Here is a “cook‐book’’ for getting the equilibrium temperature that a little planet in circular orbit around a Kerr (or Schwarzschild) black hole will settle to, when the only “heater’’ is the cosmic microwave background at 2.7 K as seen from infinity. In outline you do precisely the same thing you would do in special relativity (blue‐shift plus Doppler beaming of an otherwise isotropic bath) except that now you must combine

• the gravitational red–(or blue–)shift from infinity into your orbit
• the aberration / Doppler–beaming from the fact that the planet is moving
• the fact that part of the sky is covered by the black-hole shadow
• the usual black–body equilibrium condition

Below I give both the full GR recipe (numerical ray‐tracing) and a much simpler analytic approximation (no lensing, no frame‐dragging) that already captures the correct scalings.

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1. Full general‐relativistic method

1. Choose your black hole
   – Mass (M), spin (a). Write down the Kerr metric in Boyer–Lindquist coordinates ((t,r,\theta,\phi)).

2. Pick your orbit
   – A prograde circular orbit in the equatorial plane (\theta=\pi/2) at radius (r). From Bardeen – Press – Teukolsky (1972) one has the conserved energy per unit rest mass
   [ E ;=;-p_t/m ;=; \frac{r^{3/2}-2M,r^{1/2}+a,M^{1/2}}{r^{3/4},\sqrt{r^{3/2}-3M,r^{1/2}+2a,M^{1/2}}},, ]
   and the angular velocity
   [ \Omega ;=;\frac{d\phi}{dt} ;=; \frac{M^{1/2}}{r^{3/2}+a,M^{1/2}},. ]
   – From these you build the four–velocity of the planet
   [ u^\mu ;=;\bigl(u^t,,0,,0,,u^\phi\bigr) \quad\hbox{with}\quad u^t=\frac{dt}{d\tau};=;\frac{1}{\sqrt{-,g_{tt}-2,\Omega,g_{t\phi}-\Omega^2,g_{\phi\phi}}} ;,\quad u^\phi=\Omega,u^t. ]

3. Set up a local tetrad
   – Let (e_{(0)}{}^\mu=u^\mu). Choose three mutually orthonormal spacelike vectors (e_{(1)},e_{(2)},e_{(3)}) that span the local rest space of the planet. One convenient choice is to start from the ZAMO (zero angular–momentum observer) tetrad and then boost it by the local orbital velocity.

4. Shoot rays into the sky
   – Parameterize the planet’s local sky by two angles ((\vartheta,\varphi)). In the tetrad frame a photon leaving the planet has momentum
   [ p^{(a)} ;=;h\nu_\infty\bigl(1,;\sin\vartheta\cos\varphi,;\sin\vartheta\sin\varphi,;\cos\vartheta\bigr),. ]
   – Convert back to coordinate components
   [ p^\mu ;=;e_{(a)}{}^\mu,p^{(a)},. ]
   – Integrate the null geodesic backwards until it either hits the hole (so that direction is part of the “shadow’’) or goes out to infinity.

5. Compute the redshift factor
   – Along each geodesic you have two conserved quantities: the energy at infinity (-p_t) and the photon’s local inner product with the planet’s four–velocity
   [ \nu_{\rm obs} ;=; -,p_\mu,u^\mu,, \quad \nu_\infty ;=; -,p_t ] – Hence the usual frequency shift
   [ g(\vartheta,\varphi) ;=; \frac{\nu_{\rm obs}}{\nu_\infty} ;=; \frac{-,p_\mu,u^\mu}{-,p_t},. ]

6. Use Liouville’s theorem
   – A black–body at infinity has specific intensity (I_\nu^\infty=B_\nu[T_{\rm CMB}]). Along a geodesic
   (,I_\nu/\nu^3) is conserved. Hence at the planet
   [ I_{\nu,\rm loc}(\vartheta,\varphi) ;=; g^3(\vartheta,\varphi) ;B_{\nu}[T_{\rm CMB}]\Bigl(\nu/g\Bigr),, ] – and integrating over frequency shows that the bolometric intensity at the planet is
   [ I_{\rm loc}(\vartheta,\varphi) ;=; g^4(\vartheta,\varphi) ;I_{\infty} ;=; g^4(\vartheta,\varphi),\frac{\sigma,T_{\rm CMB}^4}{\pi},. ]

7. Compute the absorbed flux
   – The total power falling on the planet is the cross‐section (\pi R^2) times the integral of the incoming intensity over all directions that clear the hole’s shadow.
   [ P_{\rm abs} ;=; \pi,R^2 ;\int_{\hbox{\scriptsize sky,,shadow}} I_{\rm loc}(\vartheta,\varphi);d\Omega ;=; \pi,R^2;\frac{\sigma,T_{\rm CMB}^4}{\pi} ;\int_{\hbox{\scriptsize visible}} g^4,d\Omega. ] – Spread that power over the entire surface (4\pi R^2) to get the average absorbed flux per unit area:
   [ F_{\rm abs} ;=; \frac{P_{\rm abs}}{4\pi,R^2} ;=; \frac{\sigma,T_{\rm CMB}^4}{4\pi} ;\int_{\hbox{\scriptsize visible}} g^4(\vartheta,\varphi);d\Omega. ]

