A galaxy with a halo and a stupendous central black hole
Click for original image.
Cool image time! The picture to the right, cropped, reduced, and sharpened to post here, was taken by the Hubble Space Telescope as part of a survey of galaxies that have what astronomers call active galactic nuclei (AGNs). This galaxy, dubbed IC 4709, is about 240 million light years away.
If IC 4709’s core were just filled with stars, it would not be nearly so bright. Instead it hosts a gargantuan black hole, 65 million times the mass of our Sun. A disc of gas spirals around and eventually into this black hole, with the gas crashing together and heating up as it spins. It reaches such high temperatures that it emits vast quantities of electromagnetic radiation, from infrared to visible to ultraviolet light and beyond — in this case including X-rays. The AGN in IC 4709 is obscured by a lane of dark dust, just visible at the centre of the galaxy in this image, which blocks any optical emission from the nucleus itself.
To get a very vague sense of scale, this supermassive black hole is more than sixteen times more massive than the relatively inactive supermassive black hole in the center of the Milky Way. This imagery and data from Hubble will help astronomers better understand the interaction between the black hole and its surrounding galaxy.
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Click for original image.
Cool image time! The picture to the right, cropped, reduced, and sharpened to post here, was taken by the Hubble Space Telescope as part of a survey of galaxies that have what astronomers call active galactic nuclei (AGNs). This galaxy, dubbed IC 4709, is about 240 million light years away.
If IC 4709’s core were just filled with stars, it would not be nearly so bright. Instead it hosts a gargantuan black hole, 65 million times the mass of our Sun. A disc of gas spirals around and eventually into this black hole, with the gas crashing together and heating up as it spins. It reaches such high temperatures that it emits vast quantities of electromagnetic radiation, from infrared to visible to ultraviolet light and beyond — in this case including X-rays. The AGN in IC 4709 is obscured by a lane of dark dust, just visible at the centre of the galaxy in this image, which blocks any optical emission from the nucleus itself.
To get a very vague sense of scale, this supermassive black hole is more than sixteen times more massive than the relatively inactive supermassive black hole in the center of the Milky Way. This imagery and data from Hubble will help astronomers better understand the interaction between the black hole and its surrounding galaxy.
The support of my readers through the years has given me the freedom and ability to analyze objectively the ongoing renaissance in space, as well as the cultural changes -- for good or ill -- that are happening across America. Four years ago, just before the 2020 election I wrote that Joe Biden's mental health was suspect. Only in the past two weeks has the mainstream media decided to recognize that basic fact.
Fourteen years ago I wrote that SLS and Orion were a bad ideas, a waste of money, would be years behind schedule, and better replaced by commercial private enterprise. Even today NASA and Congress refuses to recognize this reality.
In 2020 when the world panicked over COVID I wrote that the panic was unnecessary, that the virus was apparently simply a variation of the flu, that masks were not simply pointless but if worn incorrectly were a health threat, that the lockdowns were a disaster and did nothing to stop the spread of COVID. Only in the past year have some of our so-called experts in the health field have begun to recognize these facts.
Your help allows me to do this kind of intelligent analysis. I take no advertising or sponsors, so my reporting isn't influenced by donations by established space or drug companies. Instead, I rely entirely on donations and subscriptions from my readers, which gives me the freedom to write what I think, unencumbered by outside influences.
Please consider supporting my work here at Behind the Black.
You can support me either by giving a one-time contribution or a regular subscription. There are five ways of doing so:
1. Zelle: This is the only internet method that charges no fees. All you have to do is use the Zelle link at your internet bank and give my name and email address (zimmerman at nasw dot org). What you donate is what I get.
2. Patreon: Go to my website there and pick one of five monthly subscription amounts, or by making a one-time donation.
3. A Paypal Donation:
5. Donate by check, payable to Robert Zimmerman and mailed to
Behind The Black
c/o Robert Zimmerman
P.O.Box 1262
Cortaro, AZ 85652
You can also support me by buying one of my books, as noted in the boxes interspersed throughout the webpage or shown in the menu above. And if you buy the books through the ebookit links, I get a larger cut and I get it sooner.
Hah! I can be a scientist! That looks like a galaxy in the process of becoming a donut, just as I predicted here a while back.
It also raises questions about the definition of “singularity”. The word conjures up a point-like image. I highly doubt that 65 million Sols can be squashed down to a nearly dimensionless point. Whatever is in there, I think it’s rather large.
Mark–
General relativity falls apart inside the event horizon.
That being said, neutron stars are something like 7 miles across (and something like 1-2 solar masses) and are the densest objects before black holes.
