A Death Star (140 - 160 km diameter - Wikipedia, 120 km diameter Wookiepedia) is very large, yes. The moon (3474 km diameter) is much larger. The first thing we can do is look at the forces involved. How much force does the moon exert on the earth, and how much force this "Death Star" exerts on earth.
The moon is about 384,400 km from Earth and weighs about 7.34767309 * 10^22 kg. The equation for force due to gravity is $F_g = \frac{Gm_1m_2}{r^2}$ where $G$ is the gravitational constant $6.67408 * 10^{-11} \frac{m^3}{kg*s^2}$. Earth is about $5.972 * 10^{24}$ kg. From this we can calculate the force on the Earth due to the moon. Plug in all the numbers and it spits out a force on Earth of $1.98 * 10^{20}$ Newtons $\frac{kg*m}{s^2}$. However, for a 1km block on the surface of earth (earth has a radius of 6371 km), the force is $3.43 * 10^{-5}$ Newtons.
Now, instead of using the moon, lets use this Death Star. Low Earth Orbit is between 160 km and 2000 km in altitude. So let's take the happy medium and say that it's at 1000 km in altitude. Now the hard part. How massive is this Death Star? If we say this Death Star is like an aircraft carrier, we can say that the density of the aircraft carrier is about the density of the Death Star.
Wikipedia lists the Nimitz-Class aircraft carrier at 100,000 long tons. A long ton is about 1016 kg. So the aircraft carrier has a mass of about 101,600,000 kg, or $1.016 * 10^8$ kg. Lets say the volume of the aircraft carrier can be approximated with a triangular prism. The only problem is that I don't know how tall the aircraft carrier is, so this is likely to be a order of magnitude approximation, because I'm going to approximate it using an equilateral triangle. A triangle is $A = \frac{1}{2}bh$ and a prism is $V = Bh$ (big B because the base in this case is an area). A Nimitz-Class aircraft carrier is 1100 ft long and 252 ft wide. That's around 334 meters by 77 meters. Since the triangle is equilateral, the height is $77 * \sin(\frac{\pi}{3}) = 77 * \frac{\sqrt{3}}{2} \approx 66.7$ meters. The area of the triangle is ~2567 m^2 and the volume of the prism ~857500 $m^3$. Now that we have mass and volume we can find density.
Density is mass divided by volume. So the density of the aircraft carrier is ${1.016*10^8 \, kg \over 857500 \, m^3} = 118.5 \frac{kg}{m^3}$. Now we need to find the volume of the Death Star. The Death Star is a sphere, so the only parameter we need is the radius. We stated at the very beginning that the Death Star is about 140 km in diameter, so its radius is about 70 kilometers. The equation for the volume of a sphere is $V = \frac{4}{3}\pi r^3$, so plug in $r = 70$, and the volume of the Death Star is about $1437000 \, km^3$ or $1.437*10^15 \, m^3$. Since the mass-density of an aircraft carrier is a good approximation for the mass density of the Death Star, then we can use the aircraft carrier density times the volume of the Death Star to get its mass.
The mass of the Death Star is $118.5 \frac{kg}{m^3} * 1.437*10^15 \, m^3 = 1.7*10^17 \, kg$. That's pretty massive. Now let's say this Death Star is orbiting at 1000 km above earth's surface. The total force on the earth is $1.25*10^18$ Newtons. That's about two orders of magnitude away from the force on earth from the moon. However the force on a 1 km block on the surface of the earth is $1.14*10^{-5}$.
Huh. That's less than the moon. But we can move it closer. Let's move it in as close as we can, 160km (LEO lower bound) + 70km (radius of Death Star) = 230km. At that height, the force on a 1 km block on the surface of the earth is $2.15*10^{-4}$. Now that's about an order of magnitude above the moon, which we already know causes tides. But is it enough to cause earthquakes?
There are sites on the internet that link the moon cycle and earthquakes. So I would say that it is indeed plausible that the Death Star could cause an earthquake simply by existing.
tldr: the death star could probably cause earthquakes.
Now, how to get rid of it. The easiest thing to do would be to lift its orbit until it no longer causes earthquakes. The easiest way to do that would be to infiltrate the spaceship and get it to put itself in a higher orbit. If there is intelligent life aboard this spaceship (very likely) then getting rid of it is going to be the least of your problems. The simple fact that they are causing earthquakes with little regard for the life on the planet is the first warning bell. They're not here to make friends. And they're here because they want something. They may want us as slaves (unlikely, robots are cheaper) or our minerals (star miners?). In any case, we need more information about who these people are and why they're here. So the first thing to do would be to try and establish contact. Examine the signals that the Death Star gives off. Then, build a plan appropriate to the information provided. If they request a representative of our race, that's the perfect way in. If their ship gives off no signals at all, that means that there is a vulnerability that we can try to take advantage of.