Physics Asked on October 3, 2021
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.
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:
$$ V(r)=-{frac {GMm}{r}}+{frac {L^{2}}{2mu r^{2}}}-{frac {G(M+m)L^{2}}{c^{2}mu r^{3}}}. $$
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.
I'm also interested in the answer to this question, this is how far I got:
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: $$frac{dt}{dtau}=frac{E}{mc^2}frac{1}{1-frac{r_s}{r}}$$ $$frac{dtheta}{dtau} = frac{L}{M}frac{1}{r^2}$$ $$left(frac{dr}{dtau}right)^2 = frac{E^2}{m^2c^2} - left(1-frac{r_s}{r}right)left(c^2+frac{L^2}{M^2}frac{1}{r^2}right)$$ 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 $mll M$.
The effective potential is $$V(r)=-frac{GMm}{r}+frac{L^2}{2GMr^2} -frac{L^2}{c^2r^3}:$$ the particle moves as $$frac{1}{2}mleft(frac{dr}{dtau}right)^2=left[frac{E^2}{2mc^2} - frac{mc^2}{2}right] + V(r).$$ 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.
Answered by Anders Sandberg on October 3, 2021
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