[Illustration: _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
(_mv_²)/(_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_)/(_r_²),
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_)/(_r_²) = (_mv_²)/(_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_ = (2_πr_)/(_v_).
For the Echo satellite this time, _T_, turns out to be just about two
hours.
[Illustration: _Figure 3_]