Last time, I described the vast Galactic habitat of the Deep Future – a Super Massive Black Hole, of the order of 10^{10} Solar Masses, with a Event Horizon 30 billion km in radius (200 AU) and probably not a huge spin factor, so an Innermost Stable Circular Orbit of 600 AU.

To last a trillion trillion years, the orbital planet rings, assuming their orbital energy is being radiated away via gravitational waves, when the planets mass 5 Earth masses (about 3.0E+25 kg), they must start at 262,000 AU.

If we use solar masses and astronomical units, the Inspiral Time becomes:

$$t = \left[\left(\tau\right)\frac{r_o^4}{(M_1 M_2)(M_1+M_2)}\right]$$

Where \(\tau\) is a time constant equal to 1.013445269E+25 seconds, ~3.2E+17 (320 quadrillion) years.

So how many years does the innermost boundary of 600 AU represent? \(\tau\) x (600/262,000)^{4} = 2.75 x 10^{13} years

By that point, their relative proximity shrinks from 29 AU to 0.0664 AU (10 million km). A 5 Earth mass planet would should as small disk from that distance, if there was a light source.

At the ISCO any object is orbiting at 0.5 c – and if it drops towards the black hole it’ll hit c at the event horizon. The infall time is something under the speed of light travel time. About 200,000 seconds. From the Event Horizon to the Singularity is \(\pi.M/c\) which is about 157,000 seconds, or 44 hours.

One final factor to ponder is the tidal forces. The Roche Limit, as defined by the masses of the objects involved, shows that a self-gravitating mass \(m\) of radius \(r\) will find its surface gravity equal to the tidal forces caused by a larger mass \(M\) at distance of:

$$D = r.\left(\frac{m}{2.M}\right)^\frac{1}{3}$$

which, for \(M\) = 10^{10} solar masses and \(m\) = 5 Earth masses (and 1 solar mass = 332,950 Earth masses), gives a distance of \(r\) x 110,000. With a heavy iron core the planet’s radius is about 9,150 km. Meaning when the planet is within about 1 billion kilometres of the Singularity it’ll begin breaking up. As the Event Horizon is 30 billion km in radius, it’s a long way in the “Dark Side”.

Incidentally, space-time moves towards the Singularity at faster than the speed of light when it crosses the Event Horizon. At 1 billion km the infalling planet is moving at ~ 5.5 c. The speed is simply the Newtonian escape velocity for the radial distance from the central mass. When really close to the Singularity, the tidal forces rise with inverse 1/3 power, but Hawking Radiation with the inverse 1/6 power, rapidly vaporising any composite object as it approaches the Point of Oblivion.

As always, great thoughts, Adam!

Question about relative times vis a vis external observer time: you gave 2.75e13 years of inspiral, but for a person on a planet orbiting at the ISCO how long would they perceive that time to be? Half c only has a small gamma factor, but given the time scales involved… also, wouldn’t frame-dragging contribute to relative passage of time for this hypothetical foreverworld?

Hi Josh

As I noted, I’m assuming a low spin factor SMBH, so the ISCO is at 3 R

_{s}, rather than up close to the Event Horizon. The Thorne Limit means astrophysical black holes only have a spin-factor of ~0.998 at max, which is a time-dilation factor of about 12. If we want really extreme time-dilation, then the spin-factor must be a lot closer to 1. Miller’s Planet in “Interstellar” had a spin-factor \(\alpha\) = 1-(1.3E-14) to have a time-factor of 7 years to 1 hour.Ahh right, frame dragging wouldn’t factor into things in this scenario. I was thinking of Stephen Baxter’s Xeelee using a SMBH as a relativistic “refuge”. The SMBH (Sag A* ) in his stories must have been spinning at the high pain factors you mentioned

Thanks for the response!

Hi Josh

The Xeelee Ring around Chandra was spinning very close to c to provide centrifugal gravity and increase the tangential length of the habitat thanks to relativistic dilation.

Hi again. I’ve just added a discussion of the tidal forces and infall speed, beyond the Event Horizon. Speeds and forces get really extreme.