posted by Ann .
at what height above the earth's surface in a circular orbit have to travel in order to have an orbital period half that of a geosynchronous satellite's period?
(a geosynchronous satellite orbits above the equator so that it is above the same location on the earth's surface at all times.)
The height may be derived from the equation for the period
...T = 2(Pi)sqrt(r^3/µ)
where T = the orbital period in seconds, r = the orbital radius and µ = the earth's gravitational constant.
A geosycnchronous orbit is one with a period equal to the earth's rotational period, which, contrary to popular belief, is 23hr-56min-4.09sec., not 24 hours. Thus, the required altltude providing this period is ~22,238 miles, or ~35,788 kilometers. An orbit with this period and altitude can exist at any inclination to the equator but clearly, a satellite in any such orbit with an inclination to the equator, cannot remain over a fixed point on the Earth's surface. On the other hand, a satellite in an orbit in the plane of the earth's equator and with the required altitude and period, does remain fixed over a point on the equator. This equatorial geosynchronous orbit is what is referred to as a geostationary orbit. The orbital velocity of satellites in this orbit is ~10,088 feet per second or ~6,877 MPH. The point on the orbit where the circular velocity of the launching rocket reaches 10,088 fps, and shuts down, is the point where the separated satellite will remain.
µ = 1.407974x10^16ft.^3/sec.^2
T = 86,164 sec./2 = 43,082 sec.
...43,082 = 2(Pi)sqrt(r^3/1.407974x10^16
Squaring both sides yields
..1,856,058,724 = 4(Pi)^2(r^3/1.407974x10^16)
Solving, r = 16,506 miles.
Sutracting the earth's radius of 3963 miles yields an orbital altitude of 12,543 miles.
The orbital velovity is V = sqrt(µ/r
V = 12,710fps = 8,666mph.
Time to complete one orbit
T = 2(Pi)sqrt(87,151,680^3/1.407974x10^16) = 43,082 sec.