Consider a satellite is moving around the earth. The position and velocity vector of the satellite are 7 = 7000-20003-4000k (km) and T 31-6)+5k (km/s), respectively in earth-centered Cartesian coordinate system. The eccentricity vector e of the orbit is (Assume the earth to be non-rotating) a) 0.2888? + 0.085233- 0.3840k b) 0.1332i +0.1246j – 0.2698k c) 0.2888? – 0.08523; +0.3840k d) 0.1332 - 0.1246j + 0.2698k
Consider a satellite is moving around the earth. The position and velocity vector of the satellite are 7 = 7000-20003-4000k (km) and T 31-6)+5k (km/s), respectively in earth-centered Cartesian coordinate system. The eccentricity vector e of the orbit is (Assume the earth to be non-rotating) a) 0.2888? + 0.085233- 0.3840k b) 0.1332i +0.1246j – 0.2698k c) 0.2888? – 0.08523; +0.3840k d) 0.1332 - 0.1246j + 0.2698k
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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This question is based on satellite so I thought to keep it under
![Consider a satellite is moving around the earth. The position and velocity vector
of the satellite are T = 7000- 20003–4000k (km) and T = 3i-63+5k (km/s),
respectively in earth-centered Cartesian coordinate system. The eccentricity
vector e of the orbit is (Assume the earth to be non-rotating)
a) 0.2888å + 0.08523 – 0.3840k
b) 0.1332i +0.1246j – 0.2698k
c) 0.2888i – 0.08523; + 0.3840k
d) 0.1332i – 0.1246j + 0.2698k](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3feb552-aff1-4564-bd5a-1ec855d65f9b%2F4f9708d5-9586-4898-98e6-289a720be462%2Fv6ceg0f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider a satellite is moving around the earth. The position and velocity vector
of the satellite are T = 7000- 20003–4000k (km) and T = 3i-63+5k (km/s),
respectively in earth-centered Cartesian coordinate system. The eccentricity
vector e of the orbit is (Assume the earth to be non-rotating)
a) 0.2888å + 0.08523 – 0.3840k
b) 0.1332i +0.1246j – 0.2698k
c) 0.2888i – 0.08523; + 0.3840k
d) 0.1332i – 0.1246j + 0.2698k
![Referring to Question 1, the true anomaly of the satellite's orbit is
a) 23.42°
b) 53.35°
c) 33.32°
d) 28.12°](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3feb552-aff1-4564-bd5a-1ec855d65f9b%2F4f9708d5-9586-4898-98e6-289a720be462%2F7bynqby_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Referring to Question 1, the true anomaly of the satellite's orbit is
a) 23.42°
b) 53.35°
c) 33.32°
d) 28.12°
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