8. Impose black–body balance
   – In steady state the planet must reradiate exactly (F_{\rm abs}) in its own local black–body spectrum, so
   [ \sigma,T_{\rm eq}^4 ;=; F_{\rm abs} ;=; \frac{\sigma,T_{\rm CMB}^4}{4\pi} \int_{\hbox{\scriptsize visible}} g^4,d\Omega, ] – i.e. [ T_{\rm eq} ;=; T_{\rm CMB} ;\biggl{ \frac{1}{4\pi} \int_{\hbox{\scriptsize visible}} \bigl[g(\vartheta,\varphi)\bigr]^4 ;d\Omega \biggr}^{1/4}. ]

That integral has to be done numerically once you have the geodesics and the redshift factor (g). You also have to subtract out the solid angle of the black‐hole shadow.

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2) A simple analytic approximation

If you ignore all lensing, frame‐dragging, and the shadow (i.e.\ you pretend the BH is just a Newtonian point mass plus special‐relativistic time‐dilation), then

• Gravitational time‐dilation at radius (r) gives a static redshift factor
[ g_{\rm grav} ;=; \frac{1}{\sqrt{1 - \dfrac{2GM}{rc^2}}} ] so that a static observer at (r) would see the CMB temperature
(;T_{\rm CMB},g_{\rm grav}.)

• Orbital Doppler–beaming. The planet moves at
[ v ;=; \frac{\sqrt{GM/r}}{\sqrt{1-2GM/(rc^2)}},, \quad \beta=\frac{v}{c}, \quad \gamma=\frac{1}{\sqrt{1-\beta^2}}. ] In special relativity an isotropic bath at temperature (T) becomes angle–dependent [ T_{\rm local}(\mu) ;=; T,\gamma,(1+\beta,\mu), \quad \mu=\cos\hbox{(angle w.r.t.\ velocity)}. ] One shows easily [ \frac{1}{4\pi}\int_{4\pi} \bigl[\gamma(1+\beta\mu)\bigr]^4 ,d\Omega ;=; \gamma^4\bigl[1+2\beta^2+0.2,\beta^4\bigr]. ]

Putting them together in the same spirit (but still ignoring the shadow) gives the analytic estimate

[ T_{\rm eq} ;=; T_{\rm CMB}; g_{\rm grav};\gamma ;\bigl[,1 + 2\beta^2 + 0.2,\beta^4\bigr]^{1/4},. ] One can check quickly that for mildly relativistic orbits this reproduces the known fact that the beaming correction is only of order unity, whereas in the very deep potential the factor (g_{\rm grav}) can become enormous.

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How one gets ~890 °C for “Miller’s planet’’
In the Interstellar scenario the authors have chosen parameters (near–maximal spin, an orbit just above the prograde ISCO) so that

– the total gravitational time‐dilation* alone* would be a factor of (F\sim6\times10^4), and they treat that as boosting the total power by the same factor
– since power (\propto T^4), they set (T^4\propto6\times10^4), i.e. (T\sim2.7,\mathrm K\times(6\times10^4)^{1/4}\approx42) K from gravity alone
– then they fold in a very large orbital Doppler‐beaming (because that orbit is extremely close in and very fast), which raises the temperature by another factor of a few tens
– in the end one finds (T\approx1163) K (about 890 °C) once you do the full angular‐integral including the Doppler factor

If you want to reproduce their exact number you must

1. pick the same Kerr spin (a),
2. solve for the prograde circular orbit radius that gives the (6\times10^4) time-dilation they quote,
3. compute (\beta) and (\gamma) at that orbit,
4. do the integral
   (;\bigl[\gamma(1+\beta\mu)\bigr]^4) over the visible sky (subtracting the hole’s shadow!),
5. take the one-quarter power, and
6. multiply by (2.7) K.

In practice you can do 4.–6. either
– numerically with a little ray-tracer in the Kerr metric (this is what e.g.\ GYOTO or geokerr will do),
– or analytically to leading order in (\beta) if (\beta\ll1).

Either way, the template is always the same:

• Find the local sky distribution of CMB photons via
(g(\vartheta,\varphi)=-p_\mu u^\mu/p_t).
• Use (I_{\rm loc}=g^4,I_\infty).
• Integrate (g^4) over the non–shadowed sky.
• Set (\sigma T_{\rm eq}^4=(\sigma T_{\rm CMB}^4/4\pi)\int g^4d\Omega).

That procedure will let you {\em predict} the equilibrium temperature for {\em any} (M,a,r). Once the numbers are in hand it quickly reproduces numbers of order 1000 K for the extreme orbits needed to get a 1 hr → 60 000 hr dilation factor.