I'm considering a story where an exploration is devised to explore a binary system containing a black hole; the choice is Cygnus X-1, with its companion supergiant star HDE 226868. It would likely be dangerous to orbit the black hole, as the accretion disk means that there will be high emission of potentially dangerous radiation (it's also full of extremely hot gas). Therefore, the explorers are orbiting the secondary, HDE 226868.

I'm wondering, though: Is it safe to orbit this star? The two objects may only be 0.2 astronomical units apart - one fifth Earth's distance to the Sun - and the star is undergoing heavy mass loss, as many supergiant stars do. Can the explorers orbit HDE 226868 without being exposed to high levels of radiation or other damaging effects? If so, how close could they orbit (the orbit must be stable over a period of at least one year, with minimal maneuvering by the ship)?

I'm assuming that the protective technology for the spacecraft they're traveling in is similar to our own, although obviously some advanced form of propulsion was required to get there. Additionally, the ship can orbit the entire system, if it has to, but preferably just the star.

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    $\begingroup$ I was going to ask something like this sooner or later, and I promise I didn't ask this because my username is HDE 226868. I simply chose a well-known binary system (Cygnus X-1) that contains a large, massive companion star. $\endgroup$ – HDE 226868 Feb 22 '17 at 14:53
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    $\begingroup$ So to clarify. 1) Ignore travel time and 2) At what range could they orbit given modern radiation shielding for space craft? $\endgroup$ – James Feb 22 '17 at 14:58
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    $\begingroup$ Admit it, you just want to learn how to dance. $\endgroup$ – PatJ Feb 22 '17 at 14:58
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    $\begingroup$ My question: does the original HDE 226868 also have a diamond? Orbiting a supergiant star with mod privileges requires a great deal of care. Alternatively, perhaps the original star doesn't have mod privileges, but HDE 226868 in his spaceship does have mod privileges? I want to see someone mod-hammer a star! $\endgroup$ – Cort Ammon Feb 22 '17 at 16:24
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    $\begingroup$ @CortAmmon I assume that the original HDE 226868 has a planet hanging around nearby that has all the right conditions for diamond mines. The star itself wouldn't have anywhere to store its diamonds, right? $\endgroup$ – Monica Cellio Jul 31 '17 at 3:11

X-ray radiation at the orbit of HDE226868

Cygnus X-1 is famous as one of the most powerful X-ray sources in the sky. According to the US Naval Observatory, the max flux of Cygnus X-1 (near the bottom of the last page on the link) is 1.2672 Crabs in the 2-10 keV range. 1.2672 Crabs is equal to $3.04\times10^{-11} \text{ W/m}^2$.

I have the distance to Cygnus X-1 at ~1900 parsecs or $5.9\times10^{19} \text{m}$, while the distance of HDE 226868 is $3.0\times10^{10} \text{m}$ away. Using inverse square law ratios, I calculate that the x-ray flux at the distance of HDE 2268868 is $$\frac{(5.9\times10^{19})^2}{(3.0\times10^{10})^2} = 3.9\times 10^{18}$$ times stronger, or 117 MW / m$^2$. That doesn't sound so good.

How extremely fast that will kill you

A rad is equal to 0.01 J deposited into 1 kg. A thousand rads will kill you. Lets say the human body is 100 kg with a uni-directional surface area of 0.5 m$^2$, and about 20cm thick. The human body is mostly water, which has a half-value layer (HVL) for 300 keV radiation of 5.823 cm (Data from here, last page). This means that if a person is 20 cm thick, they abosorb 4 half-layers of radiation; alternatively we can say that $1/2^4 = 0.0625$ of the radiation escapes, or about 93% is absorbed. This is for 300 keV, which is much higher than the 2-10 keV we would actually be seeing. Those lower energy x-rays are more likely to be absorbed, we can assume that all incident x-rays are absorbed in the human body.

The HVL of lead is about 0.16 cm (again for the relatively high 300 keV radiation). In order to drop 117 MW to 0.117 W per meter squared, you need nine orders of magnitude, or 1.4 cm of lead. This will get your only about a quarter of a rad per second. That will get you to the radiation poisoning level of ~200 rad in about 13 minutes. Six more orders of magnitude gets you about 24 years before radiation poisoning with about 2.4 cm of lead.

