By Marcel J. Sidi

Explains uncomplicated conception of spacecraft dynamics and keep watch over and the sensible elements of controlling a satellite tv for pc.

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4. 6) (Bate et aI. 1971, Bernard 1983). The inverse transformation of Eq. (WW{ ~]: x, Y, and Z are the inertial coordinates of the moving satellite. 4 Transformation from orbit parameters to Cartesian inertial frame coordinates. 6/ Kepierian Orbits in Space To calculate the velocity vector, dr dr dit v = dt = dit dt ' we must first find dit Idt from Kepler's time equation: , dM dt ' "1 dv =n = dt dit -ecos(it) dt ' from which, together with Eq. 4, we derive dit dt ,. 8) = l-ecos(it) =r· Differentiating Eq.

It is interesting also to notice that this escape velocity is larger by a factor of only V2 than the velocity of a circular orbit at the same distance r from F. 3 Geometry of the hyperbolic orbit. ~ .. '" I Hyperbolic Orbits For this important class of orbits, the total energy E is positive, E> O. This means that the kinetic energy of the satellite is larger than its potential energy, so that the sic is able to leave the gravitational attraction field of the central body. A satellite moving on a hyperbolic orbit does not revolve about the central body.

2 and the mean anomaly M from Kepler's time equation (Eq. 8). We are left with one more unknown parameter, the argument of perigee w. To find it, we use the transformation of Eq. 16) y = rsin(8). The components of r in space are known: X, Y, and Z. Having already found the inclination angle i and the right ascension of the ascending node D, we can find the argument of perigee w from the transformation in Eq. 7. 17) (Bernard 1983). From the equalities in Eqs. 17 we can finally calculate w, since 8 can be found from Eq.

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