How much of the sky would Betelgeuse fill if its nearest "edge" was placed at the distance of the moon?

Question

I know very little about astronomy, but I'm preparing a talk for a class that will involve Betelgeuse. Assume that we could cool Betelgeuse so that it wouldn't burn us up and "freeze" its size at its mean size and then place its closest "surface" 384,400 km from Earth directly "above" a viewing position (ignore other factors like gravity). If it was generating enough light to be clearly seen (rather than blackening the sky), how much of the sky would it fill?

I realize that at this distance its opposite "edge" would be far past the sun and most of the planets in our solar system.

I've seen comparisons of the size of a period (Earth) to a 20-story building (Betelgeuse), but I'm trying to think of some other ways of visualizing the size of these immense objects which are visible without a telescope.

Please feel free to mention other ways of getting our brains around how big these objects we're seeing actually are.

Thanks.

If Betelgeuse would completely fill the sky from horizon to horizon at the moon's distance, how far away would we have to push it until we could start to see some "sky" at each horizon?

Some great answers here, thanks. That "amount of area outside the airplane window" example is especially helpful.

I'm still surprised that it wouldn't take up more than half the sky looking straight up and I'm sure it's because of faulty logic on my part.

Here is what I thought: A full moon directly overhead occupies about .52 degrees of the sky. If you doubled the size of the moon, wouldn't you *nearly* double this amount? I realize a law of diminishing returns here because the moon doesn't wrap around the earth, but since Betelgeuse is 270,000 times the size of the moon, I would think that there would just be a small amount of visible sky (much less than 45 degrees) at each horizon. Feel free to help me see where I'm off-base here.

Answer 1

Well, how is this:

The sun's diameter is 109 times that of the Earth. Betelgeuse is 650 times the size of the sun. Therefore Betelgeuse is 70850 times larger than the earth.
The distance to the moon is (as you wrote) averaging 484400 km. Divide 484400 by 70850 and you get 6.837 km, or 22430 feet.
So, imaging you are in an airplane flying at 22000 feet or so (turbopropeller airplane, as jet would cruise at 31000 ft and above) and imagine looking out the window to the earth. The earth fills as much of your field of view as Betelgeuse would be if its edge was where the moon is.
To notice the curvature of the Earth, it is said you need to be about 80000 feet high, 3.5 times as high as the commuter airplane in the example above.

Answer 2

Betelgeuse raius is 650 times the Sun radius, that is about 700.000 kilometers. So it will be better for you to suppose Betelgeuse at the Sun distance. It will cover 650 times the area covered by the Sun. The angular (visual) diameter of the Sun is 30' (half degree). So Betelgeuse with its center coincident with the center of the Sun will include the Earth inside it, for its radius is 455 million km. and the Sun-Earth distance is 150 million km.
It will pass Mars orbit and reach the asteorids belt.

At the moon distance it will cover half of the sky.



For more info about Betelgeuse take a look at

http://en.wikipedia.org/wiki/Betelgeuse

Answer 3

I'd imagine that because of its size and the sheer amount of light it would create, there would only be a few hours of night on the equator. There would be perpetual day at the poles, regardless of the seasons. Twilight would be the norm outside of the tropics, and the tropics would probably be measured by the portion of the earth that gets a "night." It is possible that even at night, the amount of light would still be so great that it could never really be pitch black outside because of the sheer size difference.

The size is pretty hard for me to wrap *my* head around. But imagine this: if you put a grain of sand on a basketball, and then put that basketball on the earth, would there be any place that piece of sand could be where it would not be able to see the earth? I'd imagine that there may be one teeny tiny line around the middle of the ball, but that would also change with the seasons, wouldn't it?