So far so good! We can shield our people!

Now lets look into what 117 MW is doing to that lead. The density of lead is 11340 kg / m$^3$ and a square meter of lead, 2.4 cm thick is 0.024 cubic meters. The heat capacity of lead is 128 J/kg K. Each 1 m$^2$ surface are of lead hull weighs 272 kg and takes 35 kJ to raise by 1 degree kelvin. In order to protect us from the x-rays, this lead barrier must absorb all the x-ray energy. So the hull of our ship will increase by about 3342 K per second as it provides enough shielding to protect us from X-rays.


Without wasting any more time calculating black-body radiation and such, one can assume that the x-ray heating from the black hole will slag (and plasma!) anything in the orbit of HDE226868.

If the most powerful x-ray source in the sky is over 6000 light years away, it is good to continue being 6000 light years away from it.

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    $\begingroup$ That's . . . not great for my would-be explorers. I haven't checked all your calculations yet, but most seem to be spot-on. Thank you. $\endgroup$ – HDE 226868 Feb 24 '17 at 14:29
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    $\begingroup$ OK, but is this source radiating equally in all directions? If this is directional, maybe orbiter could just avoid it? $\endgroup$ – Mołot Feb 24 '17 at 14:44
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    $\begingroup$ @Mołot Possible, but Cygnus X-1 is variable when observed from Earth. According to the link above the lowest x-ray flux is about an order of magnitude lower than the highest. If you assume some directions get an order of magnitude less radiation, that will only change the number of seconds before your space ship evaporates. $\endgroup$ – kingledion Feb 24 '17 at 14:47
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    $\begingroup$ @Mołot From Earth, the x-ray source varies between 'hard' and 'soft' states, and in x-ray luminosity by about an order of magnitude. The more dangerous soft x-rays have the higher luminosity. I got most of my background info here. Basically, it is possible that Cygnus X-1 has a narrow beam, but if that is the case we have been in that beam since the 1964 when Cygnus was first detected. Otherwise, X-1 emits in all directions and varies by an order of magnitude or so. $\endgroup$ – kingledion Feb 24 '17 at 15:05
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    $\begingroup$ I am still thinking about this but there are enough comments under my answer. I see from these numbers that there is too much radiation at the distance of the star. Could the star itself block this radiation, shadowing a ship orbiting the dual system at the Lagrange point of the star? $\endgroup$ – Willk Apr 23 '17 at 18:55

I think you can orbit the star. I do not think the jets from the black hole are hitting the star and I do not think the accretion disk reaches that far; in any case it is a plane.

You can orbit such that the mass of the star is interposed between you and the black hole, if you are worried about unpredictable radiation from the black hole or distant consequences of the accretion disk. This orbit would be at the L2 Lagrange point (if the blue one in the picture is the star) or the L3 (if the yellow one is the star).

enter image description here

The mass stream from the star to the black hole occurs between those 2 bodies: stay out of the way of that. Radiation from the black hole would be bad, but from the viewpoint at the L2 Lagrange point, the black hole is eclipsed by the star, which shields you.

ADDENDUM In my rediscovered enthusiasm for this concept, and after reading the deleted answer by @Youstay Igo I wondered, notwithstanding the hole, how hot it would be from just the star at the Lagrange point.

I found a Lagrange Point Calculator. http://orbitsimulator.com/formulas/LagrangePointFinder.html Here are the values I put in and the distances of the various points.

enter image description here

I put the star at 23 and the hole at 14. That means the L3 point would be shaded from the hole by the star. That L3 is only 0.35 AU from the star. Mercury is 0.39 AU from our much less energetic sun.

I found an article estimating how close the space shuttle ("similar to our own") could come to our own sun without cooking.
from http://www.popsci.com/science/article/2010-07/how-close-could-person-get-sun-and-survive

Riding in the space shuttle, though, someone could get much closer to our star. The ship's reinforced carbon-carbon heat shield is designed to withstand temperatures of up to 4,700° to ensure that the spacecraft and its passengers can survive the friction heat generated when it reenters the atmosphere from orbit. If the shield wrapped the entire shuttle, McNutt says, astronauts could fly to within 1.3 million miles of the sun.

