Solar eclipses from elsewhere in the solar system

[Dawson]

I’ve just watched a fascinating talk from Australia on solar eclipses from elsewhere in the solar system, and am surprised it’s only had just over 100 views:

https://m.facebook.com/astronomysa/videos/3438550399584185/

The speaker has done some fascinating calculations.

It made me think, how bright (or not) is it at midday on each of the planets? Tricky concept for the gas and ice giants but assuma the measurement was taken standing on top of the clouds. And how far would you have to be from the Sun before it was safe to look at it with a telescope without any filters?

Do watch that talk though.

James

[Bill Barton]

Light radiating from the Sun follows the inverse square law (ie intensity = 1/d*d)

The benchmark value for the Earth at one Astronomical Unit (AU) from the Sun is thus 1/1*1 = 1.

For Mercury 1/0.3*0.3 = 11. The Sun is more than ten times brighter than on Earth.

For Venus 1/0.7*0.7 = 2. The Sun is around twice as bright as on Earth.

For Mars 1/1.5*1.5 = 0.44. The Sun is less than half as bright as it appears on Earth.

For Jupiter 1/5.2*5.2 = 0.036. The Sun is only around 4% as bright as it appears on Earth.

For Saturn, 1/9.5*9.5 = 0.01. The Sun is only around 1% as bright as it appears from the Earth.

For Neptune 1/30*30 = 0.001. The Sun is only 0.1% as bright as it appears from the Earth.

Solar filters typically reduce light levels by 99.999% (ie allow 0.00001 pu through).

We need to rearrange the inverse square law formula to make distance the object (ie d = sqrt (1/i))

sqrt (1/0.00001) = 316 AU or 29 374 000 000 miles.

At this distance the Sun would subtend an angle of arc tan 865 300/29 374 000 000 = 0.0017 deg. or 6 arc seconds (hopefully I haven’t made any mistakes in my maths).

[Peter Anderson]

Hello Bill, Whilst your mathematics are good, the problem as I see it is that the visible disk will remain the same brightness and so has the potential for the same sort of damage if you look at the sun. However, and obviously the heat factor transmitted within the eye will be much reduced if the solar disc has a reduced angular size.  Taking my argument ad absurdium, you could reason it is dangerous to even look at the stars, but in those cases their stellar disc is not resolvable and instead you are examining a disc produced by your eye or telescope optics that is many times the actual stellar disc diameter and so a mere tiny fraction of its true surface brightness.

I think we need an expert to provide some considered and professional advice on the subject.  I feel like an ancient Greek trying to form a hypothesis just by reasoning alone.

[Dr Paul Leyland]

The resolution of the naked eye is roughly 1 arc minute and the diameter of the sun at 1 AU is roughly 30 times larger. Simple arithmetic suggests that beyond 30 AU looking directly at the Sun might be dazzling but likely not otherwise harmful. The blink reflex will protect you long before retinal burns occur, though there will be localized bleaching which will fix itself within a few minutes.

[Peter Anderson]

Thanks Xilman, This is another aspect and there is valuable information there, an aspect I had not directly considered.  On the basis of this it would appear that it would be safe (if dazzling) to look at the Sun from the distance of the Orbit of Neptune. (30AU). But are there other factors in play? 

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