How big and what brightness would the sun appear to be from, say, Uranus? Is it nothing more than a function of the distance? From a planet twice as far from the sun as earth, would the sun appear half as large and half as bright?
Close. Half as large and one quarter as bright - the inverse square law, and all that.
diameter would be a linear-inverse function with respect to distance… brightness would be almost precisely varying with the area that the sun takes up in the sky… like all radiation, sunlight follows the inverse square rule.
So… since Uranus is about 19 times as far from the sun as we are, it would appear about 5% as wide as we see it (or a little under 2 arc minutes), and be about 1/360th as bright.
Now, what does 1/360th as bright mean in terms of objects we’re familiar with?? To use the (somewhat confusing) magnitude system, the sun from earth is at a magnitude of -26.7, the full moon under optimal conditions at -12.7, and venus at -4. (lower is brighter in terms of magnitude, by exponential amounts.)
My calculations are that going out to the orbit of uranus would decrease the magnitude of the sun by amout 6 point 4 magnitudes, leaving it around 20.3 So, it would be noticeably less bright than around here, but still much brighter than the full moon, for all that it would be tiny in comparison.
I remember some astronomy books with ‘hold this page out one meter from yourself and see how big the sun would appear from various planets.’ Don’t remember much, except that the mercury one was HUGE lol.
Apparent size is a trigonometry problem, actually. The Sun is 865,000 miles in diameter (1.4 million KM). That’s the “short” leg of an isosceles triangle, the other two legs of which are equal, and the distance of the planet from the Sun. The apparent size in the sky is the measurement of the angle between the two equal legs – from Earth, 0.5 degrees (AKA 30 minutes of arc). Plug the sun’s distance from any planet into the diagram, and calculate that angle.
Brightness falls off (or increases) as the square of the distance. Mars, 1.8 times the distance of earth, sees the Sun as 1/3.24, or about 30%, the brightness that we do. Venus, about 70% of Earth’s distance, gets just over twice the insolation. And so on.
At what distance would a discernable (to the naked eye) disc dissappear?
The angular resolution of the human eye is normally taken to be about 1 arcminute (1/60th degree). Anything smaller than this appears pointlike. For the Sun to subtend 1 arcminute, we’d have to be about 30 times farther from it than we are now, or about 27,900,000,000 miles, assuming I’ve done all the math right.
Well, I found a cite on the web that you need about 20x magnification for venus to show a visible disk. Taking a long estimate of venus’ distance from earth (twice its orbital radius,) and doing a lot of back-of-the-envelope math, I came up with a figure of 8 arc minutes for that, after magnification, which seems on the large side, so maybe I made a mistake.
According to that, you’d be somewhere between the orbit of jupiter and saturn when you can no longer really make out the sun as a disc. YCMV
on preview, QED’s figure of 1 arc minute sounds a lot better, though I think his miles are somewhat off. I get 4.5 billion kilometers as 30 times our distance from the sun, or almost exactly the orbit of Neptune. 
Yeah, I think I typed an extra zero when copying the figure from my calculator. When I knock one off my answer and multiply by 1.6, I get your figure.
I know this may sound like an idiotic question but I’ll ask it anyway…
Have any of the craft that we sent out there (Voyagers, Pioneers, Cassini, Gallileo) ever turned around and snapped a picture of the sun from “out there” so we can see what it would look like from the orbit of a particular planet?
Zev Steinhardt
I’d like to know that answer too, Zev.
Well the Spirit and Opportunity rovers image the sun daily. Here’s a shot from Opportunity of the moon Diemos crossing the face of the sun.
The Cassini raw images thus far contain no pictures of the sun. Likely, pointing a low light planetary camera at the sun would cause bad things to happen.
I don’t recall any voyager pix of the sun however, there is this shot of Earth from 4 billion miles. The home planet is a little more impressive from the orbit of Mars
Just because nobody mentioned this pretty obviuous thing; as the science chap says, the sun becomes a point, being why stars generally look fairly similar from a distance, because from a long enough distance, a really bright light won’t have any discernible width, really.
<insufferable pedant>
Technically, that’s not quite true. The triangle you want to draw has as its vertices the center of the sun, the point of the observer, and the point where the observer’s line of sight is tangent to the surface of the sun. The angular diameter of the sun is then twice the angle at the observer’s vertex. But as long as the distance to the sun is much greater than the Sun’s radius, Polycarp’s method will work fine. (And this holds for any of the planets — even for Mercury, the difference between my method & Polycarp’s is only about 1%.)
</insufferable pedant>
If your distance to the Sun is less than that, you’ve got more pressing matters than a mere math problem to attend to. Like working out how best not to burst into flames. 
This “pale blue dot” photo was part of a montage taking in much of the Solar System that gets close to what zev and Liberal were looking for in the case of Voyager. But the Sun was deliberately obscured from it for being too bright.
Incidentally, this caption to that Voyager montage is relevant to the question:
Another thing to consider - the response of the eye is non-linear, and we have an iris to stop down too much light. So just because the sun is only 1/160 as bright does not mean it will look that way to us.
For example, a brightly lit room like an office might have 200 foot-candles of light. But it seems pretty bright. On the other hand, at noon on a sunny day outside there can be over 10,000 foot-candles of light - 50 times brighter. And the light from a reading lamp is only about 25 foot-candles.
For a wonderful freeware download try Celestia. Honestly, it’s an amazing waste of time.
My favourite is watching Mars swing behind Phobos.
Surely you know that Uranus is where the sun don’t shine… 
Celestia is great. You can also “travel” to other nearby stars and see where our Sun appears against the backdrop of constellations.