I was wondering if anyone has ever attempted to estimate how much solar radiation would be received on a rogue planet floating in between star systems close to the center of the milky way, compared to say how much is received on Earth from our sun.

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    $\begingroup$ Wouldn't that put it in danger from the super-massive black hole cluster in that area? Not sure a rogue planet could even exist in such an environment. $\endgroup$ Commented Oct 18, 2022 at 15:25
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    $\begingroup$ Do you want to limit this to solar radiation or include radiation from the supermassive black hole which all the suns are orbiting? $\endgroup$
    – Beanluc
    Commented Oct 18, 2022 at 23:06
  • $\begingroup$ @Beanluc good question, I guess both would be interesting to know in of themselves. $\endgroup$ Commented Oct 19, 2022 at 21:50

3 Answers 3


One can do a quick Fermi estimate:

According to wiki, the densest parts in the centre and globular clusters have about 70 solar masses per cubic parsec. Let's assume it's all solar mass and thus solar luminosity stars as the Sun is a totally average star.

Now we need the average distance of stars: $4^3 = 64$ is a reasonable close guestimate for equal distribution, thus one star every 0.25 parsec in a cube, filling it completely with stars and their spheres of influence.

Now if I place this rogue planet at the worst place inside this volume, half-distance between two stars, it will be $d=0.125$ parsecs away from a solar type star. There are about 200000 AU in one parsec, and the solar constant at 1 AU is $S_0 = 1300 W/m^2$. So the illumination by the solar-mass star in 0.125 parsecs is $W = S_0 / d^2 = 2.08\cdot 10^{-6} \mathrm{W/m^2}$. We now can consider that there will be about six stars in similar distance so that we actually get slightly more illumination. Regardless, it's less than 1 in 100 million of what we see on Earth.

This actually is a worst case assumption and the likelyhood that this rogue planet is nearer to one star will make sure that it gets quite a bit more given the inverse square law for the illumination intensity. But whereever we place it - it is extremely unlikely that it ends up receiving more light than any far far out solar system body in the Kuiper belt (30...50 AU), more like remote objects in the Oort cloud (which is assumed to extend from ~2000 AU to ~100000 AU). This guestimate will change slightly to higher values of illumination if you assume a mass distribution for the stars as the luminosity scales like $L/L_0 \propto (M/M_0)^{3.5}$, thus is highly mass-dependent.

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    $\begingroup$ I'm not sure if this estimate is accurate; WolframAlpha says that "irradiance from the stars (no moon) at night on the earth surface ≈ 3×10^-6 W/m^2", which is more than your estimate gets for the galactic center. $\endgroup$
    – TLW
    Commented Oct 18, 2022 at 1:34
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    $\begingroup$ @TLW thanks for the hint. Can you provide a link for reference? I'm unable to obtain that information. That said, if it quotes brightness of the nightsky: most of the sky's brightness at night on Earth (in the absence of light pollution) is actually NOT star light, but airglow and zodiacal light. Only 7% of the night sky's brightness is due to scattered star light: en.wikipedia.org/wiki/Sky_brightness. I do not take into account any other light sources than stars. $\endgroup$ Commented Oct 18, 2022 at 2:06
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    $\begingroup$ You need to look at more stars than just the closest ones. As distance doubles the brightness of individual stars at that distances decreases by the square of the distance, but the radius, and therefore the total number of stars, increases by the cube of the distance (volume = 4/3*pi*r^3) $\endgroup$ Commented Oct 18, 2022 at 16:16
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    $\begingroup$ @JasonGoemaat yes indeed... if the universe were infinite and infinitely old and homogenuous, the sky would be sund bright everywhere - the Olber's Paradox. For this purpose here the stars of the Milkyway will suffice... yet this is only a very lower approximation in this form $\endgroup$ Commented Oct 18, 2022 at 20:16
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    $\begingroup$ "There are around 10 million stars within one parsec of the Galactic Center," -- en.wikipedia.org/wiki/Galactic_Center -- that is way more than your 70 solar masses per cubic parsec. $\endgroup$ Commented Oct 19, 2022 at 14:48

Edit: there was a dumb mistake in my code (I had left out the square root in the distance calculation), which I noticed immediately after posting the answer. This changed the result downwards by approximately one order of magnitude.

