Innermost stable circular orbit

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

${\displaystyle r_{\mathrm {ms} }=6{\frac {GM}{c^{2}}}=3R_{S},}$

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 the Schwarzschild radius, ${\displaystyle R_{S}}$, suggesting that only black holes and neutron stars have innermost stable circular orbits outside of their surfaces. As the angular momentum of the central object increases, ${\displaystyle r_{\mathrm {isco} }}$ decreases.

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

${\displaystyle r_{\mathrm {mb} }=4{\frac {GM}{c^{2}}}={2R_{S}},}$

but they are unstable. Between ${\displaystyle r_{\mathrm {mb} }}$ 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 a photon, the only possible but unstable circular orbit is exactly at the photon sphere.[2] Inside the photon sphere, no circular orbits exist. Its radius is

${\displaystyle r_{\mathrm {ph} }=3{\frac {GM}{c^{2}}}={1.5R_{S}}.}$

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 orbital effective potential which is lowest at the ISCO.

The case for rotating black holes is somewhat more complicated. The equatorial ISCO in the Kerr metric depends on whether the orbit is prograde (negative sign below) or retrograde (positive sign):

${\displaystyle r_{\mathrm {ms} }={\frac {GM}{c^{2}}}\left(3+Z_{2}\pm {\sqrt {(3-Z_{1})(3+Z_{1}+2Z_{2})}}\right)\leq 9{\frac {GM}{c^{2}}}=4.5R_{S}}$

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 ${\displaystyle \chi =2a/R_{S}=cJ/(M^{2}G)}$.[3]

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 the event horizon radius at the so called gravitational radius, still logically and locally distinguishable though. [4]

${\displaystyle r_{\mathrm {ms} }\to r_{\mathrm {mb} }\to r_{\mathrm {ph} }\to r_{E}\to r_{G}=GM/c^{2}}$

The retrograde radii hence increase towards

${\displaystyle r_{\mathrm {ms} }\to 9{\frac {GM}{c^{2}}}=4.5R_{S}}$

${\displaystyle r_{\mathrm {mb} }={\frac {GM}{c^{2}}}(1+{\sqrt {1\pm \chi }})^{2}\to {\frac {3+{\sqrt {8}}}{2}}R_{S}\approx 5.828427{\frac {GM}{c^{2}}}\approx 2.9142R_{S}}$,

${\displaystyle r_{\mathrm {ph} }=2{\frac {GM}{c^{2}}}(1+\cos({\tfrac {2}{3}}\cos ^{-1}(\pm \chi )))\to 4{\frac {GM}{c^{2}}}=2R_{S}}$.

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.[5]

• Leo C. Stein, Kerr calculator V2