Staggering Stories Forum!

"I can't hear you, Bob"

"Read the bubbles then"

LOL

Try this for size

http://www.mcsweeneys.net/2002/01/10deathstar.html

Lol..too much time on his hands....well, that and the premise is wrong from the outset...the death star wouldn't eject into space - its so large it has a noticible gravity well, I should expect, so anything ejected would just orbit it and ruin Grand Moff 'Moffy' Tarkin's view on a nice day...

um...look, I had a bad night last night so leave me alone!!!!!

macfadyan wrote:its so large it has a noticible gravity well, I should expect

Not that noticable, it's nowhere near the size of a planet (being 550 miles in diameter at the largest estimate) and mainly hollow. In fact, even if it were massive enough to have any appreciable gravity, being mainly hollow it would probably collapse.

<Anorak Removed>

More to the point, surely an evil and, I have to say, very inefficient empire would design an evil and very inefficient method of trash disposal.

<Anorak Back On> Now, what with me being a sad science geek, I've worked it out. What can I say .... I was intrigued by what the answer might be

I'm assuming the average density of the Earth is 5.5 x 10^3 kg per cubic metre and its radius is 6.378 x 10^6 metres (all accepted values) and I'm giving the Death Star some very generous estimates, that its average density is 7.87 x 10^3 kg per cubic metre (that of solid iron) and its radius is 4.95 x 10^5 metres.

You get the mass of the each body by calculating the volume of their spheres as being 4/3 (PI*r^3) and multiplying that by the average density per cubic metre.

Righto, surface gravity is calculated by g = GM/r^2 where G is the gravitational constant (6.67 x 10^-11 Newtons per square metre), M is the mass in kilogrammes and r is the radius in metres. Plug in the numbers and you work out that the force due to gravity acting on a body on the surface of .....

The Earth is 9.74 Newtons/kg (pretty close to 9.8, the accepted value)
The Death Star is 1.08 Newtons/kg

.... giving the Death Star something of the order of 10th the surface gravity of the Earth, assuming it's composed of solid iron and as large as its largest estimate allows. I'm also assuming the effect due to the gravity of the object on the surface is negligable which it would be if it were a bit of compacted trash, a spacecraft or a human being for example.

So a solid iron Death Star would have a decent gravitational field, From here onwards though, it's all guesswork. Oddly enough, there are no published values for the true mass of the Death Star but I did find some anoraky estimates suggesting it might be about 1/10th of the values I used, bringing the gravity down to 1/100th of the Earth, which isn't quite so impressive but still something to be reckoned with.

I need to get out more

if you EVER call me a sad old geek again I shall remind you of this.......

Well, if you're going to be like that.... next time we're on a space ship heading for certain death in the crushing heart of a black hole and you need someone to calculate an escape trajectory, don't come crying to me

Actually, you might be better off not coming crying to me

Checking my calculations I find a mistake, The density values should both be x 10^3. This didn't affect the figure for the Earth as I already knew its mass but it did affect the calculation for the Death Star. To cut a long story short the final answer should also be x 10^3 meaning a Death Star of solid iron would have a gravity of 1/10th that of the Earth. I'll edit the original post so it's correct.

This is probably why it's all done by computers on spaceships

Well, it's a good job we WEREN'T on a space ship heading for certain death in the crushing heart of a black hole, isn't it? ISN'T IT!!?? Tch...can't get the help these days.

Now go and sit in the corner and think long and hard about what you've done....GO ON!!!

I think MacFadyan is the captain, so I'm probably the least of your worries