Transformers 3 - Dark of the Moon

Saturday night I went to see Transformers 3 - Dark of the Moon. Did I like it? Well, it was loud and I was out of the house for three hours.

As I was watching it though, I began to wonder about some physics (SPOILER ALERT) when Sentinel Prime began transporting the remains of Cybertron. So I said to myself, "Self. What sort of gravitational effects would that have on the Earth?"

Let's start with Cybertron. I will need to find its mass. So I'll need its radius and an estimate of its density. I need an image to make measurements, but I couldn't find an image from the movie, so working from memory, I made one. It appeared to me that Cybertron would be much bigger than the Earth. By the time Earth was saved, my estimate is that about 1/16 of Cybertron (by volume) had been transported. Then the Transformers' planet had a radius five times larger than Earth's. Since the war had destroyed much of Cybertron, only the cubic structural elements were left, so instead of a solid, we have a lot of empty space. Judging from the width of the structural beams, I'd judge that 105/125 of the planet is empty.

Now for geometry. The volume of a sphere is

The radius of the Earth is about 6400 km. So the volume of the transported Cybertron is approximately 2 X 10^22 cubic meters.

We can get from the mass from the density and the volume. Aluminum has a density 2700 kg/m^3, steel has a density of 7500 kg/m^3, and 4500 kg/m^3 for titanium. Now maybe the Transformers are using graphene for their structures, but doesn't their technology look metal-based and not organic? So I'll use the density of titanium as an average value. Plugging in the numbers yields for Cybertron's mass 1 X 10^26 kg. Yikes! Seventeen times more mass than the Earth.

Before I start with the gravitational effects, I need an estimate of how far Cybertron is from Earth. It wasn't very far - maybe 1 Earth radius away; that's 6400 km (4000 miles).

The biggest effect would be that the Earth would in effect become a moon of Cybertron; technically though, they would be a planetary binary system. Binary systems orbit about their common center of mass. If we calculate the center of mass, (it's okay to think center of gravity) it would be located 470 km beneath the surface of Cybertron,

How long would it take for the Earth to orbit Cybertron? Just under 8 minutes. Let's start with Newton's 2nd law of motion. Substitute Newton's law of universal gravitation, and the expressions for the centripetal acceleration and the speed of an object in circular motion. Now solve for the period.

Night time every 4 minutes.

Now let's consider the tides. Tides occur because of the Moon, the Sun, and the Earth's rotation. The major effect is called the principal lunar semidiurnal constituent and is due to the Moon. In short the Moon's gravitational pull is slightly larger on the side of the Earth nearer the Moon than on the far side; physicists call this a gravitational gradient.

So all I have to calculate is the gravitational force on opposite sides of the Earth and subtract the two to estimate the gradient. I find that Cybertron's gradient is three hundred million times stronger than the Moon's. I wouldn't be living near any large body of water.

It's a good thing Optimus Prime is on our side.

P.S. I noticed that when Decepticons came through the portal, they maintained their motion. That's what Newton's first law of motion would demand. But when the transportation of cCybertron occurred, it didn't maintain its motion. Violation!!!