13 million miles is 0.015 AU. I had a hard time finding how much more energy than the sun HDE 226868 puts out; O type blue supergiants are very hot. 20,000x the sun is the low end. Maybe multiplying it out is too simplistic, but 20,000 * 0.015 = 300 AU. So 300 AU proximity to this giant star = 0.015 AU to the sun. That is 1000x farther than the L3 Lagrange point!

Maybe the explorers would be better off at L2 in the cool shade behind the black hole. At least the hole does not kick out the thermal energy like that. They can bring osmium shielding against the hard radiation.

ADDENDUMUM How to orbit at L2 in full view of the hole when, as per @kingledion, "So the hull of our ship will increase by about 3342 K per second as it provides enough shielding to protect us from X-rays.". I am thinking aikido - redirect your opponents momentum. Let us use xrays to negate xrays.

Xray diffraction turns on the priniciple that some crystals absorb and re-emit xrays such that there is constructive and destructive interference between the rays. Areas of constructive interference have much more radiant energy. Areas of destructive interference, much less. Ideally, there is no net loss of xray energy (as heat!) - it is just a reallocation of energy.

enter image description here

enter image description here

I propose that shield made of a crystal with these xray diffracting properties could be used to reroute the xray energy, allowing it to travel by without heating up the shield, ship or explorers. The explorers and ship, needless to say would be hiding in one of the dark areas of destructive interference.

This would take advantage of the shade of the black hole, absorbing the radiance of the star. It sidesteps the problem of xrays from the black hole by routing them around the ship.

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    $\begingroup$ "You can orbit such that the mass of the star is interposed between you and the black hole" - could you maybe make a drawing of this? As far as I can imagine it's not possible, but maybe you have in mind something I can't imagine... Also, if Earth was loosing mass to the Sun, it would be from the equator, as far as we know. Shortest path, least resistance, changed only by rotation (and indeed these two probably rotate roughly in the same plane) $\endgroup$ – Mołot Feb 22 '17 at 15:53
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    $\begingroup$ @Molot If you orbit at the L3 point of the Star-BlackHoleOfDeath system you would always have the bulk of the star between you and the BHOD. $\endgroup$ – Werrf Feb 22 '17 at 16:38
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    $\begingroup$ Can you back up some of your assertions? Mass transfer will not necessarily happen along the equator; it is likely to happen in a sort of bulge in the Roche lobe. Additionally, mass loss may be isotropic. $\endgroup$ – HDE 226868 Feb 22 '17 at 18:00
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    $\begingroup$ @Mołot I'd be interested in seeing exactly the shape of such a wind. At the moment, though, the answer doesn't address any of this but just focuses on the disk, which is not the only issue. $\endgroup$ – HDE 226868 Feb 23 '17 at 0:25
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    $\begingroup$ Maybe this will help undo those pesky minus votes. en.wikipedia.org/wiki/Lissajous_orbit which elaborates on what @Werff said up there: use star as shield. A star equals many, many cm of lead. I was pleased to read that Arthur C Clarke used this in a story: an orbit at the Lagrange point of earth at which the sun is permanently eclipsed. If the minus votes are because I do not know the shape of the wind that is harder. But I cannot think of gravitational mechanics in which gravity from the black hole pushes wind from the star out past the L2 Lagrange point. $\endgroup$ – Willk Feb 26 '17 at 13:27

I like both of the provided answers, but I'm going to add some information about the star and the stellar wind itself and build on the these two answers.

To start I think it is safe to say, based on kingledion's answer, that the x-ray emission from the black hole would roast any spacecraft that orbited near the star with an unobstructed path to the black-hole. This means that the only "safe" orbit, with respect to x-ray emission is going to be the corresponding Lagrange point, as mentioned by Will's answer. This point will protect the spacecraft from the black-hole's x-rays by always keeping the star between it and the black hole (basically a 6 million mile shield). But what about the stellar wind?

Well first let's put up some numbers;

Our sun's stellar wind (mostly called the solar wind) typically varies between 400-750 km/s, depending on if it's the fast or slow solar wind. Type O and B stars stellar wind is much faster, approaching ~2000 km/s. For simplicity sake, let's call it 4x faster.

The mass loss rate for a typical O/B type star is on the order of $10^{-6}$ solar masses per year (roughly $2\times 10^{24}$ kg/year or ~$6\times 10^{16}$ kg/s). That is 100,000,000x more than our sun! For comparison sake, that is even more than a typical coronal mass ejection from our sun.

