7
$\begingroup$

Assuming that the only fancy available technology is FTL drives (to get there) plus anything needed to protect and sustain a crew during the journey from Earth to Sagittarius A (whether it takes a few days or centuries is irrelevant to the question).

Ship and crew have no way to counteract or protect themselves from the temporal and gravitational effects of the black hole on the spacetime continuum.

How close could a ship come to Sagittarius A before being in danger of destruction?

$\endgroup$
  • $\begingroup$ I got a good answrr for this but it's going to take a lot of time to type. In short, I got the tensile strength for skin and tendons from this XKCD What If. We need to find out Sagittarius A*'s mass and the distance from it in which tidal forces will rip human skin apart (supposing an average human's measures). If anyone runs the calc and answers, they'll have my upvote and a bounty. $\endgroup$ – Renan Nov 2 '18 at 11:43
  • $\begingroup$ I'm going to answer this assuming that you are want to know how close you can get to Sagittarius A*, the compact radio source within Sagittarius A. $\endgroup$ – kingledion Nov 2 '18 at 13:43
  • $\begingroup$ @Renan The other important question would be how well the ship would hold together. We'd need to know its tensile strength and how big it is in the direction perpendicular to the black hole. $\endgroup$ – David Thornley Nov 2 '18 at 17:40
  • $\begingroup$ The tidal forces aren't a problem -- the bigger the black hole, the weaker the tidal force near its event horizon. And Sagittarius A* is several millions of times more massive than the Sun. $\endgroup$ – TimeTravellyParadoxySciFiSmeg Nov 3 '18 at 0:33
  • $\begingroup$ @DavidThornley that is unanswerable without k owing the ship's shape and what it is made of. $\endgroup$ – Renan Nov 3 '18 at 2:01
9
$\begingroup$

Note: The numbers hae been edited due to access to a better luminosity source

How powerful an emitter is Sgr A*?

First, check out this answer. I'm going to lean on those calculations here.

Second, Sagittarius A is a large object with many components. The part that will kill you is the (probably) supermassive black hole Sagittarius A*. Like Cygnus X-1 in the other answer, the most relevant emissions from our perspective will be hard x-rays. Peak energy flux from Barriere, et al., 2014 is in the 2-20 keV range, which is consistent with the energies emitted by Cygnus X-1, and with the estimates I made in that post.

Luminosity in the X-ray range is $~1\times10^{25} \text{ W}$, but peaks during flare period around $~1\times10^{28} \text{ W}$. We will us the higher number for safety. The challenge with X-rays is that they tend to penetrate things, they are hard to reflect and they kill humans in low doses. We need the flux to be low enough not to kill humans.

X-ray flux is the power emitted by the object divided by the surface area of a sphere at a distance $r$ from that object. Its equation is (letting $4\pi \approx 10$)

$$\Phi=\frac{1\times10^{27}\text{ W}}{r^2}.$$

What hard X-ray energy will not kill humans?

Radiation exposure is tricky, since some body parts are more sensitive than others. In the US, the Department of Energy radiation worker annual dose limit is 50 mSv, roughly equivalent to 5 rads of energy. Lets say it is allowable for the crew of our vessels to get the whole annual dose in just 5 days; so 1 rad per day. 1 rad is 1 J of hard X-ray energy into a 100 kg person. On a per second rate, this means a person can be exposed to a maximum of about $1\times10^{-5} W$ of continuous hard X-ray radiation to remain safe from radiation poisoning for a week.

How far away do you have to be to avoid this dose, unshielded?

Lets say a person has a surface area of $1 \text{ m}^2$. We then solve

\begin{align}\Phi=&\,\frac{\text{allowable dose}}{\text{surface area}}\\ \frac{1\times10^{27}\text{ W}}{r^2} =&\, \frac{1\times10^{-5}\text{ W}}{1 \text{ m}^2}\\ r =&\, 1\times10^{16} \text{ m} \end{align}

This is about 70,000 AU or 1 light year. Yowsers!

How far do we have to be to avoid this dose, shielded?

Lets say that we can reflect 90% of the incident X-ray radiation. Now, lets say that we can shield ourselves from much more of it with a thick hull. The problem (as noted in the other answer) is hat the thick hull will be heated by the X-ray flux that we can absorb. So we have to estimate how much heat energy we can re-emit as waste heat, to figure out how much energy we can continuously absorb from Sgr A*.

An expression of the Stefan-Boltzmann law in terms of power per unit area of emission is $$\frac{P}{A} = \epsilon\sigma T^4$$. The emissivity of a material is its ability to emit thermal radiation as blackbody radiation. Lets set $\epsilon=0.99$ (ice is $\epsilon=0.97$; ship made of ice?). The Boltzmann constant is $\sigma=5.67\times10^{-8}\text{ W m}^{-2}\text{K}^{-4}$. Lets say that the ship's hull and heat sinks can handle being heated to 1000 K and still maintain that high emissivity.

For every unit of surface area the ship receives of incoming radiation, there is approximately one unit of area for blackbody radiation out. So we can set up an equality to see what X-ray flux the ship can handle. Let $a$ be the ship's 'albedo', which is the amount of X-ray energy reflected.

\begin{align}a\Phi =& \, \epsilon\sigma T^4\\ \frac{1\times10^{26} \text{ W}}{r^2} =&\, 0.99\cdot5.67\times10^{-8}\cdot(1000)^4\text{ W m}^{-2}\\ r=&\,4\times10^{10} \text{ m} \end{align}

This is 0.3 AU, or a little less than distance from the Sun to Mecury. Note that the estimated mass of this black hole is 4 million solar masses. The Schwarzchild radius of the black hole is only a little less than this at abot $1\times10^{10} \text{ m}$. Since the energy from a black hole comes from its accretion disk, you can't consider the black hole to be a point source of X-ray energy at this close distance.

Conclusion

An unshielded human being can't get anywhere near a luminous X-ray source like Sgr A*, but with an assumption of moderately advanced technology, a ship that is designed to shed hat efficiently and that has X-ray reflection built in could get as close as Mercury is to the Sun. The geometry of the emissions from he accretion disk and the event horizon (which should definitely also be avoided) means that it is hard to pin down exactly how close you could get.

Also, this would be at the upper limit of what the ship could handle. A safe orbital distance around such a supermassive black hole would be much farther away.

$\endgroup$
  • 3
    $\begingroup$ Nice answer. I think you could also argue that the emission (coming from the accretion disk) might not be isotropic (and honestly probably isn't), so depending on the angle of approach, things might be more conducive or less conducive to a close pass. $\endgroup$ – HDE 226868 Nov 2 '18 at 15:31
  • $\begingroup$ Very true! I was under the impression that an accretion disk's radiation is highly directed, so if that is the case, my calculation is either an undershoot or overshoot by orders of magnitude. But I don't really know any way to correct it. $\endgroup$ – kingledion Nov 2 '18 at 15:35
  • $\begingroup$ Revnivstev, et al., 2004 estimate seems to reflect Sagittarius A*'s state in the (surprisingly recent) past, when it was an active galactic nucleus. "Normal" activity for Sgr A* seem to be orders of magnitude lower (3.6×10e33) NuSTAR DETECTION... $\endgroup$ – Alexander Nov 2 '18 at 17:19
  • $\begingroup$ @Alexander Super! I updated my luminosity source and the resulting calculations. $\endgroup$ – kingledion Nov 2 '18 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.