The Resistance is trying to make a quick getaway before the First Order arrives. Then—boom! It's too late. They're already here—two Star Destroyers just arrived near the planet in space.
This is the scene that opens Star Wars: The Last Jedi, which just came out on DVD. Having the whole movie means I get to do lots more fun physics analysis, including answering this question: Could you actually see those destroyers from the surface of the planet?
The first question to consider: How far can a human see? Well, that's a simple question with an easy answer. I can see the moon—that's more than 200,000 miles away. Better yet, I can see a galaxy (if it's really dark) and that is over 2 million light years away. So, humans can see really far. Distance isn't the problem.
Really, the key idea here isn't "how far" but rather "angular size." The angular size of an object depends both on the actual size and the distance from the observer: The reason you can see something like the moon even though it is super far away is because it is big. If you put the moon much farther away, you couldn't see it. You can calculate the angular size as just the object size divided by the distance (in radians). If you want to convert to degrees, you need to also multiply by 180 and divide by π.
Angular size is the reason that you can hold out your thumb and cover up the moon. At the length of your arm, your thumb has an angular size of about 2° and the moon is only 1/2° even though we all know the moon is actually bigger than your thumb. So, the question becomes: What is the smallest angular size a human can see? How about an angular resolution of around 0.02°? Of course every person is different—so that's just an approximate value.
Here's a more practical example. Look up in the sky and look for a commercial airliner. If it's directly overhead, you can usually make out the actual shape of the plane at an approximate altitude of 30,000 feet (9.1 kilometers). This average aircraft could perhaps be a Boeing 737 with a length of 35 meters. With this distance and size, a 737 would have an angular size of 0.2°. Although this is bigger than the listed resolution of the human eye, this is the value I'm going to use. I just feel that if that plane appeared to be smaller, I couldn't really tell what it was.
Now back to Star Wars. I want to calculate the angular size of a Star Destroyer as seen from the ground. But how big is a Star Destroyer? I'm going to assume the spacecraft are actually Resurgent-class Star Destroyers with a length of 2.9 km (according to Wookiepedia). If they are in something like low Earth orbit (although they don't appear to be orbiting) it would be about 400 km away. This would give it an angular size of 0.4°. That's easily visible.
But wait! Why can't you see the International Space Station from the surface of the Earth? It's mainly because of the size. The ISS is at the same altitude but has a length of about 100 meters.This gives it an angular size of only 0.01°—much too tiny to see with the naked eye. Now, you can indeed see the ISS when it passes overhead because of the reflecting sunlight, but you can't see any details of its structure. Note: You should check out something like heavens-above.com to see when the ISS will pass over you (trust me, it's awesome).
OK, one more thing. Based on the scene from The Last Jedi, how far away are the Star Destroyers when they show up? There is always a problem with movies in that the angular field of view for the camera isn't always something that is known. In this case, there is something to help out. It's a great shot with a Resistance person in the foreground and the Star Destroyer in the distance. If I estimate the location of the camera and the person, I can get a value for the field of view and then measure the angular size of the Star Destroyer.
Let's say the Resistance person has a head diameter of 18 cm and is a distance of 2.5 meters from the camera. Using this to set the field of view, I get 30.9°. This puts the angular size of the Star Destroyer at 0.75° at a distance of 221 kilometers. That's actually pretty low altitude since the view is closer to the horizon and not directly overhead. Still, it seems pretty plausible overall. I'm pretty happy this works out since the scene is pretty cool.
