How Accurate Must the Orbit Radius Be for Geosynchronous Satellites?

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For geosynchronous satellites, the maximum acceptable drift in longitude over a 10-year lifespan is 10 degrees, which translates to a margin of error in the orbit radius. The initial calculation suggested a radius error of 320 meters, but the correct value is 210 meters. This discrepancy can be resolved by applying Kepler's third law, which relates the errors in radius and time period. Specifically, the allowed error in the radius is two-thirds of the relative error in the orbital period. Accurate calculations are crucial for maintaining the satellite's position in orbit.
Dorothy Weglend
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A satellite in synchronous Earth orbit has a life of 10 years. The maximum acceptable east or west drift in the longitude of the sat during its lifetime is 10 degrees. What is the margin of error in the radius of its orbit?
I am having some trouble with the above problem. I reasoned that the maximum error would be 1 degree a year, and therefore 1/365 degree a day. Since the circumference of the orbit varies directly with the radius, I thought I could do:
(delta r)/r = (1/365)(360),
Since I know r for a synchronous sat, (42,000 km) from an earlier problem, I can get the numerical value of delta r. Unfortunately, this gives me 320 m as the error in the radius, and the book gives 210 m as the answer. A bit too large for round-off error, I think :smile:
Can anyone help me on this?
Thanks,
Dorothy
 
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Dorothy Weglend said:
A satellite in synchronous Earth orbit has a life of 10 years. The maximum acceptable east or west drift in the longitude of the sat during its lifetime is 10 degrees. What is the margin of error in the radius of its orbit?
...
Since the circumference of the orbit varies directly with the radius, I thought I could do:
(delta r)/r = (1/365)(360),
Do not forget Kepler's third law. If Ro is the radius of the synchronous orbit and To is the time period around this orbit (1 day), and R and T are the same for the actual orbit,
R/Ro=(T/To)^{2/3}.
You get the relation between the error of R and the error of T by differentiation.
DelR/ Ro = 2/3 *(T/To)^{-1/3}\ DelT/To
The allowed error in T is 1/(365*360). (Well, not quite so, look after the accurate length of year.)
As T=To = 1 day, the corresponding relative error of R is two-third of the relative error of T.
Since I know r for a synchronous sat, (42,000 km) from an earlier problem, I can get the numerical value of delta r. Unfortunately, this gives me 320 m as the error in the radius, and the book gives 210 m as the answer. A bit too large for round-off error, I think :smile:
Can anyone help me on this?
Dorothy
Just multiply it by 2/3...
ehild
 
Thanks, Ehild. That was a big help. I appreciate it.

Dot
 

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