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By Yazdani M., Gelling R. (eds.)

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Example text

Since the equation is transcendental, we cannot invert it directly to solve for E. Instead, we must resort to a technique such as the Newton-Raphson successive approximation method to solve for the desired quantity. Assuming for the moment that E is determined, the final step then is to find 0 (true anomaly) from the following equations: sin 0 -- x/1 - e 2 sin E cos 0 = 1 - ecosE cos E - e 1 - e cos E sin 0 ~ 0 = tan -1 \ c o s 0 / Now, let us see how Kepler's equation is solved. 2 Solving Kepler's Equation First, we move mean anomaly M in Eq.

Prussing, J. , and Conway, B. , Orbital Mechanics, Oxford, 1993. Roy, A. , Adam Hilger, Bristol and Philadelphia, 1988. Taft, L. , Celestial Mechanics, Wiley, New York, 1985. Wertz, J. , and Larson, W. , Space Mission Analysis and Design, Kluwer Academic Publishers, 1991. 1, The period of revolution of a satellite is 106 min. Find the apogee altitude if the perigee altitude is 200 km. 2. Find the period of revolution of a satellite if the perigee and apogee altitudes are 250 and 300 km, respectively.

Find the apogee altitude if the perigee altitude is 200 km. 2. Find the period of revolution of a satellite if the perigee and apogee altitudes are 250 and 300 km, respectively. 3,3. Find the maximum and minimum velocity of the Earth if the eccentricity of the Earth's orbit about the sun is 1/60. 6 x 10 6 kill? 4. Show that for a satellite moving on an elliptical orbit the velocity at the time of passage through the minor axis is equal in magnitude to the local circular velocity (vc). 5. A satellite in a circular orbit at an altitude hc above the Earth's surface is given a velocity v0 with a flight-path angle y.

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