What are the restrictions on the orbit of a communications satellite if it is to appear motionless in the sky as viewed from any location on earth?
The satellite must be in a cicular orbit with a period of one sidereal day (23 hours and 56 minutes). Use Kepler's third law to compute the distance from the center of the Earth.
There is a clear distinction between a geosynchronous orbit and a geostationary orbit. The early recognition of a geostationary orbit was made by the Russian Konstantin Tsiolkovsky early this century. Others referred to the unique orbit in writings about space travel, space stations, and communications. It was probably Arthur C. Clarke who was given the major credit for the use of this orbit for the purpose of worldwide communications.
The geostationary orbit is one where a spacecraft or satellite appears to hover over a fixed point on the Earth's surface. There is only one geostationary orbit in contrast to there being many geosynchronous orbits. What is the difference you ask? 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 ~22,238.64 miles, or ~35,787.875 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 ~10,088.25 feet per second or ~6,877 MPH. The point on the orbit where the circular velocity of the launching rocket reaches 10,088.25 fps, and shuts down, is the point where the separated satellite will remain. The point on the Earth's surface immediately below the satellite is moving with a velocity of 1525.85 ft./sec.
How do you determine the altitude at which a satellite must fly in order to complete one orbit in the same time period that it takes the earth to make one complete rotation?
The force exerted by the earth on the satellite derives from
...................................................F = GMm/r^2
where G = the universal gravitational constant, M = the mass of the earth, m = the mass of the satellite and r = the radius of the satellite from the center of the earth.
GM = µ = 1.407974x10^16 = the earth's gravitational constant.
The centripetal force required to hold the satellite in orbit derives from F = mV^2/r.
Since the two forces must be equal, mV^2/r = µm/r^2 or V^2 = µ/r.
The circumference of the orbit is C = 2Pir.
A geosycnchronous orbit is one with a period equal to the earth's rotational period, which, contrary to popular belief, requires 23hr-56min-4.09sec. to rotate 360º, not 24 hours. Therefore, the time to complete one orbit is 23.93446944 hours or 86,164 seconds
Squaring both sides, 4Pi^2r^2 = 86164^2
But V^2 = µ/r
Therefore, 4Pi^2r^2/(µ/r) = 86164^2 or r^3 = 86164^2µ/4Pi^2
Thus, r^3 = 86164^2(1.407974x10^16)/4Pi^2 = 2.647808686x10^24
Therefore, r = 138,344,596 feet. = 26,201.6 miles.
Subtracting the earth's radius of 3963 miles, the altitude for a geosynchronous satellite is ~22,238 miles.
To appear motionless in the sky as viewed from any location on Earth, a communications satellite must be in a geostationary orbit.
Here are the step-by-step restrictions for a geostationary orbit:
1. A geostationary orbit is a specific type of orbit around 35,786 kilometers (22,236 miles) above the Earth's equator. This altitude is where the satellite's orbital period matches the Earth's rotation period, resulting in the satellite appearing motionless relative to an observer on the ground.
2. The satellite must be positioned directly above the Earth's equator. This is because the geostationary orbit aligns with the Earth's equatorial plane, allowing the satellite to stay stationary in relation to a point on the ground.
3. The satellite's orbital inclination, which represents the angle between the plane of the satellite's orbit and the Earth's equatorial plane, must be zero. This means that the satellite's orbital path should be perfectly aligned with the equator.
4. In addition to being positioned above the equator and having a zero inclination, the satellite must also orbit in the same direction as the Earth's rotation, which is from west to east. This is necessary for the satellite to maintain its relative position with the ground.
By meeting these requirements, a communications satellite can maintain a fixed position in the sky as seen from any location on Earth, thus appearing motionless.
To appear motionless in the sky as viewed from any location on Earth, a communications satellite needs to be placed in a geostationary orbit (GEO). Here are the restrictions and requirements for a GEO:
1. Orbit Height: A communications satellite must be placed at an altitude of approximately 35,786 kilometers (22,236 miles) above the Earth's equator. This is known as the geostationary altitude.
2. Orbital Plane: The satellite's orbit should lie on the same plane as the Earth's equator. This means it must orbit around the Earth's equator at a fixed inclination of 0 degrees.
3. Orbital Speed: The satellite must orbit the Earth at the same rotational speed as the Earth itself. This is approximately 3.07 kilometers per second (6,855 miles per hour). By matching the Earth's rotational speed, the satellite will remain synchronized with a specific point on the Earth's surface, appearing motionless from a fixed location.
4. Angular Momentum: The satellite's angular momentum must be equal to the Earth's angular momentum. This ensures that the satellite will maintain its position relative to the Earth.
By satisfying these criteria, a communications satellite can maintain a stationary position in the sky as observed from any location on Earth.