Exam (elaborations) TEST BANK FOR Fundamentals of Dynamics and Control of Space Systems By Krishna Dev Kumar (Solution Manual)-Converted
2.1 The coordinate frames used in studying the dynamics of a spacecraft are as follows: a) Inertial reference frame, b) Orbital reference frame, c) Perifocal reference frame, c) Satellite body-fixed reference frame. 2.2 The inertial frames are those coordinate frames that are nonrotating and nonaccelerating frames. The inertial frames are relevant because in applying the Newton’s second law of motion ~F = m d~V dt (2.1) to derive the equation of motion of a system, the velocity ~V and the corresponding acceleration d~V /dt in the right-hand side of the above equation are to measured with respect to an inertial frame of reference. An Earth-fixed frame is not an inertial frame as it is spinning about its axis with a period of 24 hour. When viewed from space, the point on the surface of the earth moves in a circle as the earth spins on its axis. Thus, it is accelerating with an centripetal acceleration of r!2, 2 CHAPTER 2. KINEMATICS, MOMENTUM AND ENERGY where r is the position of the point of the Earth center of mass and ! is the rate of spin of the Earth. With the earth a point on its surface also orbits the Sun. With the solar system, it orbits the center of the galaxy. Thus, the Earth-fixed frame is an accelerating frame and not an inertial frame. We consider just the effect of the spinning motion of the Earth and therefore the inertial acceleration can be written as d~V dt inertial = d~V dt body + ~! × ~Vbody (2.2) The corresponding error in considering an Earth-fixed frame as an inertial frame is Error = d~V dt inertial − d~V dt body = ~! × ~Vbody (2.3) The Earth’s spin rate ! is ~! = !kˆk = − 2 T ˆk = − 2 24 × 3600 ˆk = 7.275 × 10−5ˆk (2.4) where ˆk is a unit vector along the z-direction as taken for the aircraft body-fixed frame. The order of magnitude error would be 10−4×Vbody. As this magnitude is usually very small when compared to the magnitude of other relevant accelerations like the gravitational acceleration, which is 9.81 m/s2, and we often treat the Earth-fixed frame as an inertial frame. when solving problems. 2.3 The inertial position vectors for spacecraft m1 and m2 are ~ R1 = ~R − ~L (2.5) ~ R2 = ~R + (1 − )~L (2.6) where = m2/(m1 + m2). The corresponding magnitudes are R1 = [R2 + 2L2 − 2 ~R · ~ L]1/2 (2.7) R2 = [R2 + (1 − )2L2 + 2(1 − )~R · ~ L]1/2 (2.8) where L = L0 + vt. The nomenclature Lo defines the initial length of the cable while v is the speed by which the length of the cable varies. 3
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test bank for fundamentals of dynamics and control of space systems by krishna dev kumar solution manual