So we are dealing with a stellar wind that is 4x faster with $10^8$x more mass than our solar wind.

Now let's add the black-hole. Typically a star's stellar wind is "roughly" spherically symmetric. This is different for a G type star (our sun) because of the corona, but for an O/B type star we can call it symmetric. the gravitational pull acts as a focus, producing a non-spherically symmetric stellar wind (with more wind being directed towards the black-hole, and less towards the spacecraft). The x-rays will also energize the stellar wind, but that shouldn't play a part for our spacecraft if it is sitting at the L3 point. Unfortunately I can't think of a way to calculate the anisotropy of the solar wind due to the black-hole, but let's say that it has the affect of halving the solar wind characteristics in the L3 direction (highly doubt it has that much of an affect, but for arguments sake let's go with it).

So now let's see where we are at. We have a spacecraft at approximately the same distance from HDE226868 as Mercury is from our sun. With a stellar wind that is 2 times faster than our own, with 50,000,000 more mass. Now luckily we just had a satellite (MESSENGER) that took some great data from the EPPS instrument about the energetic plasma and from the GRNS instrument for galactic rays. Looking at the data from the satellite, it seems that a stellar wind with an average energy content $2\times 10^8$ times more than our own solar wind would cause irreparable damage to both humans and instruments aboard the spacecraft. In addition the massive dynamic pressure of the stellar wind would require constant adjustments to the spacecrafts orbit to maintain a constant orbit.


It seems to me that you are going to need to orbit at quite a distance from both the black-hole and the star if you want to remain alive with today's available technology. One caveat is that stellar wind can be a lot easier to deflect than x-rays. Instead of large shielding, if your spacecraft were able to generate a significantly large magnetic field it could shield itself in the same way that Mercury's magnetic field shields it.

  • $\begingroup$ Thanks for weighing in on this, Miles. I like your answer especially because you walk thru the math and background reading. @Kingledion's is good too but the 500 will get you more that it will him. $\endgroup$ – Willk Aug 2 '17 at 19:23

Being closer doesn't necessarily mean seeing better. Not only will you be affected by Glare, the high-energy radiation will cause a phenomenon similar to Photokeratitis to your sensors as well.

Of course, the first approach would be to filter the incoming rays, but simply moving to the right orbit in a far distance has almost the same effect and is easier. You'd have to account for the light diffraction by gravity in this case.

With no interference to distort the light but the creating objects themselves, distance should be no hurdle (unless of cause you want resolutions down to the kilometer scale), and since you probably want to observe the whole system and its interaction, not being bound to the sun will probably be a positive factor as well.

At any rate, I don't see any remotely realistic approach to survive in an orbit around the star. The final distance for which you settle is determined as follows:

  • maximal possible distance: magnification power of your telescope
  • minimal possible distance: radiative equilibrium that puts you on a reasonable temperature
  • optimal distance: Between parameter 1 and 2, based on how resilient your sensors are, and determined by how long you want your measuring instruments to remain functional
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    $\begingroup$ What has a telescope to do with safety and survivability? $\endgroup$ – L.Dutch Aug 2 '17 at 5:34
  • $\begingroup$ “cause”? I think typo for “course” but you used it repeatedly. $\endgroup$ – JDługosz Aug 2 '17 at 6:30
  • $\begingroup$ If I had a telescope capable of infinite magnification, why would I even bother setting foot out of my solar system? $\endgroup$ – Sudix Aug 2 '17 at 7:00
  • $\begingroup$ @Sudix People are capable of viewing any country in the world via Google Maps, but they still go on holiday to those countries, because looking at them through a screen isn't the same as actually being there. $\endgroup$ – F1Krazy Aug 2 '17 at 13:44
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    $\begingroup$ @F1Krazy Yet I doubt anybody would willingly go wade through lava if you could instead watch it from a few meters away. If you want to observe that system for a longer time span, you'll need the maximum possible distance between you and it, which is merely limited by the fact you actually want to observe it. I've added two more problems to the huge list, and as, given the prerequisites, there is no solution, advised an altered base scenario $\endgroup$ – Sudix Aug 2 '17 at 14:00

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