Based on 2 degree separation (some are closer) approximately 920 miles
If a satellite appears to move in the sky, then the ground station antenna has to keep moving in order to follow it, and the satellite will be below the horizon, invisible, and unusable for half the time over the long run. That would really put a crimp in the operation of all the little 2-ft satellite TV dishes on top of garages.
There is only one geostationary orbit because in order for any mass m to orbit the Earth (ME) the gravitational force: EQ1: Fg = GmME/r^2 has to be such that it is equal to the required centripetal force for uniform circular motion: EQ2: Fc = mv^2/r where v is the velocity of m at radius r (distance from the center of the Earth) and: EQ3: v = 2(pi)(r)(f) f is the frequency of rotation in revolutions per second. For geostationary orbit the satellite must be in a fixed position (it must have the same frequency of rotation or angular velocity as the Earth's rotation) relative to the Earth and orbit above the Earth's equator. The necessary velocity to satisfy Fg = Fc is a specific value, therefore (since pi and f are fixed values) r is the only variable in EQ3. There is a specific orbital radius for geostationary orbit of any mass m.
the speed of light = 299792458 m / sGEO (Geostationary Earth Orbit) = 35863000 m above the Earth's surfaceround trip time = 2*(35863000)/299792458 =0.239 sthe time needed for an RF signal to reach a GEO satellite and gets retransmitted back to a ground station on earth is approximately 240 milliseconds(assuming zero signal propagation/processing time in the satellite and equatorial ground station location with the same longitude as the satellite slot)
Answer: 1.communication satellite 2.navigational satellite 3.weather satellite 4.millitary satellite 5.scientific satellite 6.satellite launches. It composed of 6 satellites...........i hope.....my answer can help you.....
In theory, 2 satellites in diametrically opposite geosynchronous orbits could cover the planet. In order for the satellites to communicate, a minimum of 3 would be needed, each at a 60 degree angle to the others. At this point, the strength and quality of coverage increases proportionally to the number of satellites.
this is if it is not a geostationary orbit, in which case it is always in the same place relative to earth. imagine a circle of radius 42250km+radius of earth(6,356km). it's perimeter is 2 x pi x 48606km = satellite's journey. then think of a really fast speed, which is the speed the satellite is moving at. divide the distance by speed and you have the time of one orbit However, by the height being 42250 , it makes me think the satellite is a geostationary satellite and so it would take 24 hours moving at approximately 12725 kmph does that answer your question?
If a satellite appears to move in the sky, then the ground station antenna has to keep moving in order to follow it, and the satellite will be below the horizon, invisible, and unusable for half the time over the long run. That would really put a crimp in the operation of all the little 2-ft satellite TV dishes on top of garages.
The Philippines has multiple satellites in orbit for various purposes such as communication, weather monitoring, and earth observation. Some examples include DIWATA-1 and DIWATA-2, the Philippine Scientific Earth Observation Microsatellite (PHL-Microsat), and Agila-2, a geostationary communications satellite.
1962 - satellite tv was relayed from Europe to the Telstar satellite over North America 1963 - Syncom 2 a geosynchronous communication satellite was launched 1965 - Intelsat I a commercial communication satellite was launched 1967 - Orbita a national network of satellite television was created in Soviet Union 1972 - Anik 1 a domestic North American satellite of Canada to carry television satellite was launched 1974 - ATS-6 an experimental educational and direct broadcast satellite was launched 1976 - Ekran a Soviet geostationary satellite to carry Direct-to-Home television was launched
What is the distance between (4, -2) and (-1,6)?
What is the distance between (4, -2) and (-1,6)?
There is only one geostationary orbit because in order for any mass m to orbit the Earth (ME) the gravitational force: EQ1: Fg = GmME/r^2 has to be such that it is equal to the required centripetal force for uniform circular motion: EQ2: Fc = mv^2/r where v is the velocity of m at radius r (distance from the center of the Earth) and: EQ3: v = 2(pi)(r)(f) f is the frequency of rotation in revolutions per second. For geostationary orbit the satellite must be in a fixed position (it must have the same frequency of rotation or angular velocity as the Earth's rotation) relative to the Earth and orbit above the Earth's equator. The necessary velocity to satisfy Fg = Fc is a specific value, therefore (since pi and f are fixed values) r is the only variable in EQ3. There is a specific orbital radius for geostationary orbit of any mass m.
1) The coverage area of a satellite greatly exceeds.2) Transmission cost of a satellite is independent of the distance from the center of the coverage area. 3) Satellite to satellite communication is very precise. 4) Higher bandwidths are available for use.
What is the distance between (4, -2) and (-1,6)?
It first depends on the type of orbit the satellite is in. If it is in a geostationary orbit, you can determine the speed by using the speed of the Earth's rotation at the equator (465m/sec), because a geostationary satellite orbits above the equator at 22,300 miles above. If it uses a geosynchronous orbit, that is, anywhere else but above the Equator, your distance above the Earth's surface is the same but your speed will differ as the inclination of the satellite is below 90 degrees. If the orbit is a LEO or MEO, your speed will obviously be faster, but the altitude of the satellite has a broader range, so knowing the altitude is essential to your calculation. If the orbit is elliptical, that is an entirely different set of equations, as satellites in elliptical orbits are 300 miles away from the Earth at their fastest to catapult them into their next pass and skyrocket up to 23,000 miles.
Just subtract the lowest number from the greatest number. For example, the distance between 3 and 8, is 8 - 3 = 5 units, the distance between -2 and 3, is 3 - (-2) = 3 + 2 = 5 units, the distance between -4 and -2, is -2 - (-4) = -2 + 4 = 2 units.
Distance= (r - R)/1000 Reference= http://lennyconundrumsolutions.blogspot.com/2008/07/round-271.html r= the distance between the Neopia's center of mass to the Kreludor moon's geostationary orbit (unsure of actual value) R= d/2 = 800 km = 800,000 m (radius) We use 1000 factor in the above equation because we want to get in kilometers. Round that distance number and we get the answer.