![]() Since is regressing down the ecliptic, the ecliptic longitude of the star X is increasing. Apply the cosine formula (Equation 3.5.2) to triangle PKX to obtain. Its angular momentum is along its axis and currently points at Polaris, the North Star. It is therefore essential for a practical observer to know how to correct for precession. This is seen in a slight bob up and down as the gyroscope precesses, referred to as nutation.Įarth itself acts like a gigantic gyroscope. This allows us to find the period of precession. ![]() The precession period is thus T 1 3. We use Figure to find the precessional angular velocity of the gyroscope. Calculating the precession period Since the Earth’s angular momentum vector precesses about the axis normal to the plane of the ecliptic, the precession rate is given by C Isin where is the Earth’s rotational angular velocity and I its (polar) moment of inertia. What is the precessional period of the gyroscope Strategy. In order to calculate C, the precession period has. The precession angular velocity adds a small component to the angular momentum along the z-axis. Its center of mass is 5.0 cm from the pivot and the radius of the disk is 5.0 cm. x Axis 90 V Figure 4.3 The basis of calculating a moment of inertia. If the distortion is along the axis of rotation of earth itself (bit unphysical I guess), then a moon need not be out of plane.\) << \(\omega\), that is, that the precession angular velocity is much less than the angular velocity of the gyroscope disk. The moon has to go out of plane of the trajectory of earth (assuming the distortion is equatorial). a distorted sphere (either in terms of non uniform density or in terms of the actual shape) Here is a picture, to help visualize the out of plane trajectories. Hence at some point in time (or maybe always) there will exist a torque perpendicular to the earth's axis of rotation, as the moon is out of plane and so attracts the bulge that's closer to it more than the bulge away.
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