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I'm imagining a situation that we can see a star change color periodically:

  1. A star and Earth orbit around a black hole

  2. The star is very near to the black hole so that it has high orbital speed, while the Earth orbits the black hole at far away

  3. As the star has a high orbital speed, when the tangential direction points towards the planet, it causes blue shift and appear as a blue sun, and when it moves away from the planet, it looks like a red sun due to red shift.

Is such a system possible?

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  • $\begingroup$ Red or blue shifts move the spectrum of a star but don't generally make it actually look red or blue, because there's also plenty of infrared or ultraviolet radiation that gets shifted into the visible spectrum to replace the visible light that's been shifted. $\endgroup$ – Mike Scott Jun 23 '17 at 5:38
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    $\begingroup$ @MikeScott Red shifted stars and galaxies really do look red. Look at the curve for 6400K — it has a distinct peak and you can see that shifting it will be very visible. $\endgroup$ – JDługosz Jun 23 '17 at 5:39
  • $\begingroup$ Is the star within the accumulation disc of the black hole? $\endgroup$ – L.Dutch Jun 23 '17 at 5:42
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    $\begingroup$ Your question boils down to “how fast can the sun orbit a BH without running into its Roche limit”. A few regulars here might be able to answer that. then see how it combines with the visible gravitational red shift. $\endgroup$ – JDługosz Jun 23 '17 at 5:44
  • $\begingroup$ @L.Dutch you mean accretion disk ? $\endgroup$ – JDługosz Jun 23 '17 at 5:47
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Please correct me if I am wrong, I just did some back-of-the-envelope calculations without thinking everything through. I do not know how humans would perceive the shifted spectrum, but let's assume it's a linear relationship which I think is the best we can hope for:

Doppler effect: (f/f_0 - 1 ) * c = v .

So if you want to notice anything, let's say f/f_0 = 0.8 ~ yellow to red, you need speeds close to light speed, in that example 0.2 * c.

How fast is your star? https://www.space.com/20303-black-hole-star-speed-record.html

2 million km/h. That means: 555,555 m/s = 0.001853 * c.

Unless you are ok with very subtle color changes ...

Oh, yes, maybe in theory a star could rotate faster, but would an "Earth" exist there? Calculating under which circumstances a faster rotating star could exist and a planet and so on is I think beyond this forum, I would go with what kind of data we have

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  • $\begingroup$ Is there another way that a star could have a color shift change, preferably with a frequency easily observable by a human being? $\endgroup$ – Neil Jun 23 '17 at 8:25
  • $\begingroup$ @Neil This goes beyond the scope of the question, but please note that the sun changes color drastically over the day. It's mostly yellow, but then in the evening, it turns into this beautiful red. $\endgroup$ – Raditz_35 Jun 23 '17 at 8:28
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You may use the relative speed between Earth and star, whether the star is moving toward Earth, to determined the color people on Earth see. According Doppler effect, if the light source (in this case, light also wave) move toward Earth, then on Earth, people will see the light with higher frequency (toward violet). But when the star move away from Earth, people see the star orange or red (low frequency)

You may refer to Visible spectrum to define it based color, a color a person can see the star if they are on the spaceship which relative station to the star. A color when star move further earth (reduce base freq) and a color when a star move toward earth (increase base freq)

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  • $\begingroup$ We know doppler is a thing. Can the star orbit fast enough to have the needed amount of shift? Also read the comments on the question. $\endgroup$ – JDługosz Jun 23 '17 at 21:54
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The short version: yes, but you might need a really big black hole.

In order to get a large spectral shift, you want as high an orbital velocity as possible for the star. To get the highest possible orbital velocity, you want it to orbit the black hole as close as possible. For a rotating black hole, you can actually set up stable circular orbits at arbitrarily small distances from the event horizon if the BH's angular momentum is large enough, but to keep things simple we'll just consider a basic Schwarzschild BH with no angular momentum. In that case, the innermost stable circular orbit occurs at a radius of $R = \frac{6GM}{c^2}$ - twice the radius of the photon sphere, and 3 times the radius of the event horizon. Calculating the proper orbital period and velocity is a bit trickier, but it turns out that for an external observer (which is what's relevant for the view from the planet), close circular orbits around a non-rotating black hole still obey Kepler's laws exactly, so the orbital velocity is given by $v = \sqrt{\frac{GM}{R}}$. Substituting in the previous equation for $R$, the factor of $GM$ cancels out, and we get $v = \frac{c}{\sqrt{6}}$. Thus, the maximum orbital velocity that the star can attain around a non-rotating black hole is independent of the black hole's mass! And it's a little bit more than $0.4c$, which is enough to shift yellow light way off into the infrared or ultraviolet. Half that velocity, at double the minimum orbit radius, would be plenty to produce obvious visible effects, with a yellow star shifting between red and violet over the course of half an orbit (assuming that the observer is in or near the star's orbital plane).

Now, we just need to make the black hole big enough to make these fast orbits wide enough that a star will fit, without overlapping the black hole or being torn apart by tides. For that, we can use the fluid satellite approximation for the Roche limit, $d = 2.44R_{BH}(\frac{\rho_{BH}}{\rho_S})^\frac{1}{3}$ We need that value to be smaller than the orbital radius, so $2.44R_{BH}(\frac{\rho_{BH}}{\rho_S})^\frac{1}{3} < \frac{6GM}{c^2}$. We can then simplify that to get:

$2.44\frac{2GM}{c^2}(\frac{\rho_{BH}}{\rho_S})^\frac{1}{3} < \frac{6GM}{c^2}$

$2.44(\frac{\rho_{BH}}{\rho_S})^\frac{1}{3} < 3$

$2.44(\frac{3c^6}{32\pi G^3M_{BH}^2 \rho_S})^\frac{1}{3} < 3$

And then solve for the mass of the black hole:

$2.44(\frac{3c^6}{32\pi G^3 \rho_S})^\frac{1}{3} < 3 M_{BH}^\frac{2}{3}$

$2.44\sqrt{\frac{3c^6}{32\pi G^3 \rho_S}} < \sqrt{27} M_{BH}$

$2.44\sqrt{\frac{c^6}{288\pi G^3 \rho_S}} < M_{BH}$

If we plug in the average density of the sun ($1410 kg/m^3$) for $\rho_S$, we end up with $M_{BH} > 1.07e38 kg$, or about 54,000,000 solar masses, with an orbital radius of about 5.5AU.

If we use double the innermost stable circular orbital radius, still producing plenty of Doppler shift, that cuts the necessary black hole mass down to a mere 19,000,000 suns, with an orbital radius of 2.25AU

For subtler color changes, you could get away with an even smaller black hole, and a smaller orbital radius for the star, but it's still going to be in the supermassive range.

We also need the distance between the event horizon and the orbit to be larger than the radius of the star, but that's pretty well covered. The sun's diameter is much less than 1AU. :) For reference, current estimates of the mass of the supermassive black hole at the center of the Milky Way are around 4,100,000 solar masses, but the SMBH at the center of Andromeda weighs in between 110 and 230 million solar masses, so this sort of object is well within the capabilities of our universe to produce.

Arranging for that star to be the primary source of life-giving heat and light for a planet would be pretty tricky, though, seeing as how the distance between them will vary by between 5 and 11AUs, and the minimum distance between the orbits of the star and the observing planet would likely need to be greater than 1AU to ensure the planet's orbital stability. Much more feasible to make the color-changing star a bright light in the sky, comparable to the Moon (which also exhibits cool Einstein-ring lensing effects when passing behind the BH), and provide another star for warmth.

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