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I assume everybody is tired of reading questions derived from the movie Interstellar, I will try to keep this short and simple:

In general in movies, in order to have stunning visuals, celestial objects are displayed as if they were extremely close to the observers, having a enormous apparent angular size.

One striking example of this that has seems counter-intuitive, is the angular size displayed for the black hole "Gargantua" in the movie Interstellar. The characters in the movie are so close to it that it can be fully displayed in great detail (see attached image: Black hole seems to cover almost/over 90 degrees of the field of view..!)

Now, we can fully experience gravitational pulls from much smaller objects than a supermassive black hole in reality. The angular size of the sun and the moon are so small that we need telescopes to appreciate details, despite their gravitational effects dominating our everyday life.

It seems to me that a supermassive black hole would have such immense gravitational effects, that in order to be able to "freely" navigate around it (as done in the movie), observers would be so far away that the angular size of the black hole would be tiny (which would result in very non-spectacular cinematography).

The question is: Realistically, for a supermassive black hole ("with a mass of 100 million times of the sun"), what would be roughly its angular size when standing at a "safe" distance from it?

enter image description here

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Assuming the spin parameter of the black hole is close to the critical value, it could be quite large.

The scenario of heating a planet orbiting close to a black hole via blue-shifted cosmic background radiation has been explored. The shadow of the black hole would take up a large fraction of the sky on such a planet, see for example figure 1 in Bakala et al. (2020). They find that for an Earth-like planet, the region where this can be achieved is located beyond the Roche limit (where the planet would be broken up) for black hole masses $M_{\rm BH} > 1.63\times10^8 M_\odot$. As shown in figure 6 of the paper, the bulk of this radiation will be in the ultraviolet, but there's probably enough visible light in the mix to make a bluish patch near the edge of the black hole shadow. Bear in mind that this would need to be a very isolated black hole, because stars would also be blue-shifted and therefore be extremely bright sources of high-energy radiation, which would be an extremely unpleasant environment.

The scenario in the movie includes an accretion disc, this scenario is considered in Schnittman (2019), who notes that for accretion rates low enough to produce the Sun-like colour of the radiation depicted in the movie, you'd have to be a long way out:

Regardless of the specific details—whether the planet is tidally locked or not, embedded in the disk or not—it is clear that for any planet to be habitable while also receiving its primary heating flux from a Novikov-Thorne type accretion disk, the planet either has to be extremely far from the black hole, or the mass accretion rate has to be an infinitesimal fraction of Eddington. This leads to an important inconsistency: The Novikov-Thorne model is predicated on an optically thick, radiatively efficient disk. At accretion rates down near $10^{-12} \dot{M}_{\rm Edd}$ the optical depth is unlikely to be large enough to allow the gas to cool, thus leading to a very hot, low-density accretion flow.

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  • $\begingroup$ Thank you for the answer. However I am not sure I understand the message: This seems to relate to a habitable planet orbiting a supermassive black hole, but in the question I am just referring to a moving observer (on a ship), no need for a heating planet in orbit? At last sentence claims "a long way out". What does this mean is the purpose of the question: How far out? And after answering that: How big would the black hole be after bein that "far out" ? $\endgroup$ – hirschme Oct 19 at 0:22
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First, to define the 'safe distance'. The most logical safe distance would be defined as being just outside the event horizon of the black hole which is the distance from the black hole at which not even light can escape. Sub-light speed orbits can still be stable outside the event horizon.

For a black hole, the event horizon's physical extent is defined as the Schwarzschild radius, $r_s$, given by the equation:

$$r_s = \frac{2GM}{c^2}$$

where M is the black hole mass, c is the speed of light, and G is the gravitational constant.

Therefore, given the above equation, the angular extent (in radians) of the black hole's event horizon is given by:

$$\theta = 2tan^{-1}(\frac{GM}{c^2d})$$

where the factor 2 comes from moving from radius to diameter.

Since you have fixed the mass of the black hole to $10^8M_\odot$, its angular extent is purely a function of observer distance from the black hole, as plotted below (in units of degrees):

Angular extent against observer distance

Therefore, with the black hole centred on our sun, the event horizon would be just inside the Earth's orbit. Meaning if the Earth was in the correct orbit, you'd be safe to observe the black hole. To put the angular extent further into context, if we were sat on Pluto at the farthest point in its orbit, the event horizon would be over 4 times larger than the full moon as seen from Earth!

So I'd say, given the apparent planetary orbits of Interstellar's exoplanetary system, Gargantua's extent is pretty realistic... Why there'd be a stable planetary system in orbit about a supermassive black hole in the first place is probably the more unbelievable bit!

PS - This answer disregards the accretion disc seen around the black hole and only concentrates in the angular extent of the event horizon.

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  • $\begingroup$ Thank you for an excellent answer. I am still missing the "safe distance" aspect to it. This answer implies then that the distance the observers are from the supermassive black-hole is nearer, than the distance from Pluto to a supermassive black-hole in place of the Sun. Is this a realistic distance to orbit? (your comments about a stable planetary system inside makes me think it is not) $\endgroup$ – hirschme Oct 15 at 16:55
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    $\begingroup$ The safe distance would be defined as any distance outside the event horizon, since inside the event horizon not even light can escape and therefore escape is impossible! I've edited the answer to include this. However, this doesn't include the safe distance from the accretion disc which is heated to millions of degrees! I'd have to do a few more calculations to establish the safe distance from that... Though accretion discs around black holes may be transient, appearing and disappearing as matter falls within the black hole's gravitational influence. $\endgroup$ – simonp2207 Oct 15 at 17:05
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    $\begingroup$ Re Sub-light speed orbits can still be stable outside the event horizon: Even photons can't orbit just outside the event horizon. The photon sphere for a non-spinning back hole is 1.5 times the Schwarzschild radius. For a particle with mass, the innermost stable circular orbit for a non-spinning back hole is 3 times the Schwarzschild radius. $\endgroup$ – David Hammen Oct 15 at 20:55
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    $\begingroup$ @DavidHammen - the black hole in the movie is supposed to have a spin parameter close to 1, which enables stable orbits much closer to the horizon. $\endgroup$ – antispinwards Oct 15 at 21:07
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    $\begingroup$ The spin of the BH in Interstellar is inconsistent. According to physics.stackexchange.com/q/148567/123208 Kip Thorne used a spin parameter of $a=1-10^{-14}$ (where 1 is the maximal spin), with the planet very close to the event horizon, to get the desired time dilation. (At such an extreme spin, the photon sphere is almost at the event horizon, and there are stable orbits for massive bodies that are almost as close). However, for visualisation purposes, they used a much less extreme spin of $a=0.6$. $\endgroup$ – PM 2Ring Oct 16 at 1:54

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