<p>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.</p>
<p>For a black hole, the event horizon's physical extent is defined as the Schwarzschild radius, <span class="math-container">$r_s$</span>,  given by the equation:</p>
<p><span class="math-container">$$r_s = \frac{2GM}{c^2}$$</span></p>
<p>where M is the black hole mass, c is the speed of light, and G is the gravitational constant.</p>
<p>Therefore, given the above equation, the angular extent (in radians) of the black hole's event horizon is given by:</p>
<p><span class="math-container">$$\theta = 2tan^{-1}(\frac{GM}{c^2d})$$</span></p>
<p>where the factor 2 comes from moving from radius to diameter.</p>
<p>Since you have fixed the mass of the black hole to <span class="math-container">$10^8M_\odot$</span>, its angular extent is purely a function of observer distance from the black hole, as plotted below (in units of degrees):</p>
<p><a href="https://i.sstatic.net/Hf8nQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Hf8nQ.png" alt="Angular extent against observer distance" /></a></p>
<p>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!</p>
<p>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!</p>
<p>PS - This answer disregards the accretion disc seen around the black hole and only concentrates in the angular extent of the event horizon.</p>

CC BY-SA 4.0
2020-10-15T16:55:56.997Z
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)
hirschme
CC BY-SA 4.0
2020-10-15T17:05:05.440Z
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.
simonp2207
CC BY-SA 4.0
2020-10-15T20:55:15.097Z
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.
David Hammen
CC BY-SA 4.0
2020-10-16T01:54:53.820Z
The spin of the BH in *Interstellar* is inconsistent.  According to https://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 $a=0.6$.
PM 2Ring
CC BY-SA 4.0
2020-10-19T00:20:00.553Z
@simonp2207 I am still not sure if this answer defends the sizes seen in the movie/image. According to the graph, even for an angular size of 20 degrees (much less than the image) we would still be around 10 AU far away. For an angular size of ~90 degrees we would be maybe even < 1 AU which seems unrealistically close. Also in the answer you give the example fo standing on Pluto, a BH on the sun would be 4 x larger than the moon, which is still a very small angular size. Does this all hint towards the black hole in the image being way too large for a realistic distance?
hirschme
CC BY-SA 4.0
2020-10-15T16:47:19.403Z
2020-10-16T09:56:11.747Z
2020-10-16T09:56:11.747Z
simonp2207
<p>I assume everybody is tired of reading questions derived from the movie Interstellar, I will try to keep this short and simple:</p>
<p>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.</p>
<p>One striking example of this that has seems counter-intuitive, is the angular size displayed for the black hole &quot;Gargantua&quot; 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..!)</p>
<p>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.</p>
<p>It seems to me that a supermassive black hole would have such immense gravitational effects, that in order to be able to &quot;freely&quot; 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).</p>
<p><strong>The question is: Realistically, for a supermassive black hole (&quot;with a mass of 100 million times of the sun&quot;), what would be roughly its angular size when standing at a &quot;safe&quot; distance from it?</strong></p>
<p><a href="https://i.sstatic.net/9Sykl.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9Sykl.jpg" alt="enter image description here" /></a></p>

<black-hole><supermassive-black-hole><angular-diameter>
Is the angular size of the black hole in the movie "interstellar" completely overblown?
hirschme
# Is the angular size of the black hole in the movie "interstellar" completely overblown?

(Asked by: username_0 on 2020-10-15)

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](https://i.sstatic.net/9Sykl.jpg)](https://i.sstatic.net/9Sykl.jpg)

**Tags:** black-hole, supermassive-black-hole, angular-diameter


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## 1 answers

### Answer by username_1 on 2020-10-15. Score: 1

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](https://i.sstatic.net/Hf8nQ.png)](https://i.sstatic.net/Hf8nQ.png)

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.

> **Comments:**
> **username_0**: 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)  
> **username_1**: 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.  
> **username_2**: 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.  
> **username_3**: The spin of the BH in *Interstellar* is inconsistent.  According to https://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 $a=0.6$.  
> **username_0**: @username_1 I am still not sure if this answer defends the sizes seen in the movie/image. According to the graph, even for an angular size of 20 degrees (much less than the image) we would still be around 10 AU far away. For an angular size of ~90 degrees we would be maybe even < 1 AU which seems unrealistically close. Also in the answer you give the example fo standing on Pluto, a BH on the sun would be 4 x larger than the moon, which is still a very small angular size. Does this all hint towards the black hole in the image being way too large for a realistic distance?  

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