While the IC 4709 supermassive black hole is indeed a big “smbh”—at 65 million solar masses it’s perhaps 15 times the mass of the Milky Way’s ‘A*’ with 4.3 solar masses—but even that is easily outclassed by more massive smbh’s elsewhere. The Andromeda [M31] Galaxy’s supermassive black hole ‘M31*’, for instance, has a mass of at least 110 million solar masses—or more than 25 times the mass of the Milky Way’s A*. But that’s hardly even the start, as one sees below in the case of the extraordinary smbh known as ‘TON 618’.
Let’s put some numbers on this hitherto qualitative assessment. It isn’t hard—there’s no calculus, no complex numbers—just a purely ordinary formula, much like E = Mc²!
The general relativistic (i.e., the modern “law of gravity”) formula for the so-called “Schwarzchild” radius of a black hole—the theoretical distance from the center of the black hole to its “event horizon” (that is, the invisible, immaterial, purely gravitational surface located at that distance away from the center, from whence—if one were to descend closer than which—there can be no subsequent backing out afterwards)—would be as follows:
R = 2GM ⁄ c²
For this equation, R = the resulting radius, M = mass of black hole, G = gravitational constant, c = speed of light. All units are “mks” (meter-kilogram-second). Thus, the Schwarzchild radius varies linearly by mass M.
Constants and units possess these values:
G (gravitational constant) = 6.6743 · 10⁻¹¹ {m³⁄s²·kg}
c (speed of light) = 299,792,458 {m⁄s} = ~3 · 10⁸ {m⁄s}
AU (astronomical unit) = 149,597,870,700 {m} = ~150 million {km}
1. Thus, for instance, if the sun were to somehow find itself compressed down into an extremely compact ball smaller than its theoretical event horizon—so that it would thereby become a black hole—the following computation would find that “stellar-mass” black hole’s size. The mass M in the foregoing black-hole radius equation would therefore be the sun’s, to wit:
Msun = 1.9885 · 10³⁰ {kg}
Thus, the calculation for the resulting Schwarzchild event-horizon radius for the black-hole-ified sun might proceed as follows:
R = 2 (6.6743 · 10⁻¹¹) (1.9885 · 10³⁰) ⁄ (3 · 10⁸)² {units as above}
R = 2.6544 · 10²⁰ {m³⁄s²} ⁄ 9 · 10¹⁶ {m²⁄s²}
R = 2,953 {m} = 2.953 {km} = ~3 {km} (or ~2 miles)
In light-speed units that distance would be:
R = 0.00000985142 {light-second} = ~10 {milli-light-seconds}
Thus, the event-horizon size (radius) of a sun-mass black hole would be just shy of 3 km, or about 2 miles. Compare that limited size with the sun’s actual radius of 6.957 · 10⁸ = ~700 million km—a size which is thus some 233 million times larger than the radius of its (equivalent-mass) black hole.
Interestingly, if the sun were to somehow, someday, quietly turn into a black hole—except for the lack of light, the sun’s planets (such as earth) wouldn’t notice any change—gravitationally. They would continue in their orbits circling what was formerly the sun, now a stellar-mass black hole, indefinitely. This is because beyond the former surface of the now-defunct sun, the now-black hole’s gravity (1 solar mass worth) propagates through space exactly as the sun’s did before. Black holes are therefore not a kind of cosmic vacuum cleaner(s), sucking up anything in sight—at least no more than gravitating bodies ever were.
2. Now, let’s do the same thing for the Milky Way galaxy’s central supermassive black hole, known as ‘A*’, and see what that looks like.
The mass of A* is:
M(A*) = 4.3 million · Msun
M(A*) = (4.3 · 10⁶) (1.9885 · 10³⁰) {kg}
M(A*) = 8.55055 · 10³⁶ {kg}
Ergo, according to our formula the event-horizon size calculation for A* would be:
R = 2 (6.6743 · 10⁻¹¹) (8.55055 · 10³⁶) ⁄ (3 · 10⁸)²
R = 1.1414 · 10²⁷ {m³⁄s²} ⁄ 9 · 10¹⁶ {m²⁄s²}
R = 12.6996 · 10⁹ {m} = 12.6996 · 10⁶ {km} = ~12.7 million {km}
Converting to light and AU units:
R = 42.3 {light-seconds} = ~⅔ {light-minute}
R = 0.0847 AU = ~¹⁄₁₂ AU
Thus, the event-horizon size (radius) of the Milky Way’s supermassive black hole A* is about 12.7 million kilometers (or 7.9 million miles)—which is about 18 times larger than the sun’s radius. In terms of light-speed, that’s 42.36 light seconds, or about ⅔ of a light-minute. In astronomical units it’s about ¹⁄₁₂ AU.
As a result, if the Milky Way’s supermassive black hole A* were located in the position of the sun, the earth (and all the planets) would orbit located far outside its event-horizon bounds—though, no doubt, its enormous gravity (the gravity of 4 million suns in place of one) would no doubt suck them all quickly in, pronto.
