# Would the Star Wars Death Star in a low Earth orbit be torn apart by Earth's gravity?

I got the question from this reddit comment:

Well the moon is a quarter million miles from earth. The death star would only need to be in low earth orbit to stay aloft (which according to Wiki is 100-1200 miles in altitude, not 250k miles). I would like to see what the death star looks like at say 200 or 500 miles and not a quarter million.

I think - but my physics is very very rusty so it's just a hunch - that the Death Star would be torn apart so close to a big planet.

What is the physics of this? Gravity depends on distance, square of the distance. Given the huge size of the Death Star I think the difference in force between the near-planet point of the space ship and the far point is too big for such a structure.

Strange enough, I can only find the opposite question when I google, how the Death Star would affect a planet.

• How big is the Death Star? Jan 11, 2017 at 7:03
• Found this: "Various sources state the first Death Star has a diameter of between 140 and 160 kilometers. There is a broader range of figures for the second Death Star's diameter, ranging from 160 to 900 kilometers." -- apparently there is no exact number. EDIT: Apparently the first Death Star was 160 km and the second one 900 km according to this. Everybody has a different number it seems. Jan 11, 2017 at 7:04
• The Roche limit tells you how close you can orbit, but it's for bodies held together by gravitation only, and the Death Star, I guess, is welded and bolted. The tidal force is $F_T = 2GMur/d^3$, so you should compare that to some measure of the strength of steel. Jan 11, 2017 at 7:23
• The Death Star cannot stay indefinitely in an orbit below the geosynchronous orbit at a height of 35,786 kilometers no matter how strong it is. Below this height, the tides on the Earth caused by the Death Star will fall behind the Death Star due to friction, this will cause the Death Star to spiral inward. Jan 11, 2017 at 8:16
• @CountIblis: Nice comment, but I suppose if the DS can zip around the galaxy it's got enough "puff" to keep from falling in :) Jan 16, 2017 at 14:58

Probably not.

Consider the Death Star as two hemispheres, one closer to Earth and one further. The center of mass of a hemisphere is $\frac38$ the way from the center of the sphere to the edge. Let us suppose the Death Star is orbiting 300 km above Earth and has a radius of $r$ of 80 km. Then the bottom hemisphere is 270 km above Earth and the top hemisphere is 330 km above Earth.

The Death Star would have an acceleration of $G M_{earth} / R^2$, where $R \approx 6700 km$, the radius of the Earth plus the height of the Death Star. The acceleration of the bottom half due to gravity is $G M_{earth} / R_{bot}^2$, where $R_{bot} \approx 6670 km$, because it is slightly closer to Earth. The delta is approximately $\frac{G M_{earth}}{R^3} \Delta R$ with $\Delta R \equiv R - R_{bot}$, and this delta must be made up for by internal tension in the Death Star; there is a force from the top half on the bottom half pulling it upwards away from Earth.

We can approximate the force as $g \frac{\Delta R}{R} m$ with $g$ gravitational acceleration in low Earth orbit (which we take to be 10 m/s^2) and $m$ the mass of the lower half of the Death Star. This is about $.05 m/s^2 * m$

Giving the Death Star a density of $1 gm/cm^3$, we get a mass of about $10^{18} kg$, or a force of $5*10^{16} N$ between the two halves. That gives a tension of about 2.5 million Pascals, about two orders of magnitude below the strength of steel. (Note that in giving the Death Star a mass of 1 gm/cm^3, it would be roughly 20% structural if made of steel, so there is a safety factor of about 20.) The Death Star would feel a lot of stress in low Earth orbit and would deform a noticeable amount, but wouldn't necessarily be torn apart.

Let's also consider the gravitational pull between the two halves, to see how much that helps hold it together. Using the same numbers as above and modeling the Death star's hemispheres as point located at their centers of mass, the force between them comes to $2*10^{16} N$. This is off by a factor of 3/4, but suffices to show that while the gravitational pull is significant, it wouldn't hold the Death Star together; it has to be held together structurally.

• In other words, the strength of the force is insignificant next to that required to destroy a planet! Jan 11, 2017 at 7:43
• You've assumed the whole equatorial plane in the Death Star transmits the force, no? So this calculation is pretty optimistic, but given that we've got at least an order of magnitude margin on our side, the conclusion remains the same - i.e. that one could come up with a design that would work. I'm really surprised - my gut feeling without your calculation would be that any "built" thing coming much nearer than the Roche limit would be shredded! so +1 for teaching me something! Jan 11, 2017 at 10:21
• The calculation is right, but the Death Star had a diameter of 120km. The results are though much the same. Jan 11, 2017 at 11:18
• @WetSavannaAnimalakaRodVance Roughly, that's what I did, yeah. I giving a density of 1 gm/ cm^3 I made the Death Star roughly 15% - 25% structural, depending on the material. So I needed the stress to be at least 4 - 7 times lower than the tensile strength. I could have spelled that out in more detail! Jan 11, 2017 at 12:11