Science Fiction... or Science FACT?

So it turns out that Elder-Geek.com was interested in my calculations, so I wrote a feature article on it:

Is Science Fiction Science FACT? - Death Star Physics

You should definitely check that out! It's got all you need in there! The story behind my quest for knowledge, the math, cool pictures, THE WORKS!

Alright, the moment you've all been waiting for! Here's the physics behind the Death Star:

I wanted to find the minimum amount of energy needed to destroy Alderaan. I thought that a good way to do this would be to find the amount of energy needed to tear the planet apart, chunk by chunk, at escape velocity (meaning just fast enough to pull free of the gravitational field). This is nowhere near as dramatic as it is in the film, but it’s a start. First, we’ll use the equation for the potential energy of an object in the gravitational field of the planet:

where G is the gravitational constant, M is the mass of Alderaan, and m is the mass of the chunk being torn off. We want to make this easier on ourselves, so we’re going to substitute an equation for both masses. We’ll do this by using:

where m is mass, ρ is the density, and V is volume. First up, the mass of Alderaan! We know the volume of a sphere to be:

so our new equation for M is:

We’re going to take a slightly abstract approach to finding an equation for m. We’re going to imagine that we’ll “peel” a layer off of the planet that’s infinitesimally small (really, really, really, small). We’ll find the volume of this layer by multiplying the surface area of the spherical layer (4πr²) and the width of the layer (dr). This will give us the following equation for the mass of the layer:

Now we’re ready to calculate the energy. We’re going to take the integral of our formula so that we can add up every layer from the surface to the core. Assuming Alderaan has a radius R:

Since most of the values in the equation are constants, we can take them out front and simplify the integral:

After integrating, we see that plugging lower boundary term results in zero, so we now have our final equation:

We can now plug in our values:

• G = 6.673 x 10^-11 m³ kg^-1 s^-2
• Diameter of Alderaan = 12,500 km → R = 6250 x 10³ m
• ρ_Alderaan*= 5515.3 kg m^-3

Plugging in these numbers gives us:

E = 2.038 x 10³² J

*Note: I assumed that the density of Alderaan in equal to that of the Earth. I did this for several reasons: (1) Both Alderaan and Earth are terrestrial and support human life, (2) Alderaan’s diameter is only 250 km shorter than the Earth’s, and (3) Alderaan does not have any major oceans, just rivers and lakes. This means that it would have less water than the earth and more dense land mass.