freely falling observer's frame this will give you $$\frac{d^{2} \chi^r}{d \tau^{2}}=-R_{{c} {b} {c}}^{{r}} \chi^{{b}}$$ which on substituting the appropriate values of the Riemann tensor $${R^a}_{bcd}$$ will give $$\frac{d^{2} \chi^{{r}}}{d \tau^{2}}= \frac{2GM}{r^{3}} \chi^{{r}}.$$

So, the tidal force depends on the radial value $$r$$. Let us say we want to find out the tidal force at the event horizon ($$r_{\rm hor} = 2GM/c^2$$) of this black hole. Then $$\frac{d^{2} \chi^{{r}}}{d \tau^{2}}= \frac{c^6}{8 G^2 M^2} \chi^{{r}}.$$ So we find that the larger the $$M$$, the smaller the tidal force at the horizon becomes. On doing some calculation, I find that for $$M = 10^4 M_{\odot}$$ the tidal force is $$51$$ N (see pic below)

• " ... freely falling observer's frame..." What freely falling observer are you talking about here? Commented Oct 7, 2023 at 20:23
• I mean the frame of the observer that is falling into the Black hole and hasn't yet turned on any rocket thrust.
– S.G
Commented Oct 7, 2023 at 20:45
• A radially infalling observer? I.e. what geodesic is this observer on? Does this matter? Commented Oct 7, 2023 at 22:05
• Yes, you are right I think that it won't matter what path the observer takes, as long as it is a geodesic. Although I have only focused on the radial part of the tidal acceleration here.
– S.G
Commented Oct 7, 2023 at 22:52
• It's not obvious that tidal forces should be independent of path, given that (unlike Newtonian gravity) relativistic gravity is velocity-dependent.
– Sten
Commented Oct 8, 2023 at 3:18