William, from Honolulu, wrote in this week to ask:

If there was a space station/city the size of San Francisco in geostationary orbit, what would it look like from ground level with the naked eye? Would it cast a noticeable shadow?

It'd be a big project, to say the least.Image Credit: NASA |

This is a fun one! But to properly explore it, we need to talk a bit about orbits, because there's a surprising amount of information implied in the question.

For our readers who are unfamiliar with the concept, a

*geostationary orbit*is a circular orbit where a satellite is moving at just the right speed to stay over one spot on the planet's surface as the Earth rotates. The interesting thing is that there's only one height where this is possible, and it's pretty far away.

Although astronauts on the International Space Station

*look*like they're in zero-gravity, the reality is that they appear weightless because they're in free-fall. 250 miles up, Earth's gravity is nearly as strong as it is here at the surface, so the ISS has to move horizontally at 17,000 miles an hour in order to remain in orbit, like a cannonball launched so fast it goes over

*the horizon before it falls to the ground. This means that the ISS orbits Earth about 16 times a day, rather than the once-a-day of a geostationary orbit.*

So how far up do we have to go to reach geostationary orbit? The answer is pretty incredible—about 22,000 miles from Earth's surface (or 26,000 from the core), almost a tenth of the way to the moon. To make the journey around in one day at that height, the satellites have to move at about 2,170 mph, more than twice as fast as we're moving here on Earth's surface.

So now that we've got some ground work laid down, we can get to the fun part of answering William's question: the math!

San Francisco has an area of about 50 square miles, according to the city's page on Wikipedia. For convenience, we'll assume our space station is a circular disc facing Earth, with a square area of 50 miles.

We can use the formula for the area of a circle to find the radius of that circle:

Dividing both sides of the equation by pi tells us that r

^{2}is about 16, which yields a radius of 4 miles.

Knowing that, and the distance to a geostationary orbit, all that's left is a bit of trigonometry to figure out how big our space station would look!

When we talk about an object's apparent size, we use the term

*angular diameter*to talk about how much of our visual field it takes up—for example, if you're standing in front on an infinitely large flat wall, it takes up half of your 360° field of view, so it has an angular diameter of 180°. The sun and moon, though they're wildly different sizes, both have an angular diameter of about 0.5°, thanks to their different distances from Earth.

This is what we're trying to find here—what portion of the 180° night sky this 8-mile-across space station would take up, if it's 22,236 miles away. To do this, it helps to recall some high-school trigonometry, and build a mental triangle between the space station and a hypothetical observer below it on the surface of Earth. That way, we can use the trigonometry functions (sine, cosine, and tangent) to figure out the angle that relates these numbers.

A straight line between your eye and the center of the space station would be 22,236 miles long, while the distance from the space station's center to its edge is 4 miles. These two figures already comprise the

*adjacent*and

*opposite*legs of a right triangle, as shown below. If we wanted to use the sine or cosine functions to solve this problem, we could calculate the length of the hypotenuse using the pythagorean theorem, but it's not really necessary for this—the tangent function already relates the opposite and adjacent legs of a triangle to the angle we're looking for.

The

*tangent*function, given an angle as its argument, tells us the ratio of the opposite to adjacent legs of a right triangle with that angle as its most acute. Correspondingly, we can plug in the same ratio to the*inverse tangent*function and find the angle of interest:
As shown above, that's only half the total angular diameter, meaning that in total this disc would take up 0.0206° of the visual field. Remember: for comparison, the moon is about 0.5°. So what would this look like?

Well, we can see geostationary satellites in the night sky above us, and all of them are less than 8 miles across. That means that we'd easily be able to pick out the San Francisco Satellite—but it would likely appear as a bright point in the sky, like a star or planet, rather than having resolvable features like the moon. Like other satellites in geostationary orbit, it would appear to move very slowly against the background of stars—not actually moving, but remaining in place as the Earth turns and the visible sky shifts.

Well, we can see geostationary satellites in the night sky above us, and all of them are less than 8 miles across. That means that we'd easily be able to pick out the San Francisco Satellite—but it would likely appear as a bright point in the sky, like a star or planet, rather than having resolvable features like the moon. Like other satellites in geostationary orbit, it would appear to move very slowly against the background of stars—not actually moving, but remaining in place as the Earth turns and the visible sky shifts.

Although such a small space station wouldn't cast a visible shadow, we'll be talking more about apparent size and distance in another post soon, exploring the upcoming solar eclipse—so stay tuned!

**—Stephen Skolnick**

One would have to presume that any designers of said city wouldn't be stupid enough to make it flat (or go to the trouble to orient it perpendicular to Earth and keep it that way), so in reality this should be based on a spherical or near-spherical design.

ReplyDeleteThis lessens the visual impact significantly. I won't go through the math, but presume an average height of your 50 square miles is inhabited volume, and use a sphere of that volume. The diameter is a LOT smaller (but at least it would always look the same from Earth as it goes round and round, oriented to absolute rest frame coordinates).