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Science
by Jerry Oltion

How to Calculate an Orbit

 

Science fiction writers, like science fiction stories, come in many varieties. There's the soft kind, whose science—and scientific understanding—is, shall we say, soft. There's the far-out kind, whose speculations are so unusual it's hard to tell if they're being scientific or not. There are gritty realists who insert just a science-based tweak here and there. And there are the hard sf types, who pull out all the stops but try to get everything as scientifically accurate (or at least as plausible) as possible.

Among the practitioners of the latter, one litmus test has become something of a cliché: Can the author calculate his/her own orbits? It's one thing to know that something must go very fast to orbit the Earth, but it's another thing entirely to know how fast that is. Or how fast you have to go to orbit Mars.

It's surprising that this has become such an entrance barrier to the exclusive hard-sf club, because calculating orbits is easy.

For those of you in a hurry, the formula is v = √rg. That's all you need to know. For those of you who want to really master this one, there's a little more to it than that.

 

What Is an Orbit, Anyway?

 

If you drop a rock, it will hit your foot. If you toss it a little bit outward as you drop it, it will miss your foot and hit someone else's nearby. If you really give it a shove, it'll fly farther forward than it does downward, and it'll cause an owie several feet away. With unreasonable strength behind the throw it might go a mile or two before it hits the ground (and woe to any foot it lands on there). And if you were Superman and gave it all you've got, the rock would turn to gravel, but if it didn't, it might fly sideways so fast that the Earth's surface would curve away from it at the same speed that it falls. Atmospheric friction wouldn't let it continue for very long, so we have to imagine doing this without air in the way, which we can actually do a couple hundred miles up.

 


It's easy to visualize what would happen: Our rock zips away from us and starts falling toward the center of the Earth, but the Earth curves away just enough that the rock stays at the same height above the ground. Without air resistance, that rock would continue moving exactly as before, traveling the same distance per second and falling toward the center of the Earth the same distance as before. But once again, the Earth's curvature would move that surface downward enough that the rock would miss. And it's falling toward the center of the Earth, which means the angle of its path would keep changing to match the surface as it moves forward. So it would just keep on flying all the way around the Earth until it hit you in the back of the head.

Okay, so how fast does that need to be? The formula above will tell you that, but it doesn't tell you why. So let's figure it out ourselves.  

Somewhere, Over the Horizon

 

The Earth is roughly 8,000 miles in diameter. The horizon looks like it's a long ways off, but for a person whose eyes are 6 feet above the ground, the horizon is actually only 3 miles away. That's the point at which the Earth's horizon is six feet below your tangent line, the line you'd get if you laid a really long, really straight ruler flat against the ground at your feet. So if you threw your rock straight outward at exactly orbital speed, then that rock would have to drop six feet in order to stay the same distance off the ground when it got three miles out.

How long does it take a rock to drop six feet? That's easy: drop a rock six feet and time it!1 You need a really good timer, like maybe a video camera running at 100 frames per second, but however you measure it, if your precision is fine enough you'll get 0.61 seconds.

So if your orbiting rock has to drop six feet in three miles, and it takes 0.61 seconds to drop six feet, then it stands to reason that your rock has to travel three miles horizontally in 0.61 seconds. Divide 3 miles by 0.61 seconds and you get 4.9 miles per second. That's how fast your rock is orbiting. There are 3,600 seconds in an hour, so multiply that out and you get 18,000 miles per hour.2  

Double Check

 

That's really fast. Is it right? Let's use our formula above, the one that's been tested by Kepler, Newton, and probably even Einstein. v(elocity) = √r(adius) times g(ravity). Our radius is 4,000 miles (half the Earth's diameter), and gravity pulls at 32 feet per second per second. We have to use the same units for both numbers, so let's multiply 4,000 miles times 5280 feet in a mile to get 21,000,000 feet for the Earth's radius. Multiplying r times g gives us a big number, which my calculator's square root button tells me is 26,000 feet per second. There are 5,280 feet in a mile, and 3,600 seconds in an hour, so we multiply our 26,000 by 3,600 and divide by 5,280 to get...18,000 miles per hour.

