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.
 A: 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.
