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
| km |
| r² |
where k is a constant that essentially represents the mass of the earth.[1] Newton’s law also tells us that this force will be pointing toward the center of the earth if the earth is spherical. When the satellite is in circular motion, the centrifugal force and the gravitational force must balance each other. Hence we have
| km |
| r² |
| mv² |
| r |
and from this we can solve to find that the velocity of the satellite must be equal to
v = √
| k |
| r |
In the case of the Echo I satellite, which was designed to have a radial distance of r = 5000 miles, this velocity amounts to about 4.4 miles per second. The time for one revolution in orbit is obtained with the formula
T =
| 2πr |
| v |
For the Echo satellite this time, T, turns out to be just about two hours.
Figure 3