Friday, July 29, 2011

Geostationary orbit: Are satellites faster than the space shuttle?

Two colleagues and I went out for lunch today and one of them asked what the word 'geosynchronous' meant. It's a term to describe the orbit of a satellite that appears to be stationary over the Earth, we answered, and then we all three pushed our imaginary glasses up the bridge of our noses.

We were half right with our explanation. Geosynchronous is a term used to describe the orbit of a satellite that moves at the same speed that the Earth rotates about its axis. However, because this orbit can be titled over the Earth like an angel with a lopsided halo, the satellite can appear to move north and south in the sky throughout the day, though it always stays over the same line of longitude.

A geostationary orbit, the one we often think of when we hear the word "geosynchronous," is when a satellite is in a geosynchronous orbit over the equator. In this kind of orbit, the satellite appears to be stationary over the Earth.

In the same way that a square is always a rectangle but a rectangle isn't always a square, a satellite in a geostationary orbit is always in a geosynchronous orbit, but not the other way around.

Communications satellites are often in geosynchronous orbits so that the antennas of ground stations can remain constantly pointed at the same spot in the sky. Weather satellites are also common geostationary orbiters so that they can constantly monitor the same spot on the Earth.

Back to our lunch discussion: How fast, we wondered over our kebabs, would a geostationary satellite have to be moving to stay stationary in the sky? The space shuttle orbiter, we know, orbits at around 8,000 meters per second (18,000 miles per hour) but it does a complete orbit in about 90 minutes. Would a geostationary satellite be going faster or slower?

To find out, I did a little math. To find the speed of an object, we divide the distance it crosses by the time it takes to cross that distance. (Speed equals distance divided by time.) The speed of the Earth's rotation is 465 meters per second, which we get from dividing it's circumference, 40,075 km, by 86,4000 seconds (the number of seconds in a day).

To find the circumference of the geostationary satellites' orbit, we add the radius of the Earth, 6,378 km, to the height of the satellite's orbit, 35,786 km, (which we obtained from Wikipedia) to get 42,164 km. We then multiply that number by 2*pi (the equation for the circumference of a circle is the circle's radius times 2*pi) to get 264,924 km.

Because the satellite has the same orbital period as the Earth's rotation, we divide the orbital circumference by 86,400 seconds and we get 3,066 meters per second (or 6,858 miles per hour) -- quite a bit slower than the space shuttle. Still, that's way faster than the average bear.

1 comment:

  1. For my conceptual astronomy class, I often express Kepler's Third Law (P^2=a^3) not as a formula in terms of distances and times, but as a description in terms of speeds: a planet closer to the Sun orbits faster than a planet farther from the Sun. Mathematically, if you assume circular orbits or T^2=r^3, you can rearrange this to be in terms of speed (use circumf/T=v) and you get that v^2∝1/r. (I don't give my students that derived formula, but in case you wanted a proof you can check my algebra.)

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