By A. Frank D'Souza

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Since m 1 m 2 , m 1 + m 2 ~ m 2 and the equivalent mass of the satellite becomes m 1 m 2 /(m 1 + m 2 ) ~ m 1 , which is its actual mass. Eliminating va for the circular orbit, we obtain + ,+ mz)J'/2 71 to a malfunction of the control, the satellite is not projected horizontally but at an angle a: to the horizontal and, as a result, is projected into an elliptic orbit. Find the perigee and apogee of the orbit. ) First, we consider the circular orbit for which Case 4: e = 0, E = -[G(m 1 m 2 )]2/2h 2 • The orbit is a circle which is a special case of the elliptic orbit.

3 Sec. 55) (~ m~v~) 2 - (~ m1v1) ! " I; ,, M. )t This equation states that the sum of the angular impulses of the external forces about the origin is equal to the change in angular momentum of the system. 57) is zero and we get lz)Z] l 1 ). 39) with respect to time from It to 12 , we obtain l 2 ) 2 • Hence, or = mv! which is a statement of the conservation of linear momentum. ko 2 • Neglecting friction, the only force that does work is the spring force and hence the total mechanical energy is conserved.

We consider a rigid body which has angular velocity vector with respect to an inertial frame XYZ (Fig. 1). The coordinate system, Oxyz, has its origin ~t a reference point, 0, of the body and it rotates at ~e same angular velocity w as the body. Such a coordinate system, Oxyz, is called a body coordinate system. 1) The position vector R of the point P with respect to the inertial coordinate system X YZ is the vector sum of R. 1 INTRODUCTION R = R. +; This chapter is devoted to the study of dynamics of rigid bodies by the direct application of Newton's second law.