Assume that Earth is a sphere with radius R. A satellite has an elliptical orbit with the center of Earth at one focus. The lowest point of the orbit is 5R above the surface of Earth, when the satellite is directly above the North Pole. The highest point of the orbit is 11R above the surface of Earth, when the satellite is directly above the South Pole. What is the height of the satellite above the surface of Earth, when the satellite is directly above the equator?
The satellite's orbit is an ellipse with major axis of length 18*R. The earth's center lies at one of the foci of the ellipse, with the ellipse's latus rectum in the plane of the earth's equator.
The ellipse's semimajor axis is 9*R, and of course the focus at earth's center if 9*R - 6*R = 3*R from the center of the ellipse.
The eccentricity, e, of the ellipse is found as 3*R / 9*R = 1/3. The ellipse's semiminor axis, b, can be found from e = sqrt(1 - b^2/a^2):
1/3 = sqrt(1 - b^2 / (81*R^2))
By Wolfram Alpha,
b = 6*sqrt(2)*R
The distance the satellite passes above the earth's equator is half the length of the latus rectum. That half length is b^2/a = (72*R^2)/(9*R) = (72/9)*R = 8*R. Subtracting R, that's 7*R.
Revised (correcting distance of earth center to 3*R, from 4*R, after reading preceding post).
Edited on January 23, 2025, 10:17 am
Edited on January 23, 2025, 10:18 am
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Posted by Charlie
on 2025-01-23 09:59:52 |
solution
