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Question 281902: A satellite is placed in elliptical lunar orbit around the moon, with a perilune (nearest surface) altitude of 110km and an apolune (farthest from surface) altitude of 314km. Assuming the radius of the moon is 1728km and the centre of the moon is at one focus of the ellipse. Under this assumption, find the equation of the ellipse
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! A satellite is placed in elliptical lunar orbit around the moon, with a perilune (nearest surface) altitude of 110km and an apolune (farthest from surface) altitude of 314km. Assuming the radius of the moon is 1728km and the centre of the moon is at one focus of the ellipse. Under this assumption, find the equation of the ellipse
place (0,0) at midpoint of the major axis which lies in this case along the x-axis
(+- a, 0) are the ends of the major axis, its length is 2a
(+- c, 0) are locations of the foci , the moon's center is at one of them
(-a,0) is perilune and it corresponds to 110 km from moons surface
(a,0) is apolune and it corresponds to 314 km from moons surface
perilune distance = a(1-e) where e is eccentricity = 1728 + 110 = 1838
apolune distance = a(1+e) where e is eccentricity = 1728 + 314 = 2042
a(1-e) + a(1+e) = 1838 + 2042 = 3880 = 2a --> a = 1940
1940(1-e)= 1838
1940 - 1940e = 1838
102 = 1940e
0.05257732 approx. = e
1940(1+e) = 2042
1940 + 1940e = 2042
1940e = 102
again 0.05257732 approx. = e
b^2 = a^2(1-e^2)
b^2 = 1940^2 * (1 - 0.05257732^2)
b^2 = 3763600 * (1 - 0.05257732^2)
b^2 = 3753196
b approx. = 1937.31670101
foci --> c^2 = a^2 - b^2 = 3763600 - 3753196 = 10404
c = 102
x-axis number-line:
-a,-c-r,-c,0,c,-c+r,a (r is radius of moon)
-1940,-1830,-102,0,102,1626,1940
1940-1830=110
1940-1626=314
equation of the ellipse:
x^2/a^2 + y^2/b^2 = 1
so:
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