3. Now let’s repeat the process again for the far larger supermassive black hole IC 4709 mentioned in the posting above:
Mic = 65 million · Msun
Mic = (65 · 10⁶) (1.9885 · 10³⁰) {kg}
Mic = 1.2925 · 10³⁸ {kg}
Ergo, the Schwarzchild event-horizon radius calculation for IC 4709 would be:
R = 2 (6.6743 · 10⁻¹¹) (1.2925 · 10³⁸) ⁄ (3 · 10⁸)²
R = 1.7253 · 10²⁸ {m³⁄s²} ⁄ 9 · 10¹⁶ {m²⁄s²}
R = 191.97 · 10⁹ {m} = 191.97 · 10⁶ {km} = ~192 million {km}
Converting to light and AU units:
R = ~640 light-seconds = ~10⅔ light-minutes
R = 1.2798 AU = ~1¼ AU
Thus, if the IC 4709 black hole were lie in the position of the sun, the earth at its distance would be “orbiting” well below the former’s event-horizon “surface”—which situation likely wouldn’t last long, though we don’t (and can’t) know what’s happening within the bounds of the event horizon—because of the “event horizon.”
4. Finally, let’s repeat the process for the truly stupendous supermassive black hole—known as “TON 618,” located at a (“comoving”) distance of some 18.2 billion light-years—with its mass of 66 billion times the mass of the sun.
Mton = 66 billion · Msun
Mton = (66 · 10⁹) (1.9885 · 10³⁰) {kg}
Mton = 1.3124 · 10⁴¹ {kg}
Ergo, according to our formula, the event-horizon size calculation for TON 618 would be:
R = 2 (6.6743 · 10⁻¹¹) (1.3124 · 10⁴¹) ⁄ (3 · 10⁸)²
R = 1.7519 · 10³¹ {m³⁄s²} ⁄ 9 · 10¹⁶ {m²⁄s²}
R = 1.9492 · 10¹⁴ {m} = 1.9492 · 10¹¹ {km} = ~195 billion {km}
Converting to light and AU units:
R = 650,194 {light-seconds} = ~180.6 {light-hours} = ~7½ {light-days}
R = 1,299.5 {AU} = ~1,300 {AU}
Thus, the Schwarzchild event-horizon radius of the stupendously huge supermassive black hole TON 618 would be about 195 billion kilometers [121 billion miles] in size—or some 1,300 times larger than the radius of earth’s orbit round the sun. In terms of light-speed, that’s more than 650,000 light-seconds in radius size—which is to say: around 180 light-hours, or about 7½ light-days, or more than a light-week! Still, that’s minuscule compared with the distance to other stars (more than 4 light-years at least), not to speak of the scale of the Galaxy: perhaps 100,000 light-years across.
Looking through the other end of the telescope, the spacecraft Voyager 1 (earth’s most distant robotic emissary) is presently located some 164 AU (or a bit less than a light-day) away from earth and sun.
Errata:
• 1st paragraph: The mass of the Milky Way’s smbh is 4.3 million solar masses.
• Item 1: The Schwarzchild radius of a 1-solar-mass black hole is ~10 micro-light-seconds [not milli-].
• Item 1: The sun’s actual radius is about ~700 million meters [not km]—“a size which is thus some 233 thousand times larger than the radius of its (equivalent-mass) black hole.”
Michael McNeil,
That was some fun analysis. I like your comparisons of the black holes, using the solar system planets: “They would continue in their orbits circling what was formerly the sun, now a stellar-mass black hole, indefinitely.”
And: “if the Milky Way’s supermassive black hole A* were located in the position of the sun, the earth (and all the planets) would orbit located far outside its event-horizon bounds—though, no doubt, its enormous gravity (the gravity of 4 million suns in place of one) would no doubt suck them all quickly in, pronto.”
I would like to add (without computation) additional thoughts on the comparisons with the other black holes. As you noted, their gravities would cause them to fall into the black holes, but the planets would continue to orbit some of those black holes if they were orbiting fast enough (these are the speeds I’m not calculating).
However, there would be the problem of tidal forces that would pull apart some of the solar system’s planets. It seems to really happen to some objects:
https://behindtheblack.com/behind-the-black/points-of-information/strange-things-at-center-of-milky-way/
Robert wrote:
It would be interesting to know the rate at which the planets would have to be whipping around the A* black hole to maintain their present orbits if A* were to substitute for the sun—and what the tidal g-forces would be on opposite sides of the planet tearing (e.g.) the earth apart in such a case—but I too have made no attempt to try to compute it.
Damn it—M* in the foregoing should be A*.
Michael McNeil: Fixed.
Thanks, Robert. I was motivated to pursue these question(s) further.
