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A spacecraft in an elliptical orbit around earth has perigee radius of 7,900 km and apogee radius of 73,000 km, what is its specific energy? Gravitational parameter of earth is 398,600 km3/s2.

A spacecraft in an elliptical orbit around earth has perigee radius of 7,900 km and apogee radius of 73,000 km, what is its specific energy? Gravitational parameter of earth is 398,600 km3/s2. Correct Answer -4.927 km2/s2

Given, Apogee radius (ra) = 73,000 km Perigee radius (rp) = 7,900 km Gravitational Parameter (μ) = 398,600 km3/s2 Semi-major axis (a) = (ra + rp) / 2 = (73,000 + 7,900) / 2 = 40,450 km Specific Energy (ε) = -μ/(2a) = -398,600*(2*40,450) = -4.927 km2/s2

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