# A Satellite in Sun-Synchronous Orbit

## 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.

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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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D H
Staff Emeritus

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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Damn, I don't know what I am doing wrong here. We are given

$$cos(\lambda_0) = \frac{R_e}{R_e + h} = sin(\rho)$$

I got $$sin(\rho) = .9481$$ using $$R_e = 6378 km$$ and $$h = 349 km$$ calculated previously.

So $$\rho = 71.46$$ degrees, and $$D = R_e cos(\rho) = 2139 km$$

For part D, swath width, $$S_w = 2 \lambda_0 R_e$$, and since $$sin(\rho) = cos(\lambda_0)$$, and since I got $$sin(\rho)$$ to be .9481, $$\lambda_0 = arccos(.9481) = 18.54$$. Now our professor said this was in radians, so does that mean $$\rho$$ is in radians too? If so, is part C incorrect? However, $$S_w = 2 \lambda_0 R_e$$ = 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.

D H
Staff Emeritus
The swath width equation $S_w = 2 \lambda_0 R_e$ is valid only if $\lambda_0$ is in radians. Your result, 18.54, is in degrees. Convert to radians.