And in fact, satellites orbiting just outside the Earth's atmosphere travel at 17,500 miles per hour. Things orbit more slowly the farther away they are, so our calculations are pretty much dead on, within the limit of precision of our initial figures.  

Farther Afield

 

How about orbiting the Moon? Well, the Moon's radius is about a quarter of Earth's, so call it 1000 miles. It has about 1/6 of Earth's gravity, so call it 5 feet per second squared. Using v = √rg and doing our unit conversions as above, we get 3,500 miles/hour. That's way slower than Earth orbit! Well, it should be. The Moon has less gravity.

Mars? It's roughly 2,100 miles in radius, and its gravity is about 38% of Earth's, or 12 feet/second squared. You can do the math, right?3  

Non-Circular Orbits

 

What happens if you throw your rock even harder? Does it go into a higher orbit? Well, yes and no. It definitely climbs higher on the side of the Earth away from you, but it comes right back around to its starting altitude in what's called an elliptical orbit.

You might imagine that you could throw your rock into a higher orbit by aiming higher, but alas, that doesn't work either. If you throw the rock at a higher angle, even if you throw it much faster, that rock's path will circle around (actually it will ellipse around) until it's coming at you from behind at the same angle that you threw it. But you threw it upward, so the Earth is in the way! The rock never makes it all the way around. It will rise really high about a quarter of the way around the Earth and curve around and whack into the ground somewhere thousands of miles behind you.

To achieve a higher orbit, you have to throw your rock from a higher starting point. Or you throw your rock at a higher-than-usual velocity for its current orbit, wait for it to coast out to the far end of its elliptical orbit, and then give it a nudge to circularize that orbit at its higher elevation. Note that by the time the rock gets all the way out there to the end of its ellipse, it will be travelling much slower than it was at first, and your additional nudge doesn't speed it all the way up to the same 18,000 miles/hour you needed for low-Earth orbit. Things orbit more slowly the farther away they are; it just takes more energy to put them there.  

Free Lunch

 

Speaking of which, why do we launch rockets eastward instead of westward? Because the Earth rotates from west to east. Near the equator, that rotation means the surface is moving at about a thousand miles an hour. It's free motion! Our rocket is already moving at a significant fraction of its orbital velocity before we even light the fuse. If we aimed it westward, we would have to cancel out that thousand-mile-an-hour velocity, then build up another thousand miles an hour, just to arrive at the initial velocity that we get for free if we launch eastward instead.  

Escape Velocity

 

I should mention that if you throw your rock(et) too fast, the ellipse of its orbit will be so long that it will never come back. That's called escape velocity, and it's 1.4 times the orbital velocity at any given altitude. Of a circular orbit! It could be just a sneeze faster than an elliptical orbit.  

Bonus Nerdy Calculation

 

You can also use the same formula that we use for orbits to determine how fast your space station has to revolve to produce one gee of spin gravity. Suppose we have a space station 100 feet across. That's a radius of 50 feet. If we spin it to get one gee (32 feet per second squared), its rim has to move at 40 feet per second. That doesn't sound so bad until you realize the circumference of that space station is only 314 feet (pi times the diameter). That means it makes one revolution every 8 seconds, which is pretty fast. Your astronauts will probably get dizzy. If you're designing a space station, go for 1000 feet diameter; that gives you a comfy...

Ha! You figure it out.

 


 

1  No, you can't time it dropping 600 feet and divide by a hundred, because falling rocks accelerate. They move farther and faster with each passing second.

2  I'm sticking with only two significant digits here, because these are back-of-the-envelope calculations and the answer can't be more accurate than the initial numbers.

3  7,900 miles/hour using my back-of-the-envelope approximations. The actual figure is closer to 7,500, so we're well within the ballpark..
 

__________________________________

Jerry Oltion has been a science nut since he was old enough to spell "curious." He has written science fiction almost as long, and has done astronomy somewhat less. He writes a regular column on amateur telescope making for Sky & Telescope magazine, and spends many, many nights a year out under the stars.
 

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