Sunday, February 22, 2009

high-speed orbits

The news a week or so ago about the high speed crash between the Iridium communications satellite and a defunct Russian military satellite got me wondering about why orbiting satellites have to go so damn fast. The artist's impressions of these things always give you the feeling that they are just drifting around up there but clearly that ain't so. So I got out my envelope and started to doodle on the back of it...

First of all let's figure out how fast someone on the equator is moving simply due to the Earth's rotation. The radius of the Earth r0 = 6,400 km. A person on the equator covers a distance equal to the Earth's circumference = 2 π r0 = 40,000 km in 24 hours, corresponding to a speed of 1,700 km/hr! Of course we don't notice this because everything around us is moving the same way, but it gives us a reference point to work from.

Next, let's figure out what radius (r) orbit will keep a satellite over the same patch of Earth using only the effect of gravity to keep it at the right speed. Gravity must pull on the satellite just enough to change the direction of its velocity vector v through a certain angle over time, i.e. at a certain angular velocity (ω = v/r = 360o/day). This change in velocity corresponds to an acceleration = v ω = r ω2.

We know that the acceleration due to gravity at a distance r from the center of the Earth g(r) is proportional to 1/r2.

So g/g0 = (r0/r)2, where g0 = g(r=r0) = 9.8 m/s2 (the acceleration due to gravity that we experience at the Earth's surface).

Now we equate g = (r0/r)2 g0 = r ω2 (*)

which gives a geosynchronous orbit at a radius r = 42,000 km. A satellite in this orbit must travel at v = r ω = 11,000 km/hr.

What about a low orbit such as the unfortunate Iridium satellite was recently in? These satellites apparently orbit about 780 km above the Earth's surface, i.e. r = 6400+780=7180 km. Plugging this into equation (*) gives ω = 360o every 100 min, i.e. a complete circuit around the Earth every 100 min. This corresponds to a velocity of 27,000 km/hr.

Of course, if you're outside the Earth's atmosphere in nice empty space, any velocity feels like you're just drifting around. Unless it's not quite empty and you hit something.

5 comments:

Boris Legradic said...

Nitpick:

Angular velocity is commonly (SI) given in radians/second, so it would be 2pi/day. Doesn't change anything in your calculation, of course.

It's just that I have been fighting with non-standard units in the literature lately (Townsend! who has ever heard of Townsend as a unit, I ask you? Not I!), so I felt compelled to point that out.

Boris Legradic said...

Oh, and nice post! It's funny how unintuitive physics become once you leave familiar territory - even for physicist.

Mike said...

Interesting post, even though you lost me with the physics math.

Stephanie B said...

Nice derivation.

I love orbital mechanics even though I was all set to correct your values on geostationary (surely, I thought, they'd be faster than low earth orbit since you have to go so much faster to get there), but that's just the potential energy drainin' ya.)Except, of course, you were right. Too much time playing with low earth orbit - my brain isn't working right.

And, Boris, I know what you mean on SI units. I'm so damn tired of dealing with pounds (which could be mass or force) instead of something clear cut.

As a point, in my novel, everything is in SI and I refused, though requested, to include English units.

NewtonsOcean said...

Thanks Boris. I checked out your post on Townsends etc - no wonder your tolerance for awkward units is measured in Nitpicks!

So for anyone reading this who isn't sure what radians are all about, they are the natural way to measure angle. Instead of arbitrarily cutting up a circle into 360 deg, you define an angle as the length of the curved piece of circumference (measured in units of the radius). So 360 deg becomes 2π radians. And of course the equations involving the angular velocity ω only work with angles measured in radians.

Mike - thanks for dropping by. If I lost you I failed! I could have taken an extra paragraph to explain how one can understand centripetal acceleration as linear velocity x angular velocity, which I thought about while doing the dishes that evening! The dishes didn't actually help me, but trying to explain it to one of my daughters did...

Thanks Stephanie - I can't wait to see your work of fiction published with a special note printed across one corner: "SI units used throughout"!