Using some numbers from planetmaker's answer, I did a very quick simulation in an attempt to get a rough idea of a typical situation.

I'm assuming that there are 70 solar masses of stars per cubic parsec, and it's made of 70 individual solar-mass stars. I also used the value of $1300\: W/m^2$ at one AU for the illumination.

I assumed the planet was in the centre of a one-parsec cube. I then chose 70 uniformly random positions for stars inside the box, and computed the illumination to the centre. Adding these together is the total power received by the planet.

I repeated this computation for a hundred thousand different random configurations of stars, and the median result is pretty stable at $1.4 \times 10^{-5}\: W/m⁻²$.

I took the median because there are outliers where a star lands very close to the centre, making the illumination orders of magnitude higher, which skews the mean of the ensemble. I verified the result further by taking the median over a million configurations, which didn't change the result.

Here's what I wrote (in the Julia REPL):

julia> using Statistics

julia> d(x) = sqrt((x[1] - 0.5)^2 + (x[2] - 0.5)^2 + (x[3] - 0.5)^2) * 206265
d (generic function with 1 method)

julia> L0 = 1300

julia> L(star) = L0 / d(star)^2
L (generic function with 1 method)

julia> result = median(sum([L(star) for star in eachcol(rand(3,70))]) for n = 1:1000000)

The constant 206265 in the distance function converts the distance from parsec to AU.

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    $\begingroup$ A problem here is that the radiation coming from a uniform distribution of stars is limitless as the size of the neighbourhood grows. If I double the size of the box, the distances to the stars double, decreasing their brightness by a factor of 4, but at the same time the volume, i.e. the number of stars in the box increases by a factor of 8, so the result is double the original. $\endgroup$
    – Dronir
    Commented Oct 18, 2022 at 12:50

Here is an estimate that says at the core of M32 it would be like twilight, but in the Milky way it would be just bright enough to read by. Let's

It really depends on how close you are to the galactic core though since there are 10 million stars within 1 parsec of the core (7.5 million per cubic parsec), and 100 parsecs away it is just 100 per cubic parsec.

Let's use the value 100 parsecs away from the galactic center since the area is large and the density is similar for many parsecs and do some napkin calculations and stellar counts. I'll ignore things like how bright the stars are, how much mass they have, and how much dust there is. It seems like stars near the galactic center tend to be larger and hotter and there is population of giant stars created some millions of years ago in a burst of stellar generation. Things like that could really throw off the calculations since giant stars can be much more luminous than our sun. Such as Deneb which is one of the brightest stars in our sky though it's one of the furthest stars visible to the unaided eye at around 2600 light-years away.

If I just use the volume and 100 stars per cubic parsec, and I take the formula for the volume for a sphere: $v=\frac{4}{3}\pi r^3$, we can reverse it to get the radius enclosed by a sphere with the given volume: $r^3=\frac{3}{4}v/\pi$ and $r=\sqrt[3]{\frac{3}{4}v/\pi}$. Taking the 100 stars per cubic parsec and plugging in 100 for $v$ I get .6203505 parsecs for a radius of a sphere with a volume of 1 cubic parsec.

So I'll use volume as the driver since the stat we get is 100 stars per cubic parsec. That gives a volume of 0.01 cubic parsec for each star. I'll take the maximum distance for a volume, for example 0.01 cubic parsecs gives a volume with a radius of .13365 parsecs or 27567 AU. There should be a star within that distance, let's assume the max distance of any volume we're checking based on the radius increased for each star added. Assuming the stars are all the same as the sun, that star would give your rogue planet as much energy as $\frac{1}{27567^2}$ the energy that the Earth receives from the sun, a little over 1 billionth. Plugging in 0.02 cubic parsecs gives us a distance of 0.1684 parsecs or 34732, so that star would contribute $\frac{1}{34732^2}$ that of the sun to the Earth. At 1 parsec that gives us about 500 stars and the total radiance ends up being about 15 billionths that of the sun. Converting that to $\frac{w}{m^2}$ that gives $2.04 e^-5 )\frac{w}{m^2}$.

