A satellite, moving in anelliptical orbit, is 355 km above Earth\'s surface at i

Question

A satellite, moving in anelliptical orbit, is 355 km above Earth's surface at its farthestpoint and 172 km above at its closest point.

(a) Calculate thesemimajor axis of the orbit.
(b) Calculate the eccentricity of the orbit.

I tried using e=c/a, buti'm looking for two of those variables, so i couldnt solve
my physics textbook said that the perihelion distance (closest tothe earth) + aphelion distance (farthest from the earth) = 2a, butusing that I get a = 263500 m, which is incorrect for part a
for part b, i used ea = a - perihelion distance, and using theincorrect a that i found in part a, i solved e to = 0.347, whichwas also incorrect.

any help is very appreciated!!

Explanation / Answer

First of all, the terms are apogee and perigee. The termsyou're using refer to orbiting the Sun. The parameters of an ellipse are from the center of the orbitedbody. The provided numbers are from the surface of the Earth,so you must add in the Earth's radius. 2*a = Re + Ap + Re + Per = 6378km + 355km +6378km + 172 km = 13283km a = 6641km There are several ways to compute c. We know thefollowing: a + c = Re + Ap a - c = Re + Per If we subtract the second equation from the first: 2*c = Ap - Per c = (Ap - Per)/2 Eccentricity is: e = c/a = (Ap - Per)/(2*a) = (355km - 172km)/13283km = 0.0138

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