At what distance could a human look directly at the sun without eye damage?

On there was an article about the Mars explorer catching a partial solar eclipse (Phobos being too small to fully obscure the sun). I asked if I would suffer eye damage if I looked at the sun directly on Mars, since it is so far away.

Someone answered that I would but the reason given was that there is almost no atmosphere on Mars to soften the affect of the sun’s light. Then there was a followup question, which I am repeating here (because I didn’t like the answer there*). At what distance, presumably within the solar system, could I look directly at the sun for, say, one hour, without eye damage?

(*The answer by someone in the NPR comments section was that even Neptune would not be far enough away because although the amount of light was much smaller, it also came from a much smaller area, which means that staring directly at that tiny light would still cause damage. This doesn’t make sense to me, it sounds like, in space, you wouldn’t be safe looking at any star.)

I don’t think this question can be given a simple answer. It will have to describe how MUCH damage, and for how long a time. Even from Neptune, if you stare at the sun for a week it will probably do SOME damage, just as if you glance from earth on a clear day but only for a microsecond.

This is a great question.

From around Pluto’s orbit, the Sun’s apparent magnitude using a naked unaided eye would be about -17 or -18. That’s still considerably brighter than the full moon from earth by about a factor of 400 or 500 and likely still bright enough to at least be painful to watch if not outright dangerous.

To be about the same brightness as the brightest stars, you’d need to get out about 2 light years.

To be safe, I’d want to be at least at the outer reaches of the solar system, past the majority of comets and asteroids and such. It would still be about as bright as the full moon but might not cause permanent damage.

To collect the same amount of light and damage your eyes from out there, your eyes would have to be much bigger. You were right about that.

To add to this, a back of the envelope estimate gives -18 apparent brightness as about the brightness of staring at a 60W light bulb from about 2 meters out. No false modesty, that would hurt my eyes for sure.

So, a guesstimate at a safe distance? I’d say give it at least a light month out. Maybe a couple light months out to be absolutely sure. It would still be the brightest object in the sky but you might be able to stare at it without permanent damage.

For comparison’s sake … Venus has an apparent magnitude from Eath around -4.5. I don’t if one can stare at Venus indefinitely, but several minutes seems to be no problem.

OSHAgives 8 hour limits for laser light in the visible spectrum of about 10^-5 W/cm^2.

The sun has a luminosity of about 4*10^26 W. So the OSHA safe range will be on the order of 10^13 Meters or a few light days.

How about if you were on the surface of Titan? You’ve got a nice thick atmosphere to absorb much of the light.

Pluto may be 40 times further than the sun, so the amount of light reaching the eye will be 1/1600th. BUT … If one looks at the sun from Pluto, the image on the retina will be 1/1600th as large. In other words, per unit of area, the radiation reaching the retina will be the same. It will still burn the retina, but a much smaller area.

Wow, am I ever totally surprised. I wouldn’t have thought that it would be even close to the moon’s brightness, let alone hundreds of times brighter. But I suppose that if we compare it to the stars which are hundreds of times further away, I guess it makes sense.

Ignorance fought!

This is true, but… Nobody’s eyes are ever completely still, and damage from the Sun is not completely instantaneous. There will come some point at which natural eye movement will result in enough motion blur that the net effect is harmless. When precisely that point is, I don’t know offhand.

Well, color me educated. Thanks for the answers, folks, and I owe that guy an internal apology (since I didn’t write what I thought at the time) for doubting him.

Buck Godot, that looks like a nice calculation but I don’t pretend to understand it.

It seems to me that the extraterrestrial distances are purely academic; as for people here on earth, of course, there is no difference since the light source is 93 million miles away.

Sorry, I cut a few corners in the calculation, and a lot more in the explanation. Here is a slower and more accurate derivation

The sun produces about 410^26 Watts of energy. At a distance R, this energy will be spread out to cover a sphere of Radius R. The surface area of such a sphere is 4Pi*R^2

So at a distance R you will receive (410^26)/(4Pi*R^2) Watts per unit area. According to the Osha link, the greatest allowed laser light intensity that is safe for 8 hours is 10^-5 Watts/cm^2 so we want to find R such that

410^26Watts/(4Pi*R^2) = 10^-5 Watts/cm^2.

Doing a bunch of algebra this leads to R^2 = (10^31/Pi) cm^2

or R= 1.8* 10^15 cm = 1.8 * 10^13 m.

1 light year is 9* 10^15 m, so R = 0.002 Light years or about 0.7 light days. a little less than the distance of Voyager 1 to the sun (1.9 * 10^13 m).

On the Bad Astronomy site I have seen mention of the diffraction limit of the eye, which would prevent the Sun from forming an infinitely small spot on the retina. So if the Sun subtends an angle that is smaller than that limit, it will spread out on the eye to a larger and therefore dimmer spot. This limit is about 0.4 minutes of arc, the diameter of the Sun at 80 AU.

So at any distance greater than 80 AU, the Sun will never seem any smaller to the human eye (though it will appear dimmer).

I’m a bit pushed for time, so I can’t figure out how to apply that to Buck Godot’s calculation, but feel free it you want to have a go yourself.

1 AU = about 8.3 light minutes, so about 665 light minutes, or not quite half a light day.

Anyone have any data for how much Titan’s atmosphere attenuates light at the moon’s surface?

Titan’s atmosphere not kind to visible light, which has made it difficult to survey its surface from space. Not much visible sunlight gets through. As wiki puts it, the sun would appear to be a lighter patch of sky on Titan.

Another consequence of this is, until you are 80AU out, sunlight will have the same apparent surface brightness as it does from earth. So up to that distance, the Sun will still “burn” the retina as badly, it’ll just “burn” a smaller spot.

For reference, 80 AU is roughly about where the heliopause is (give or take) and roughly twice the Sun-Pluto average distance.

During a solar eclipse, just one percent of the sun’s disc direct light can very quickly cause permanent retina damage. So can the arc light of a welder.

The human eye is very intolerant of prolonged exposure of a source of light much much fainter than the sun.

Is the OP looking for the distance at which one cah gaze at the sun for an unlimited time and suffer no adverse effects at all? Obviously, the harm would be a function of the duration of the exposure. I can go out and quickly glance at the sun for a fraction of a second, perfectly harmlessly.

When the sun is setting right at the horizon with a dark orange aspect, I can stare at it until it disappears and I see the green flash, which I did nearly every day for a year.

So the safe point , at which you could look at the Sun safely, is somewhere between 80 AU and 0.7 light days; a little under 0.6 light days, presumably.

Note that this assumes that your eye is completely motionless, staring at the Sun, which would be a bright point of light. As Chronos noted, the eye makes involuntary movements all the time, so this scenario is unlikely. It also assumes a perfectly transparent visor on your spacesuit, also unlikely.