# How much gravitational force does the Death Star exert?April 14, 2015 2:42 PM   Subscribe

What would happen if Darth Vader parked his Death Star in our neighborhood?

I believe that the first Death Star is (was? will be? would be?) more massive than our moon, and therefore would exert much more gravitational force than our moon. Therefore, if Darth Vader parked the Death Star within our solar system, it would greatly affect the gravitational balance of the solar system and basically ruin everything.

Some of my friends contend that the Death Star is less massive than our moon, and would therefore only affect tidal forces on, say, Earth, if he parked near to it.

Who is closer to correct?
posted by the matching mole to Technology (13 answers total) 3 users marked this as a favorite

Well, they did call it a small moon.

For this sort of question, your go-to source is Wookiepedia, not Wikipedia.
posted by RobotHero at 2:47 PM on April 14, 2015

"Various sources state the first Death Star has a diameter of between 140 and 160 kilometers." The moon is 3,475 km in diameter. It's also solid; the Death Star is mostly trash compactors and princess storage. I say the Death Star would have a negligible effect on anything not blown up by it.
posted by Sing Or Swim at 2:50 PM on April 14, 2015 [16 favorites]

The internet says the first death star was 140-160km in diameter. The moon is 1700km. If you assumed similar density our moon would mass *way way more* because volume is proportional to radius cubed. V = 4/3 * pi * r^3

So the volume of the death star is 1608495 km^3

The volume of the moon is 1929330588 km^3

So the moon would mass 1200 times more than the death star. I don't think the death star would be a serious problem, gravity wise.
posted by RustyBrooks at 2:52 PM on April 14, 2015 [1 favorite]

Whoops, I have the numbers wrong there, I had the *diameter* of the moon at 1700 km but that's actually the radius. So the moon is actually more like 10,000 times more volume than the death star.
posted by RustyBrooks at 2:53 PM on April 14, 2015 [4 favorites]

Response by poster: Wow, I was way off on the size of the Death Star. Thanks for bringing some hard dimensions; that helps me visualize and understand the question better.
posted by the matching mole at 2:58 PM on April 14, 2015

Best answer: FYI, I did visual size comparison of the first and second Death Star to our Moon on Flickr. Click the left right arrow for some scenes of how it might look from our Earth. Notice I used the word "it" not "them" because the first Death Star was too small to bother much of anything. Same thing for the second one, but it at least it was much bigger.
posted by Brandon Blatcher at 3:17 PM on April 14, 2015 [10 favorites]

Also, it's not Darth Vader's Death Star, it's the Emperor's. Vader doesn't even run it, just drops by every now and then to scare the various troops and relax in his private suite. They were run by a Moff, a made up title definitely worth giggling about.

Its description as "the size of a small moon" really doesn't say much, as moons come in all sorts of sizes. It probably sounded impressive in 1977, when general knowledge of moon sizes was limited.
posted by Brandon Blatcher at 3:29 PM on April 14, 2015 [4 favorites]

Best answer: So the moon would mass 1200 times more than the death star. I don't think the death star would be a serious problem, gravity wise.

The thing is that the death star, made mostly out of steel, is denser than the moon. Comparing the density of the death star to an aircraft carriers (they're both massive weapons stations, effectively), estimates are about 1.08×10^15 metric tons of steel, or 1.08x10^18 kg. Meanwhile, the moon is estimated to be 7.34 x 10^22 kg.

Using newton's law of gravitation, the force of gravity at the surface of the death star would be G*(1.08x10^18)/((70x10^3)^2) while the force of gravity on the surface of the moon would be G*(7.34x10^22)/((1.7x10^6)^2

Force of gravity on surface of death star = G*(1.08x10^12)/4900 = 2.2x10^8
Force of gravity of surface of moon = G*(7.34x10^10)/2.89 = 2.54x10^10

=> The force of gravity on the moon is 115 times greater than the force of gravity on the death star. That's almost 1/700th the force of gravity experienced on earth.
posted by deanc at 5:09 PM on April 14, 2015 [2 favorites]

Sorry... "Force of gravity on surface of..." should read "Acceleration of gravity on surface of..."
posted by deanc at 5:37 PM on April 14, 2015

The thing is that the death star, made mostly out of steel, is denser than the moon.

As seen in Return the Jedi, the middle of Death Star 2.0 had a huge amount of empty space in the center for its "hypermatter reactor, which possessed an output equal to that of several main-sequence stars. Within this chamber burned a reaction of prodigious proportions, fed by stellar fuel bottles lining its periphery."

The force of gravity on the moon is 115 times greater than the force of gravity on the death star. That's almost 1/700th the force of gravity experienced on earth.

From the same link above: "The incredible energies harnessed by the station combined with its great mass gave the Death Star magnetic and artificial gravitational fields equal to those found on orbital bodies many times greater in size."

posted by Brandon Blatcher at 8:04 PM on April 14, 2015

The Death Star is mostly hollow. We see the big internal spaces all the time.
posted by w0mbat at 8:54 PM on April 14, 2015

Well, there are also ships called Interdictors in the Imperial fleet, so called because they have fairly massive gravity well projectors, used to yank passing ships out of hyperspace into normal space or, I think more properly, prevent ships from fleeing battle by escaping into hyperspace.

It would totally stand to reason that the Death Star would have some of these - I mean you need something to fill up all that space. In that case, the Death Star could cause significant havoc to the Earth or other bits of the solar system. According to the second link above, the Empire was experimenting with gravity well projectors on a planetary surface as a weapon of mass destruction that could "cause severe earthquakes or even deform the entire globe."
posted by Naberius at 10:16 AM on April 15, 2015

Yeah this is fundamentally unanswerable without knowing the detailed contents of the Death Star. I mean, a neutron star is only ~10 kilometers in diameter - the size of a small town - and it has more mass than our entire Sun (1.4-1.6 solar masses is typical).

A black hole could be even more massive and/or smaller, and if I was building a Death Star, I'd have a few of those in storage for energy production, hyperspace tunneling, general meddling with local gravity, and so on. So my Death Star would easily exert more gravitational pull than our Moon. The Emperor's? I dunno.

(As an aside, we have neutron stars that spin faster than a kitchen blender, and some have magnetic fields strong enough that if you replaced the moon with one of them, they'd erase the credit cards in your pocket... Why even bother with a Death Star?)
posted by RedOrGreen at 11:36 AM on April 15, 2015

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