satellite horizon elevation azimuth North East
If we use a telescope, we note the time of the observation and we usually take a photograph of the satellite. We locate the satellite in terms of the two angles shown in [Figure 1]. One of these is the elevation angle—the number of degrees a telescope must be tilted above the horizon to see the satellite. The second is the azimuth angle—the number of degrees between the plane in which we measure the elevation angle and the north direction. Of course, we can also point a radar antenna at the satellite in the same manner. The radar can receive a signal transmitted by the satellite, or else it can send a signal to the satellite and watch for the reflected waves that eventually return. In the latter case, the satellite must have sufficient surface area to produce an adequate reflected signal. These two kinds of precision tracking were both possible with Echo I. Radar can also do something that optical equipment usually can’t do: measure the distance out to the satellite.
The Basic Physics of Satellite Motion
Figure 2
earth north pole satellite θ = n · t
The Echo I satellite was launched into a circular orbit inclined at an angle to the plane of the earth’s equator. In [Figure 2] this equatorial plane intersects the plane of the satellite orbit along the line OPM. The point O represents the center of the earth, the point M is on the satellite orbit, and the Point P is on the equator. At any instant, the satellite may be located in its orbit by the angle θ, which is measured between the line OM and the line OQ, where the point Q is the satellite’s location. If the satellite moves in a circular orbit, as in this case, the angle θ is proportional to time. That is, we can write θ = nt. We call n the angular speed of the satellite; one way of measuring this is in degrees per second.
Thus, the satellite is whirling at a constant speed about the earth like a stone tied to a string. Let us examine the physics of this situation a little more closely with the help of [Figure 3]. If the satellite is moving with the velocity v, then we know that the centrifugal force acting on it is
| mv² |
| r |
where m is the mass of the satellite and r is its distance from the center of the earth. Obviously, no string ties the satellite to the earth, but the force of gravitational attraction between the earth and the satellite has the same effect. Newton’s law of mutual attraction tells us that this force is proportional to the product of the two masses divided by the square of the distance between their centers, or