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Question 153079This question is from textbook College Algebra
: I am having the worest luck tonigh I can not get these question to come out right. Can someone please help me!
Here is the Problem and the Question!!
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 is equal to the eccentricity of the orbit.
(3) Find the Apogee. Find the Perigee.
This question is from textbook College Algebra
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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