# Limit to the Oberth effect in a supermassive Black Hole flyby

To make this question clear, here are the details of the situation I wish to entertain.

• A spacecraft does a powered gravitational assist, where it fires engines in a near-approach to the body

• We should all be aware the rotating black holes can directly act on a passerby, and both because of the focus on supermassive, and for mathematical simplicity, I'd like to assume the standard Schwarzschild black hole:

• Best to assume the spacecraft starts with $$v_{infinity}=0$$, that is, its incoming speed before entering the gravity well is assumed to be minimal.

• The spacecraft comes as close as possible without falling in, or however close gets the maximum Oberth effect.

• I want to convert $$\Delta v$$ the spacecraft engines generate to the final $$V$$ after it has departed the gravity well.

To sum up, I want some expression for a relativistic Oberth effect that would apply in the most extreme case.

## Preliminary Thinking

Previous question:

What happens to orbits at small radii in general relativity?

I guess a logical approach would be to follow the same approach as the computation of the Oberth effect for a parabolic orbit based on energy balance. But if you go highly relativistic, the gravitation as well as kinetic energy terms can get quite complex, here is gravitational:

$$V(r)=-{\frac {GMm}{r}}+{\frac {L^{2}}{2\mu r^{2}}}-{\frac {G(M+m)L^{2}}{c^{2}\mu r^{3}}}.$$

I could also guess that the optimal approach is at the IBCO radius of 3/2 times the Schwarzschild radius. But this still leaves quite a few things to plug in, and I'm doubtful about the validity of the approach overall.

Heck, just to put it out there, let's say I use the non-relativistic Oberth equation assuming the IBCO approach distance:

$$V=\Delta v{\sqrt {1+{\frac {2V_{\text{esc}}}{\Delta v}}}} =\Delta v{\sqrt {1+{ \sqrt{\frac{GM}{3/2 r_s}} \frac {2}{\Delta v}}}} =\Delta v{\sqrt {1+ { \frac {2 c}{3 \sqrt{3} \Delta v}}}}.$$

This would give a multiplier of something like a factor of 100 for a 10 km/s burn. But this is almost certainly wrong, applied outside its range of applicability.

• The mechanical stresses that your ship will undergo are going to be ... significant. Jun 27 '19 at 9:47
• This is the same principle as the Hills' mechanism. The escape velocity is $\approx \sqrt{v \Delta v}$, with $\Delta v$ as you define it and $v$ the speed in the potential well. Nov 30 '19 at 4:20
• I don't think it's a good idea to ignore the spin of the black hole, since it's likely to be very large. See the diagram at the end of this answer: astronomy.stackexchange.com/a/20292/16685 Mar 29 '20 at 14:32

The spacecraft follows a geodesic, and if it does an impulsive boost at a point it will now follow a different geodesic from that point but with a different 4-velocity. The incoming trajectory starts with velocity $$v_0$$ at infinity and the new one ends at velocity $$v_1$$, so the overall Oberth boost is $$|v_1-v_0|$$.
The standard textbook equations for time-like Schwarzschild geodesics are: $$\frac{dt}{d\tau}=\frac{E}{mc^2}\frac{1}{1-\frac{r_s}{r}}$$ $$\frac{d\theta}{d\tau} = \frac{L}{M}\frac{1}{r^2}$$ $$\left(\frac{dr}{d\tau}\right)^2 = \frac{E^2}{m^2c^2} - \left(1-\frac{r_s}{r}\right)\left(c^2+\frac{L^2}{M^2}\frac{1}{r^2}\right)$$ where $$E$$ is the energy of the craft, $$L$$ its angular momentum, $$r_s$$ the Schwarzschild radius, $$M$$ the mass of the central body and $$\tau$$ proper time. The spacecraft mass $$m\ll M$$.
The effective potential is $$V(r)=-\frac{GMm}{r}+\frac{L^2}{2GMr^2} -\frac{L^2}{c^2r^3}:$$ the particle moves as $$\frac{1}{2}m\left(\frac{dr}{d\tau}\right)^2=\left[\frac{E^2}{2mc^2} - \frac{mc^2}{2}\right] + V(r).$$ It allows different orbit types depending on ($$E,L$$). The ones we care about are the ones that are unbounded in the past or future. $$E$$ must be larger than $$mc^2$$ (otherwise it cannot escape to infinity ).
So, to do the maneouvre we drop a craft from infinity towards the hole. It starts with velocity $$v_0=\sqrt{\frac{E^2}{m^2c^2}-c^2}$$ at $$r=\infty$$. It approaches until the righthand side of the motion equation becomes zero at $$r_{turn}(E,L)$$. At this point we change velocity to get $$E',L'$$ and the craft recedes to infinity; we calculate its velocity $$v_1 = \sqrt{\frac{E'^2}{m^2c^2}-c^2}$$ and will have our answer $$v_1-v_2$$.
The part where I get stuck is how to calculate what $$E',L'$$ different boosts imply. Also, realistic boosts will change $$m$$ to $$m'$$ if the ejected mass is significant.