Vine - La chavala que me saludó en la Biblio Mikkel
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For elliptical orbits:
| Parameter | Abbr. | Unit | Determines… | GPS example |
|---|---|---|---|---|
| Semi-major axis | a | meters | Size of the eliptical orbit | 26,559.8Km |
| Eccentricity | e | NaN | Shape of the eliptical orbit | 0 |
| Mean anomaly | M | Radians | Satellite position |
Convert to CIS:
| Parameter | Abbr. | Unit | Determines… | GPS example |
|---|---|---|---|---|
| Inclination | i | Deg | Inclination of the orbit relative to equator plane | 55º |
| Argument of perigee | \(\omega\) | Deg | Orientation of the perigee relative to ascending node | 0º |
| Right ascension of the ascending node | \(\Omega\) | Deg | Orientation of the orbit |
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Kepler Laws
- Area velocity is contant in satellites
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Mini assignment 1
¿a of GPS satellite?
$$ T = \frac{2 \pi}{\sqrt{GM}} \cdot a^{3/2} \\ a = \big ( \frac{T \cdot \sqrt{GM}}{2 \pi} \big )^{2/3} \\
\text{Result} = 26,5 \cdot 10^6 $$
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Mini assignment 2
- The distance from the Earth to Apoapsis: \(r_a\)
- The distance from the Earth to Periapsis: \(r_p\)
- Semi-major axis: a
Eccentricity is given by:
$$ e = \frac{r_a - r_p}{r_a + r_p} $$
¿Show that the distance from center of orbit to focal point is?
$$ \lvert \lvert c - F \lvert \lvert = ae $$
Solution:
$$ e = \frac{r_a - r_p}{r_a + r_p} \\ a + ae = r_a \\ a - ae = r_p \\ \Rightarrow e = \frac{ae}{a} \Rightarrow ae = ae \\ \text{Proffessor answer} \\ a = \frac{r_a + r_p}{2} \Rightarrow ae = \lvert c - F \lvert = \frac{r_a -r_p}{2} $$
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