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?

Question

(a geosynchronous satellite orbits above the equator so that it is above the same location on the earth's surface at all times.)

Answer 1

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.

Therefore,
...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.