Mission Design for the CubeSat OUFTI-1
Mission Design for the CubeSat OUFTI-1 Mission Design for the CubeSat OUFTI-1
CHAPTER 8We have now the vector ˆN s expressed into the orbit reference:⎧ ⎫ ⎧ ⎫⎨ x orb ⎬ ⎨ x ecl ⎬y orb⎩ ⎭ = R 4R 3 R 2 R 1 y ecl⎩ ⎭z orb z ecl(8.6)As the satellite is moving on the orbit plane, what we are interested in tocalculate the eclipses time is actually the projection of N s on the orbit plane.As indicated in figure 8.1, we calculate the angle β that the projection of N sgenerates with the x orb :( )Ns,yN s = atanN s,x(8.7)Figure 8.3: Sun rays direction projected on the orbit plane.So far, we know the eclipse’s central angle θ ∗ = 180 ◦ + β and therefore weknow the corresponding orbit radius.As the distance between earth and sun is much bigger than the earth’s radius,we can make the hypothesis that the lines determining the entrance and the exitfrom eclipses are tangent to the earth surface, as shown in figure 8.1. Hence,we have:¯θ out = 90 ◦ − acos(Rer out)¯θ in = 90 ◦ − acos(Rer in) (8.8)We can also exploit the relationship between radius and anomaly and wehave:Galli Stefania 76 University of Liège
CHAPTER 8.POWER SYSTEMcos ( 90 ◦ − ¯θ) R e( (out = 1 + ecos θ ∗ +P¯θ))outcos ( 90 ◦ − ¯θ) R e( (in = 1 + ecos θ ∗ −P¯θ))in(8.9)We solve this two equations and we have ¯θ out and ¯θ in . Then we transformthem into the corresponding eccentric anomalies in order to calculate the eclipsesduration.A simulation over an year orbit shows the eclipse duration shown in figure8.4: it means that, given the position of earth respect to sun, roughly correspondingto the day of the year, all the orbit taking place on that day have theindicated eclipse’s duration.Figure 8.4: Eclipse duration as a function of earth anomaly8.2 Configuration and solar cellsIn order to quantify the available power, we also need to know the satelliteconfiguration and, more specifically, the solar panels orientation.The first thing to point out, is that, as we are not planning any attitude control,we need solar panels on each face: in fact, we can’t risk to have a face withoutsolar cells watching the sun causing a fall in the power production.Galli Stefania 77 University of Liège
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CHAPTER 8.POWER SYSTEMcos ( 90 ◦ − ¯θ) R e( (out = 1 + ecos θ ∗ +P¯θ))outcos ( 90 ◦ − ¯θ) R e( (in = 1 + ecos θ ∗ −P¯θ))in(8.9)We solve this two equations and we have ¯θ out and ¯θ in . Then we trans<strong>for</strong>m<strong>the</strong>m into <strong>the</strong> corresponding eccentric anomalies in order to calculate <strong>the</strong> eclipsesduration.A simulation over an year orbit shows <strong>the</strong> eclipse duration shown in figure8.4: it means that, given <strong>the</strong> position of earth respect to sun, roughly correspondingto <strong>the</strong> day of <strong>the</strong> year, all <strong>the</strong> orbit taking place on that day have <strong>the</strong>indicated eclipse’s duration.Figure 8.4: Eclipse duration as a function of earth anomaly8.2 Configuration and solar cellsIn order to quantify <strong>the</strong> available power, we also need to know <strong>the</strong> satelliteconfiguration and, more specifically, <strong>the</strong> solar panels orientation.The first thing to point out, is that, as we are not planning any attitude control,we need solar panels on each face: in fact, we can’t risk to have a face withoutsolar cells watching <strong>the</strong> sun causing a fall in <strong>the</strong> power production.Galli Stefania 77 University of Liège
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