A Satellite in Sun-Synchronous Orbit

  • #1
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Homework Statement


A remote sensing satellite is deployed in a circular sun-synchronous orbit with inclination 96.85 degrees.

A. Calculate orbit altitude. I did this and got 349km.
B. The spacecraft coefficient is 200 kg/m^2. Solar conditions during the spacecraft launch are characterized by the solar flux F10.7 = 90. Is the orbit altitude sufficiently high for the spacecraft to operate through the estimated 2-year life time? What would be the spacecraft approximate lifetime if it is launched during solar maximum?
C. What is the maximum distance from the spacecraft to the horizon? I got 2139 km.
D. What is the maximum possible swath width? I got 12094 km.
E. What is the maximum possible time in view? I got 1560 seconds.
F. What is the ground speed of the subsatellite point? I got 7.3 km/s
G. The ground communications station is located at the satellite ground track. What is the communication time if the local conditions limit spacecraft visibility to elevation angles larger than 18 degrees?



The Attempt at a Solution



Assuming I got A,C,D,E,and F correct, I need help with B and G. For part B, I know the solar flux is related to the atmospheric density. If i get the atmospheric density I can use that, the radius of orbit and the ballistic coefficient to calculate the change in radius per revolution, and can calculate # of revs in a given time period since I know orbital velocity. I just don't know how to find the atmospheric density, rho, in the equation below which relates what I just talked about..

delta R(per rev) = -2*pi*(1/ballistic coeff)*rho*R^2

For part G, I have no clue where to start or what to do.
 

Answers and Replies

  • #2
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This picture might help. The elevation angle is epsilon. lambda_0 is the swath angle. rho is not the same as the rho in part B.

http://server6.pictiger.com/img/680776/computer-games-and-screenshots/geometry.jpg [Broken]
 
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  • #3
D H
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Check your answers for D and E. Do these make sense given your answer for C?

For B, you must have been told to use some atmospheric model. This site, http://modelweb.gsfc.nasa.gov/atmos/atmos_index.html [Broken], is a good reference point for a number of atmospheric models.

For G, use the law of sines: sin(A)/a = sin(B)/b = sin(C)/c.
 
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  • #4
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Damn, I don't know what I am doing wrong here. We are given

[tex]cos(\lambda_0) = \frac{R_e}{R_e + h} = sin(\rho)[/tex]

I got [tex] sin(\rho) = .9481[/tex] using [tex]R_e = 6378 km[/tex] and [tex]h = 349 km[/tex] calculated previously.

So [tex]\rho = 71.46[/tex] degrees, and [tex] D = R_e cos(\rho) = 2139 km[/tex]

For part D, swath width, [tex]S_w = 2 \lambda_0 R_e[/tex], and since [tex]sin(\rho) = cos(\lambda_0)[/tex], and since I got [tex]sin(\rho)[/tex] to be .9481, [tex]\lambda_0 = arccos(.9481) = 18.54[/tex]. Now our professor said this was in radians, so does that mean [tex]\rho[/tex] is in radians too? If so, is part C incorrect? However, [tex]S_w = 2 \lambda_0 R_e[/tex] = 2(.9841)(6378 km) = 12094 km. If I used degrees for lambda_0 this number would be absurdly huge. So how would this calculated answer not make sense in light of what I calculated for part C?

And what is wrong with part E? If part D is correct, isn't the time in view just the swath width divided by the speed of the satellite?

Any help with this would be greatly appreciated.
 
  • #5
D H
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The swath width equation [itex]S_w = 2 \lambda_0 R_e[/itex] is valid only if [itex]\lambda_0[/itex] is in radians. Your result, 18.54, is in degrees. Convert to radians.

Look at your own picture. The length of the circular arc from the subsatellite point to the horizon is half the swath width. How could that distance be three times the distance from the satellite to the horizon? BTW, your picture is a bit exaggerated. The satellite's altitude is 349 km, which is about 1/18 Earth radii.
 

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