How Accurate Must the Orbit Radius Be for Geosynchronous Satellites?

[Dorothy Weglend]

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
Can anyone help me on this?
Thanks,
Dorothy

 

[ehild]

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
Can anyone help me on this?
Dorothy

Just multiply it by 2/3...
ehild

 

[Dorothy Weglend]

Thanks, Ehild. That was a big help. I appreciate it.

Dot