After having a conversation with ChatGPT (a.k.a. Bing) as well as Grok on these matters, it now appears that:
1. Speaking classically, the formula for the period of an orbiting body is a (Newtonized version of) Kepler’s 3rd Law, to wit:
T = 2π √ (a³ ⁄ GM)
where:
T = orbital period
a = semimajor axis
G = gravitational constant
M = mass of central gravitating body (mass of orbiting body is deemed negligible)
Plunking in the values for the mass of supermassive black hole A* along with the length of earth’s semimajor axis into this formula:
T = 2π √ ((150 · 10⁹)³ {m³} ⁄ ((6.6743 · 10⁻¹¹) (4.3 · 10⁶) (1.9885 · 10³⁰)) {m³/s²} )
T = 2π √ ((3.375 · 10³³ {m³}) ⁄ (5.70689 · 10²⁶ {m³/s²}))
T = 2π √ (5.9139 · 10⁶) {s²}
T = 2π (2,431.85) {s}
T = 15,280 {s} = 4.2444… {hr} = 4 hr. 14⅔ min.
Earth, then, would have about a 4¼ hour-long year in such an orbit—still 150 million km distant from its central primary, just like now—which in this case would be the center of the A* supermassive black hole.
2. How fast would earth be moving in its orbit?
Presuming a circular orbit, 2π · 150 million km radius = 942.5 million km = ~1 billion km = ~1 trillion meters.
Orbital velocity v, therefore, on average would be:
v = (942.5 · 10⁹ {m} ⁄ (15,280 {s}) = 61.682 · 10⁶ {m/s} = 62,682 {km/s}
Converting to light units:
v = (61.682 · 10⁶ {m/s}) ⁄ (3 · 10⁸ {m/s}) = 0.20575
That’s more than 20% of the speed of light!
3. Lastly, we’d like to know what the tidal acceleration on earth from the A*-sized central supermassive black hole (in the location of the sun) might be. Both Bing and Grok agree that the equation to use here is:
a = 2GMR ⁄ d³
where:
a = (approximate) acceleration due to tides (from central body)
G = gravitational constant
M = mass of central gravitating body (e.g., A* or the sun)
R = radius of planet or moon (e.g. earth)
d = distance from central gravitating body (e.g., 1 AU)
a = 2 (6.6743 · 10⁻¹¹ {m³/s²·kg}) (4.3 · 10⁶) (1.9885 · 10³⁰ {kg}) (6.371 · 10⁶ {m}) ⁄ (150 · 10⁹ {m})³
a = 7.2717 · 10³³ {m⁴/s²} ⁄ 3.375 · 10³³ {m³}
a = 2.1546 {m/s²}
Converting to units of g’s, we see that:
a = 2.1546 {m/s²} ⁄ 9.8 {m/s²} = 0.21986 = nearly 22% of a g!
Thus, though the strain of tides from the central supermassive black hole would strain earth mightily (ocean tides miles high anybody? stupendous earthquakes?), the force appears insufficient to simply tear the planet apart (whew—what a relief!).
For closer worlds than earth though… Another day, perhaps, I should repeat this calculation for Mercury.
Michael McNeil,
Thank you for doing the math.
Michael-
Good stuff! After spending 8 years watching Physics lectures on Youtube, I can actually follow along!
Ref: “the force appears insufficient to simply tear the planet apart”
Yes, whew…! (first world problem, but good to know.)
Question:
Would an Earth ever initially coalesce in such a tidal environment to begin with?
Since I’m a nerd and can’t resist a challenge, I thought I’d give it a try using what I can remember from high school physics,
namely F = GMm/r^2, F=ma, and a = V^2/r
I whipped up a quick back-of-the-envelope (well, actually back-of-the-spreadsheet) workup, first using the Sun at the center:
Msun 2.00E+30 kg
Mearth 6.00E+24 kg
G 6.67E-11
AU 1.50E+11 m
Rearth 6.38E+06 m
r = AU 1.50E+11 m
Favg = GMm/r^2 3.56E+22 N
a = Favg/Mearth 5.93E-03 m/sec^2
v^2=a(AU) 8.89E+08 m^2/sec^2
V = 29822 m/sec
Speed of light (c) 3.00E+08 m/sec
V as % of c 0.01%
That jives pretty well with the 30 Km/sec approximately velocity of Earth around the Sun
Now assuming A* instead of the Sun:
MA* 8.55E+36 kg
Mearth 6.00E+24 kg
G 6.67E-11
AU 1.50E+11 m
Rearth 6.38E+06 m
r = AU 1.50E+11
Favg = GMm/r^2 1.52E+29 N
a = Favg/Mearth 2.53E+04 m/sec^2
v^2=a(AU) 3.80E+15 m^2/sec^2
V = 61659549 m/sec
Speed of light (c) 3.00E+08 m/sec
V as % of c 20.55%
Comes in pretty close to Michael’s calculations! Wow, didn’t think I’d remember all that.