Question 1. Some telescopes can be moved by computer control to a given location in the sky, say a star an astronomer wishes to observe. Such telescopes sometimes have acquisition TV monitors with a field of view of about 50 arc seconds (1 arc-second = 1/60 are minute, which in turn equals 1/60 of one degree of arc). Now, on the celestial equator, how many arc-seconds of right ascension does the rotation of the Earth cause to pass by a given point in one second of time? If the astronomer's clock is off by 10 seconds, should he expect to see the star on his TV when the telescope moves there? Do you expect the same result if the astronomer's object is near the North celestial pole?