Comparing to values in this paper, that's ten times as bright as all the stars and planets in our night sky, but sill only 1/100th the brightness of a full moon.

However, this is the density 100 parsecs from the galactic center stars get exponentially denser until near the actual fore there can be 10 million stars within one parsec. Looking at our distance however the luminosity keeps increasing as we take into account more stars, but since each one gets progressively fainter, the total brightness doesn't increase exponentially.

I wrote a little JS script to figure it out. This page says the sun could be visible 58 light years (17.86 parsecs) away. If I just look at that volume, there would be 2.38 million stars visible to the naked eye at night. Most of them would be just at the edge of vision though. Some relevant brightness counts:

  • 2 stars brighter than venus to us
  • 80 stars brighter than Sirius, the brightest star in our sky (next to the sun)
  • 215 stars brighter than Canopis (2nd brightest star)
  • 1331 stars brighter than Betelgeuse
  • 18899 stars brighter than the 92nd brightest star in our sky
  • 37854 stars naked eye visible in an urban area (~6 times visible in clear dark sky on Earth)
  • 599599 stars visible to most people (100x the stars in our sky)
  • 2.38m stars visible to people with excellent night vision

The sky would be absolutely magnificent. Stars outside that would continue to provide light, though they wouldn't be distinguishable to the naked ye, and be easily visible to binoculars, and looking out to 100 parsecs I calculate flux of about 1e-3, about that of a half-moon.

Of course as I mentioned earlier, the stars tend to be bigger and brighter near the galactic core. A single 'Deneb' would be as bright as venus at 56 parsecs away.

  • $\begingroup$ 10 million stars in 1 spherical parsec at 100 parsecs away is 2 * 10^7 times farther than the sun. If they average 100 times as bright as the sun (skewed by super-bright ones), that is 10^9 times as much light produced. Inverse square means 2 * 10^-6 as bright as the sun, which is about as bright as the full moon! (the core alone) 1 parsec radius at 100 parsec distance is 0.01 radian radius, twice as big as the moon. $\endgroup$ Commented Oct 19, 2022 at 14:59
  • $\begingroup$ @Yakk-AdamNevraumont Sorry if I wasn't clear. It's actually 10 million stars within one parsec of the galactic center. , and that's actually 4.189 cubic parsecs. I'm not using that in my calculations though, I'm using the 100 stars per cubic parsec density which is 100 parsecs away from the galactic center and is pretty consistent throughout that region, with density increasing towards the core and decreasing away from it. At the core I would use about 2.4 million instead of 100 for the count of stars in a cubic parsec, but it wouldn't be accurate as density decreases outwards. $\endgroup$ Commented Oct 19, 2022 at 15:08
  • $\begingroup$ I'm saying at 100 parsecs away from the galactic core, the 10 million stars (if they average 100x brighter per star than Sol) around the core are all by themselves as about bright as the full moon to a planet floating 100 parsecs away. And that 1 parsec radius sphere is twice the size of the full moon from the planet. Now, the core isn't discrete -- there will be a bunch of stars 2 parsecs away from it etc -- but my point is that neglecting that huge number of stars by only looking locally misjudges the brighness of the sky. $\endgroup$ Commented Oct 19, 2022 at 15:25
  • $\begingroup$ Ah, gotcha :) Yeah, my calculations only use the static density at 100 parsecs away from the center, still trying to be miserly assigning the brightness of Sol to all the stars. It ignores changes in density, differences in stellar brightness, and dust just to give an idea of how the results could be calculated. The actual values were not readily available in my google searching, but I do ignore some negative factors too like the increase in dust near the galactic core the amount of energy coming in as IR instead of visible light. $\endgroup$ Commented Oct 19, 2022 at 15:59

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