Size of object in Earth orbit visible from ground

[AtomicDog]

In the Battle of Sector 001 "Star Trek: First Contact", a Borg cube in low earth orbit fought off the combined forces of Starfleet until it was destroyed by the timely arrival of the Enterprise.

My question is,  given the size of a cube at 3000 meters per side and the approximate altitude of 370km (same as the space station), what angle would it subtend as seen from the ground? Could you even point to it during the day or twilight and see that a battle was going on?

[Allan F]

That object would be about 0.46 degrees across. If it was sunlit, it would be clearly visible. The ISS is at a comparable height, but much smaller, and clearly visible.

Edit: A Borg Cube would appear at least 1100 times bigger - depending on the angle you see the ISS from.

[AtomicDog]

That's almost as big as the moon! Could you show me how you computed that,  so I can apply it to things like the Death Star?

Edited to add: Thanks! I'm currently writing a short story about a young woman who observed the battle from the ground and how it made her decide to enroll in Starfleet Academy.

[Allan F]

Using the sine relation, a triangle with sides 370, 370.0001 and 3 kilometers, the size of the object would be (sin(90)/370 = Sin(x)/3.

Sin(90) is 1, so the equation reduces to 1/370 = sin(x)/3 and Again 1/123 = sin(x) and Again Sin(x) = 0.00802 -> x=0.46

The moon is at 1000x that distance, with a diameter of 3400 km, so yes, the objects would be visually comparable in size.

[Allan F]

The sine relation is not strictly speaking the correct way to compute this, but when the accuracy is not critical, it gives an easy way to calculate a very close number. To achieve greater accuracy, the sine relation can still be used, but then you'll have to use it on two equal right-angled triangles with sides x, 1.5 and 370.

[AtomicDog]

Just what I need. Thanks again!

[Allan F]

More accurately calculated: Side lengths 3, 370 and 370.01216

The angle would then be (sin(90)/370.01216) = sin(x)/3 -> sin(x) = 0.0081076 -> x= 0.46453635131 degrees.

The biggest error in my first calculation stems from dividing 370 with 3 and only use the integer part in the rest of the calculation, but it does not matter. There are other and bigger sources of error.

Since the precision of the original numbers aren't specified, the numbers of significant digits on the answer doesn't make any sense beyond 0.46.

[Chew]

That is the most convoluted answer I've seen that didn't come from a conspiracy theorist! Just say the answer is sin(angle) = 3/370.

[Allan F]

That was what I did, but I also showed how I reached that conclusion. It has been a long time since I did any math, and I'm at Work at the moment (doing nothing, unfortunately). So I had to Work the method out so that I myself could see I was on the right track.

[ka9q]

For such small angles you can use the small angle approximation

sin(x) ~ x

So the angle, in radians, subtended by an object is simply the size of the object divided by the distance:

3000 m / 370,000m = .008 radian = 8 mrad (milliradian)

To convert radians to degrees, multiply by 180/pi:

.008 rad * 180/pi = 0.46 deg.

or multiply milliradians by 0.0573:

8 mrad * 0.0573 = 0.46 deg.

Note that this is the maximum size of the object, when it is directly overhead. Few passes are directly overhead, and even when they are it'll only be there for an instant. For most of a pass the range will be considerably greater. A typical round number I use for the range to a satellite in low earth orbit is 2000 km, so the apparent size would be 3/2000 = 1.5 mrad = 0.086 deg.

[Allan F]

That is true for an object in an unpowered orbit. In the film, the Borg Cube was clearly under power, and able to pick a spot and stay there, but hey - it's a movie.

[ipearse]

I've only just spotted this, so my apologies for being late to the party, but I thought the battle between the Borg Cube and Starfleet was at Wolf 359, not Earth orbit.

[Allan F]

2 different battles. This one is from "First Contact".

[ipearse]

Whoops! Yes, you're right.