How close to a black hole can an object orbit elliptically? I know circular orbits are no longer stable at distance less than 3 times the Schwarzschild radius. But what about elliptical orbits? Can an object have a semi-major axis or perihelion at distance of less than 3 times the Schwarzschild radius?
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$\begingroup$ Semi-major axis and perihelion are not the same thing. I suspect that you're more interested in perihelion, since I'm not sure that semi-major axis is a well-defined concept in a curved geometry. $\endgroup$– Michael SeifertCommented Nov 16, 2021 at 14:10
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$\begingroup$ I know that, I've refrased the question. $\endgroup$– blademan9999Commented Nov 16, 2021 at 14:25
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2$\begingroup$ You mean "pseudo-elliptically"? The only elliptical orbits are circles. The rest have precession of periastron so that they aren't closed ellipses. $\endgroup$– ProfRobCommented Nov 16, 2021 at 19:46
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$\begingroup$ Related (some background information): Why are orbits 1.5rs<r<3rs unstable around a Schwarzschild black hole? $\endgroup$– Peter MortensenCommented Nov 16, 2021 at 23:39
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3 Answers
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A bound elliptical orbit around a Schwarzschild black hole must have $r > 2 r_s$ at all times (where $r_s = 2 M$ is the Schwarzschild radius). Deriving this result is a good exercise for students learning about the Schwarzschild geometry, so I won't go through all the details, but the basic sketch of the proof is as follows:
- Recall that a massive particle moving in a Schwarzschild geometry is equivalent to a particle moving in a classical "effective potential" given by $$ V_\text{eff}(r) = - \frac{M}{r} + \frac{\ell^2}{2 r^2} - \frac{M\ell^2}{r^3}, $$ where $M$ is the mass of the black hole and $\ell$ is the specific angular momentum of the particle.
- Note that for a bound orbit, we must have $V_\text{eff}(r) < 0$ at all times.
- Find the points at which $V_\text{eff}(r) = 0$ for a given value of $\ell$. This will be the closest possible value of perihelion for a bound orbit for a particular value of $\ell$.
- Find the value of $\ell$ that allows for the closest perihelion. It turns out to be $\ell = 4M$, and for that value of the angular momentum you must have $r > 2 r_s$ to satisfy $V_\text{eff}(r) < 0$.
answered Nov 16, 2021 at 14:35
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$\begingroup$ @safesphere : You have misparsed the text of the Answer. In "can get arbitrarily close to $r=2r_s$ (where $r_s=2M$ is the Schwarzschild radius) but no closer", the phrase "but no closer" is never interpreted as modifying the parenthetical; it modifies "can get arbitrarily close to $r=2r_s$", which is $4M$. $\endgroup$ Commented Nov 17, 2021 at 7:24
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$\begingroup$ Tangentially related: A mass in close orbit will radiate gravitational waves; is that energy loss relevant close to the minimal radius and does it therefore lead to a decay of such an orbit? $\endgroup$ Commented Nov 17, 2021 at 12:26
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$\begingroup$ @safesphere: I rephrased the first sentence; hopefully it's clearer now. $\endgroup$ Commented Nov 17, 2021 at 12:32
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$\begingroup$ @Peter-ReinstateMonica: You're right that this would cause a gradual decay. But the rate of decay also depends on the orbiting mass (I believe it's proportional to $m^2$). I suspect that so long as $m \ll M$, the decay would still be rather slow even for an orbit that gets to $r \approx 2r_s$. But I will admit I don't have a definitive proof to back that up. It might be a good question to ask separately! $\endgroup$ Commented Nov 17, 2021 at 16:05
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$\begingroup$ How smalle can the semi-major axis be? $\endgroup$ Commented Nov 19, 2021 at 9:49
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Yes, highly eccentric orbits can go deeper. (Kostic 2012) derives analytic solutions in terms of elliptic functions, noting (in footnote 4) that the closest approach would be 2 Schwarzschild radii out for a $l=2$ orbit. Such orbits may not be very elliptic-looking since there can be multiple turns per perimelasma approach (OK, periapse is the more common term).
answered Nov 16, 2021 at 14:29
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1$\begingroup$ Note that $l$ here is the "reduced angular momentum", which Kostic defines as the specific angular momentum divided by the Schwarzschild radius. (Just in case the reader is confused about how your answer compares with mine.) $\endgroup$ Commented Nov 16, 2021 at 14:39
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1$\begingroup$ But this ignores radiating away energy through gravitational waves(?). And circularise the orbit(?). $\endgroup$ Commented Nov 16, 2021 at 23:47
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$\begingroup$ "Such orbits may not be very elliptic-looking since there can be multiple turns per perimelasma approach (OK, periapse is the more common term)" So, it'd look like a spirograph drawing, then? $\endgroup$ Commented Nov 17, 2021 at 10:58
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$\begingroup$ @PeterMortensen - See physics.stackexchange.com/questions/458444/… for the circularisation issue. Basically, this is a pretty slow process for non-heavy bodies. $\endgroup$ Commented Nov 17, 2021 at 17:26
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I reality, in the frame of general relativity, the real elliptic orbit does not exist and all bodies will end-up falling in the black hole. The nearer you get from the event horizon, the faster you will fall inside which means the "limit you are looking for does not exist. It's more like, the farther you are, the longer you will keep orbiting.
I addition, since orbits can be more or less eccentric it is delicate to just give one distance.
