A satellite has the following orbital characteristics: T = 43,082 s (orbital period), e = 0.75, ? = 270 deg, i = 63.4 deg, ? = 45 deg, t-t p = 0 (time past periapsis) a) Determine the altitude when the satellite crosses the equator b) Determine the minimum and maximum altitudes of the satellite as well as the latitudes where they occur.

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2:59 PM 1.0KB/s X O 4G the1 18 Question An Earth orbiting satellite has the following orbit characteristics. Assume the Earth is spherical in your analysis. P 43082 sec (orbital period) 0.75 (eccentricity) 270 deg (argument of periapsis) 63.4 deg (inclination) 45 deg (right ascension of the ascending node) T. = 0 sec (time at periapsis) a. Determine the altitude when the satellite crosses the ascending node, hAN. b. Determine the minimum and maximum altitudes of the satellite, hmin and /mar, and the latitudes at which they occur, Lin and Lmar. c. Determine the amount of time in a single orbital revolution that the satellite is in the Southern Hemisphere. d. Assuming the other orbital elements remain unchanged, what is the value of eccentricity at which the minimum altitude would be 0? Answer Total answers posted by the expert is: 648 First we calculate the semi-malor axis using orbital period using, Kepler's laws G*M = (omega2 * r 3) omega = 2*pi/P where P= orbital period using above, P=sqrt(4*pi2*a3 /G*M) where, a is semi major axis G= 6.67*10-11 m3 kg-1 s-2 M=5.971*1024 kg solving a=26555788.62 m Now, for maximum and mimimum altitudes of satellite maximum, at apoapsis= a*(1+e)=a*1.75=46472630.08m minimum , at periapsis = a*(1-e)=a*.25=6638947.155 m for minimum altitude to be 0 eccentricity needs to be 1(orbit becomes a parabola) O