Page 50 - Intro to Space Sciences Spacecraft Applications
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vp =J-                                   Orbital Principles   37
                                                                                 (2 - 13)


                          Being  associated  with  the  minimum  orbital  radius,  the  periapsis
                          velocity vp is thefastest velocity in the orbit.

                        We now have enough information to analyze established orbits if given
                      just a few of the orbital parameters.

                      Example Problem:

                           Due to thrust limitations and the reaches of  the atmosphere, the
                         Space Shuttle is limited to operations between about 200 km to 800
                         km altitudes. From this information, determine the orbital parame-
                        ters associated with an elliptical orbit between these two altitudes.

                      Solution:

                           a. Semi-major axis (equation 2-1):
                             a = 6,878 km

                        Note:  altitudes must  be  converted to  orbital  radii by  adding the
                        radius of the earth, & = 6,378 km for use in equation 2-1.

                           b. Orbital period (equation 2-6):
                             T = 5,677 sec = 94.6 min

                         Slightly over an hour and a half between sunrises!

                           c. Total specific energy (equation 2-9):
                             E = -28.98  km2/sec2

                        Note: Like eccentricity, knowledge of the total energy can indicate the
                         type of orbit as well. All “closed” (elliptical and circular) orbits have
                         negative values for total energy. Zero or positive total energies indi-
                         cate the “open” (parabolic or hyperbolic) orbits introduced earlier.

                           d. Eccentricity (using equation 2-3 or equation 2-5):
                             e = 0.0436
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