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Question 153148: Just need help on the completion of this problem which is due tomorrow 8/24 Thanks
The Conic Sections, Sequences and Series
i) Problem: The moon travels an elliptical path with Earth as one focus. The maximum distance from the moon to Earth is 405,500 km and the minimum distance is 363,300 km.
(1) What is the eccentricity of the orbit?
(2) For a planet or satellite in an elliptical orbit around a focus of the ellipse, perigee (P) is defined to be its closest distance to the focus and apogee (A) is defined to be its greatest distance from the focus. Show that A-P is equal to the eccentricity of the orbit.
A+P
(3) Find the Apogee. Find the Perigee.
(4) Explain the problem step by step and what mathematical concepts are being applied to this problem?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! i) Problem: The moon travels an elliptical path with Earth as one focus. The maximum distance from the moon to Earth is 405,500 km and the minimum distance is 363,300 km.
a = 405,000 ; b = 363,300
Then c^2 = a^2-b^2 = 3.2256x10^10
So c = 179599.55
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(1) What is the eccentricity of the orbit?
e = c/a = 179599.55/405000 = 0.443456...
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(2) For a planet or satellite in an elliptical orbit around a focus of the ellipse, perigee (P) is defined to be its closest distance to the focus and apogee (A) is defined to be its greatest distance from the focus. Show that is equal to the eccentricity of the orbit.
Comment: This question is confused and meaningless.
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(3) Find the Apogee. Find the Perigee.
Focus is c = 179599.55 from the center
Apogee:distance from focus to furthest point = 179599.55+405000 = 584599.55 km
Perigee: distance from focus to closest point = 405000-179599.55 = 225400.45 km
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Cheers,
Stan H.
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