Amir Mehdi

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George Mason University *

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353

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Aerospace Engineering

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Apr 3, 2024

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docx

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Amir Mehdi IT 353 DL01 Fall 2023 HOMEWORK 2
Q1) I have read and understood all the directions for this assignment, especially Q18. Q2) I retrieved my data set from Blackboard at 2023-NOV-07 16:53:59 EDT.  1 XX227U YYNNNPPP 23282.88279351 -.00000014  00000-0  00000-0 0  999Z 2 XX227   0.0188 133.4147 0001767  58.3953 143.4051  1.00271602 5116Z Q3) XX227 XX27 Q4) 23282.88279351 2023OCT09 21:11:13.359 Q5) Inclination: 0.0188 The satellite orbit: it is an equatorial orbit because it is so close to 0 that make it close to equator. Q6) Right Ascension of the Ascending Node: 133.4147 ° Q7) Eccentricity: 0001767 The orbital path of this satellite is: the path would be circle, but this Eccentricity is so close to 0 that makes it like circle shape, or we can call it a dormant shape like close to perfect circle. Q8) Argument of Perigee: 58.3953 ° Q9) Mean Anomaly: 143.4051 ° True Anomaly: up to e 3 : 143.4171 °
Q10) Mean Motion: 1.00271602 Orbital Period: 1436.10814min Q11) .99726 days is 1436.0544 min. Q12) Satellite orbital period: 1436.1081 min Earth rotational period: 1436.0544 min Because both numbers are so close to each other it is synchronous or nearly synchronous orbit or it this satellite is geostationary satellite. Q13) The orbit of this satellite is geostationary or Clarke obit. The category I chose is https://www.n2yo.com/?s=54742&live=1 Q14) Modified satellite number= XX27 is equator it has approximately is has same rotation speed as earth (geostationary satellite), so it is close to 24 hours and it is not passing anywhere because it is stationery on equator. Q15) Epoch: 2023OCT09 21:11:13.359 GMST: 22h 24m 07.809s
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