https://obs-web.rs.umbc.edu/ 2026-04-29T16:46:25+00:00 https://obs-web.rs.umbc.edu/ https://obs-web.rs.umbc.edu/lib/tpl/bootstrap3/images/favicon.ico text/html 2024-03-18T16:10:05+00:00 Anonymous (anonymous@undisclosed.example.com) https://obs-web.rs.umbc.edu/doku.php?id=train:lectures:positionalastro&rev=1710792605&do=diff ---------- Celestial Sphere The celestial sphere is a modern take on the {\emph Geocentric Model} of the Universe that gives us a model of the sky on which we can map angular coordinates. The geocentric model placed the Earth at the center (geo-centric$ALT = 0^\circ$$ALT\in[0,90]^\circ$$90^\circ$$AZ=0^\circ$$ALT\in[0,360)\circ$$90^\circ$$ALT$$180^\circ$$AZ$$(ALT, AZ) = (120^\circ,0^\circ) \rightarrow (60^\circ,180^\circ)$$(ALT, AZ)$$~24^\circ$$\alpha$$^\circ$$1^\circ = 60'$$1' = 60''$$1^\circ … text/html 2024-03-18T16:10:16+00:00 Anonymous (anonymous@undisclosed.example.com) https://obs-web.rs.umbc.edu/doku.php?id=train:lectures:siderealtime&rev=1710792616&do=diff Sidereal Time For largely historical reasons, our time-keeping system is reckoned by the position of the Sun. We can also choose to keep time by reckoning how long it takes for a specific star on the Celestial Sphere to get back to the same point in the sky night after night. Since one rotation of the Earth is all it takes to do this for distant stars, that is the Sidereal Day. $~1/365^{th}$$~1^\circ$$~1^\circ$$\alpha$$24^h$$360/24 = 15^\circ/hr$$15''/s$$24^h$$23^h56^m$$23*3600 + 56*60 = 86160s… text/html 2024-03-24T12:54:41+00:00 Anonymous (anonymous@undisclosed.example.com) https://obs-web.rs.umbc.edu/doku.php?id=train:lectures:telescopeoptics&rev=1711299281&do=diff Telescope Optics I The Primary Optical device is the first lens or mirror that light touches. The diameter of the Primary Optical Device is also (generally) the Aperture Diameter. The light-gathering power of a telescope depends on the area of the aperture, and so this $n$$\theta_i = \theta_f$$s$$f$$\frac{s}{f}[rad]$$\sigma$$f(x;\mu,\sigma) = A\exp^{-(x-\mu)^2/\sigma^2/2}$$\mu=0$$e^0$$$A\exp^{-x_0^2/\sigma^2/2} = \frac{1}{2}A\exp^{0}$$$$\exp^{-x_0^2/\sigma^2/2} = \frac{1}{2}$$$$\frac{-x_0^2}{2\…