1.354e14 kg assuming the Death Star has a radius of 160 km. However there's conflicting information about the diameter of the Death Star (some say 120 km, others 160 km) and that's the diameter and not the radius.
So the Death Star would be considerably lighter than your friend's calculation.
My calculations give 3.385e13 kg assuming a radius of 80 km which yields a density of 1.578e-2 kg/m3 or 15.78 g/m3 which is way, way low. In fact it's much lower than the density of air.
Three possibilities: I made a mistake in my calculation, Google's gravity value for the Death Star is incorrect, we shouldn't expect consistency from a story George Lucas made up almost 40 years ago.
Wouldn't you assume the Death Star has artificial earth-normal gravity, just like all the spaceships? (Not to crash the geeky fun with geeky observations.)
However, while I haven't watched every episode of Lost I think it'd be a lot more noticeable if gravity there was 4.815162 times regular Earth gravity. Everybody would be crawling around, for starters.
How exactly did they compute that constant? I suspect the figure they give is the result of many heated debates and much flaming on mailing lists and whiteboards.
Well theres so much material in the Expanded Universe including a book called "Death Star" that tells the story of the Rebel Alliance victory from the perspective of people who worked there. So, I would just have to assume that they went over the specs/dimensions/weight of the Death Star at some point.
That page itself has conflicting reports of the size. At the top it says "Size: 160 kilometers in diameter," but lower on down the page, it says "The Death Star was 120 kilometers in diameter."
It's also only valid at a certain radius (probably the surface). So the sensor wouldn't be calibrated correctly after you went inside (it would be off by differing amounts depending how far from the center you were).
The same is true of the constants for the planets, but at least they are (mostly) solid and not designed to be lived inside of.
