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Theinnermost stable circular orbit (often called theISCO) is the smallest marginally stable circular orbit in which atest particle can stably orbit a massive object ingeneral relativity.^[1] The location of the ISCO, the ISCO-radius (${\displaystyle r_{\mathrm{i}\mathrm{s}\mathrm{c}\mathrm{o}}}$), depends on the mass and angular momentum (spin) of the central object. The ISCO plays an important role in black holeaccretion disks since it marks the inner edge of the disk.

The ISCO should not be confused with theRoche limit, the innermost point where a physical object can orbit beforetidal forces break it up. The ISCO is concerned with theoreticaltest particles, not real objects. In general terms, the ISCO will be far closer to the central object than the Roche limit.

Inclassical mechanics, an orbit is achieved when a test particle'sangular momentum is enough to resist thegravity force of the central object. As the test particle approaches the central object, the required amount of angular momentum grows, due to theinverse square law nature of gravitation. This can be seen in practical terms inartificial satellite orbits; ingeostationary orbit at 35,786 kilometres (22,236 mi) the orbital speed is 10,800 kilometres per hour (6,700 mph), whereas inlow Earth orbit it is 27,000 kilometres per hour (17,000 mph). Orbits can be achieved at any altitude, as there is no upper limit to velocity in classical mechanics.

General relativity (GR) introduces an upper limit to the speed of any object: thespeed of light. If a test particle is lowered in orbit toward a central object in GR, the test particle will eventually require a speed greater than light to maintain an orbit. This defines the innermost possible instantaneous orbit, known as the innermost circular orbit, which lies at 1.5 times theSchwarzschild radius (for a black hole governed by theSchwarzschild metric). This distance is also known as thephoton sphere.

In GR, gravity is not treated as a central force that pulls on objects; it instead operates by warpingspacetime, thus bending the path that any test particle may travel. The ISCO is the result of an attractive term in the equation representing the energy of a test particle near the central object.^[2] This term cannot be offset by additional angular momentum, and any particle within this radius will spiral into the center. The precise nature of the term depends on the conditions of the central object (i.e. whether ablack hole has angular momentum).

For a non-spinning massive object, where the gravitational field can be expressed with theSchwarzschild metric, the ISCO is located at

${\displaystyle r_{\mathrm{m}\mathrm{s}}=6\frac{GM}{c^2}=3R_S,}$

where${\displaystyle R_S}$ is the Schwarzschild radius of the massive object with mass${\displaystyle M}$. Thus, even for a non-spinning object, the ISCO radius is only three times theSchwarzschild radius,${\displaystyle R_S}$, suggesting that onlyblack holes andneutron stars have innermost stable circular orbits outside of their surfaces. As the angular momentum of the central object increases,${\displaystyle r_{\mathrm{i}\mathrm{s}\mathrm{c}\mathrm{o}}}$ decreases.

Bound circular orbits are still possible between the ISCO and the so-calledmarginally bound orbit, which has a radius of

${\displaystyle r_{\mathrm{m}\mathrm{b}}=4\frac{GM}{c^2}=2R_S,}$

but they are unstable. Between${\displaystyle r_{\mathrm{m}\mathrm{b}}}$ and the photon sphere so-called unbound orbits are possible which are extremely unstable and which afford a total energy of more than the rest mass at infinity.

For a massless test particle like aphoton, the only possible but unstable circular orbit is exactly at thephoton sphere.^[3] Inside the photon sphere, no circular orbits exist. Its radius is

${\displaystyle r_{\mathrm{p}\mathrm{h}}=3\frac{GM}{c^2}=1.5R_S.}$

The lack of stability inside the ISCO is explained by the fact that lowering the orbit does not free enough potential energy for the orbital speed necessary: the acceleration gained is too little. This is usually shown by a graph of the orbitaleffective potential which is lowest at the ISCO.

The case for rotating black holes is somewhat more complicated. The equatorial ISCO in theKerr metric depends on whether the orbit isprograde (negative sign in${\displaystyle r_{\mathrm{m}\mathrm{s}}}$) orretrograde (positive sign in${\displaystyle r_{\mathrm{m}\mathrm{s}}}$):

${\displaystyle r_{\mathrm{m}\mathrm{s}}=\frac{GM}{c^2}\left(3+Z_2\pm \sqrt{(3-Z_1)(3+Z_1+2Z_2)}\right)\le 9\frac{GM}{c^2}=4.5R_S}$

where

${\displaystyle Z_1=1+\sqrt[3]{1-{\chi }^2}\left(\sqrt[3]{1+\chi }+\sqrt[3]{1-\chi }\right)}$

${\displaystyle Z_2=\sqrt{3{\chi }^2+Z_1^2}}$

with the rotation parameter${\displaystyle \chi =2a/R_S=cJ/(M^{2}G)}$.^[4]

As the rotation rate of the black hole increases to the maximum of${\displaystyle \chi \to 1}$, the prograde ISCO, marginally bound radius and photon sphere radius decrease down to theevent horizon radius at the so-called gravitational radius, still logically and locally distinguishable though.^[5]

${\displaystyle r_{\mathrm{m}\mathrm{s}}\to r_{\mathrm{m}\mathrm{b}}\to r_{\mathrm{p}\mathrm{h}}\to r_E\to r_G=GM/c^2}$

The retrograde radii hence increase towards

${\displaystyle r_{\mathrm{m}\mathrm{s}}\to 9\frac{GM}{c^2}=4.5R_S}$

${\displaystyle r_{\mathrm{m}\mathrm{b}}=\frac{GM}{c^2}(1+\sqrt{1\pm \chi }{)}^2\to (3+\sqrt{8})\frac{GM}{c^2}\approx 2.9142R_S}$,

${\displaystyle r_{\mathrm{p}\mathrm{h}}=2\frac{GM}{c^2}\left(1+\cos \left(\frac{2}{3}{\cos }^{-1}(\pm \chi )\right)\right)\to 4\frac{GM}{c^2}=2R_S}$.

If the particle is also spinning there is a further split in ISCO radius depending on whether the spin is aligned with or against the black hole rotation.^[6]

• Leo C. Stein, Kerr calculator V2[1]

• General relativity
• Black holes
• Orbits

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