<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

"It is the business of the future to be dangerous." - Alfred North Whitehead, philosopher.