Wouldn’t the important thing be the difference in acceleration between the sun and the earth due to the slight different distance? I’m betting the orbit would still be stable, but I wouldn’t like to do the sums.
Hens; **Grey[\B]'s calculations specifically called for 14 layers.
Well, two out of three have already occurred …
With or without an echo?
We would be safe enough for a while, I think;
the stars in our local neighbourhood are at decent intervals of distance from us,
so that if you arranged them in a row arranged at a distance from us corresponding to their distance from the Earth, you would find that the closest several thousand light years would have widely spaced stars-
representing all the stars found on the surface of a sphere n light years in diameter, arranged in a line.
The stars wouldn’t start to touch each other until you reached a long way from Sol…
A few billion light years away from us at the centre the stars would be packed so close together that they would collapse into two massive black holes, each a little less than the size of the universe…
the few stars left near us would probably get peeled off into these holes, until we are the only system left (more or less)
Now try to work dark matter, quintessence and dark energy into the picture.
Sorry- no idea what would happen there.
I give up…
That should be-
A few billion light years away from us at the centre the stars would be packed so close together that they would collapse into two massive black holes, each a little less than half the mass of the universe…
Oh, and I just thought of another comment: if the stars look aligned, they’re probably not, as by the time the light gets here, they’ll have moved on. OTOH, if gravity moves at the speed of light…
Actually, the density of stars on the line would increase as roughly the cube of the distance. So, starting with one star at a distance of 4 light years, you’d have ~15000 stars at ~100LY out. The flattened shape of the galaxy (~300LY thick at earth) would eventually distort that calculation, but still, some sort of giga-Hypernova blasting away 100 LY out would not be pleasant for us earthlings.
Could you explain that? I would have thought it would be with the square of the distance (proportional to the surface area of a sphere with that radius, assuming the stars are evenly distributed (which I know they’re not…))
And hang on: if density is proportional to cube or square of distance, this cancels out the inverse square proportion of gravitational attraction. So the gravitational force would be ::calculates:: awesome.
You know, I think you’re right, it’s the square. The cube factor would be for stars in the total volume of the sphere, and this is just the stars at a given distance.
Sure enough, The High Energy Astronomer says that there are ~14,600 stars within 100LY.
Using the SQUARE for surfaces, that’d give ~625 stars at ~100LY out. Or roughly 1 star every 9.4 billion miles along the line at that distance.
See- plenty of room.
I seem to remember an Isaac Asimov mystery which relied on the same sort of calculation…
Only if you think 100 AU is roomy. 