In Need of Some (Measurable) Illumination

[Dr Wu]

< Mentor Note -- thread moved to Astro forum from the Sci-Fi Fantasy forum >

I apologise in advance if this comes across as hopelessly esoteric, but here goes: picture a 250 metre diameter sphere suspended in space (okay, in orbit round the Sun) with a surface temperature of 8,000 K. Now I'm aware that the brightness (light-signature?) of such an object would fall away quite rapidly over a given distance. Even so, I'd appreciate it very much if someone knowledgeable on the subject of optics could reveal to me how close an observer would need to be in order to perceive this radiant sphere as a naked-eye object. Also, if convention requires the answer to be in lux, lumens, watts, candle-power etc., is there some way to translate such metrics into absolute magnitude? Or is this like asking what the absolute magnitude might be of Piccadilly Circus or Times Square?

Many thanks.

 

[mfb]

WolframAlpha gives 2.1*10¹⁰ lux.
To convert this to apparent magnitude, we can use the sun: 100,000 lux gives an apparent magnitude of −26.74. The human visibility limit is roughly at +6, or 2.512^-26.74-6=8*10^-14 times this lux value: 8*10^-9 lux. Using the inverse square law, your 125m radius sphere could be seen as far as 2*10¹¹m away, which happens to be a bit more than 1 AU. Assuming the eyes are really well adapted to darkness.

There is no direct way to convert lux to intensity values, as the conversion depends on the spectrum.

 

[Dr Wu]

Many thanks again, mfb! NB. I had no idea that the Sun's apparent luminosity is equal to 100,000 lux. That's something to bear in mind should I find myself faced with a similar problem.

Incidentally, the orbital distance to the Sun maintained by this 125m radius globe is equal to that plied by the planet Venus. This means it should be readily visible (albeit dimly) during its close approaches to Earth, which is what I wished for all along :)

 

[mfb]

During close approaches to Earth it also appears to be close to Sun. You need the right angle - far away enough to be visible over the horizon at night, but not too far away.

 

[snorkack]

During close approaches to Earth it also appears to be close to Sun. You need the right angle

Venus can never be at the right angle. Its maximum elongation is just 47 degrees, not 90.

Maybe an easier derivation would be:
Sun is magnitude -26,7.
An object 30 magnitudes dimmer than Sun would be +3,3. Modestly visible, like Megrez.
It would be 10 to the power of 12 times dimmer. For example, the same surface brightness, but 10 to the power of 6 times smaller in angular linear size.
For example, at 1 AU but 1400 m across, where Sun is 1 400 000 km across.
Now, barely visible, at +6? That´s another 2,7 magnitudes dimmer. Roughly 3,5 times smaller again. So, like 400 m across.
Your ball is 8000 degrees, against 5700 of Sun. Well, the bolometric luminosity is easy to estimate by Boltzmann law - 4 times brighter. But how do you get the visual luminosity?

 

[mfb]

"right" as in "correct".

But how do you get the visual luminosity?

That's the reason I calculated it via lux values.

 

[meBigGuy]

Here are some additional perspectives
http://www.colorado.edu/physics/phys1010/phys1010_fa06/lectures/class23.pdf

 

[Dr Wu]

Hi, this is less of a question than the need for some verification. . . possibly. Again it concerns the light values of a celestial object, in this instance a 20 km diameter neutron star that has a surface temperature of 8,200 K. Now someone very helpfully explained to me that this star's luminosity output is equal to 3.22 x 10^17 Watts - or some 10 million times fainter than the Sun. I then calculated from this data that the neutron star's absolute magnitude to be +22.3. Nevertherless, I'm not altogether convinced if this is the correct answer. Unfortunately too, I need to get this right because it affects at several points the SF novel I'm writing. I would, therefore, appreciate it very much if someone can tell me if I'm on the right lines here.

Again, many thanks.

 

[mfb]

There is a difference between absolute magnitude for all radiation and for visible light, but the result looks good.

 

[snorkack]

a 20 km diameter neutron star that has a surface temperature of 8,200 K. Now someone very helpfully explained to me that this star's luminosity output is equal to 3.22 x 10^17 Watts - or some 10 million times fainter than the Sun. I then calculated from this data that the neutron star's absolute magnitude to be +22.3. Nevertherless, I'm not altogether convinced if this is the correct answer. Unfortunately too, I need to get this right because it affects at several points the SF novel I'm writing. I would, therefore, appreciate it very much if someone can tell me if I'm on the right lines here.

Making a consistency check:
Procyon B has absolute magnitude +13,0. Its surface temperature is 7740 K. And its diametre is about 17 000 km.
A neutron star with 17 km diameter and 7740 K surface temperature should thus have absolute magnitude +28,0 - nowhere near 22,3. Taking into account the 20 km diameter is trivial; the 8200 K temperature is more complex, but it still points to absolute magnitude over +27.

 

[Dr Wu]

Yes, +27 mag makes far better sense. Certainly it accords well with the absolute magnitude range of many known (non-pulsar) neutron stars. All things considered, it seems to me that a temperature of around 8,000 K is fairly modest, given how sizzlingly hot some of these little beasties can be.

Thank you very much :)