A. The hour angle (HA) is the difference between the local sidereal time (LST) and the right ascension (RA) of the star, given by HA = LST - RA. At transit, HA = 0. Since Rigel will transit in 75 minutes (1.25 hours), it is currently east of the meridian, so its HA is negative. Thus, the HA is -1.25. B. LST = RA - 1.25 hours = 05h 14m 32.3s - 1h 15m = 03h 59m 32.3s. C. A star due east on the horizon must have a declination of approximately 0° (on the celestial equator) to rise exactly due east. Such a star has HA = -6 hours, so its RA = LST - HA = LST + 6 hours = 03h 59m 32.3s + 6h = 09h 59m 32.3s. D. Rigel will set when its HA is approximately +6 hours (since its declination is close to 0°, similar to an equatorial star, which sets 6 sidereal hours after transit). Thus, LST at setting ≈ RA + 6 hours = 05h 14m 32.3s + 6h = 11h 14m 32.3s.
Question:
A. The hour angle (HA) is the difference between the local sidereal time (LST) and the right ascension (RA) of the star, given by HA = LST - RA. At transit, HA = 0. Since Rigel will transit in 75 minutes (1.25 hours), it is currently east of the meridian, so its HA is negative. Thus, the HA is -1.25. B. LST = RA - 1.25 hours = 05h 14m 32.3s - 1h 15m = 03h 59m 32.3s. C. A star due east on the horizon must have a declination of approximately 0° (on the celestial equator) to rise exactly due east. Such a star has HA = -6 hours, so its RA = LST - HA = LST + 6 hours = 03h 59m 32.3s + 6h = 09h 59m 32.3s. D. Rigel will set when its HA is approximately +6 hours (since its declination is close to 0°, similar to an equatorial star, which sets 6 sidereal hours after transit). Thus, LST at setting ≈ RA + 6 hours = 05h 14m 32.3s + 6h = 11h 14m 32.3s.
Asked by: vishal
Created at: 2025-09-11 03:25:25
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