Question

An Earth satellite’s orbit has a period of 55.555 hours, and a semilatus rectum of 72,254.4 km. What is the orbit’s eccentricity?

Equations:

a = {(ra + rp)/2}

ra = a (1 + e)

rp = a (1 - e)

p = (b2/a) = a (1 – e2)

h = rpvp = rava

ε = - μ/(2a)

P = 2π√(a3/μ)

Use: Earth radius = 6378.1363 km

Gravitational parameter μ = 3.986012 X 105 km3/sec2

Answer #1

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