For calculating the Right Ascension of the moon I need the GMST (Greenwich Mean Sidereal Time) in Degrees. I have found the formula GMST = 100.4606184 + 0.9856473662862 * D + H * 15, D = number of days, H = Universal Time in hours I think the unit of GMST is days. RA is then calculated GMST(degrees) + Longitude(degrees) converted to Hours Minutes Seconds. In the example to the above formula D = 305.209, H = 17, that results GMST(days?) = 656.288 but GMST(degrees) = 296.288° For the actual date the values are 8750.95 days? and it should be -11.54° for the actual RA. How can I convert GMST(days) to GMST(degrees)? Or do I think wrong?
6
-
1$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotMar 28 at 13:05
-
1$\begingroup$ The formula you found is in degrees. You can tell this is the case via the $15*H$ term. $\endgroup$– David HammenMar 28 at 14:09
-
$\begingroup$ BTW, JPL have an equation for GMST here. The result is in seconds, so divide by 240 to get degrees. Note that 3155760000 = 36525 * 86400, the number of seconds in a Julian century. I don't know why they include that term... $\endgroup$– PM 2RingMar 28 at 14:32
-
$\begingroup$ "RA is then calculated GMST(degrees) + Longitude(degrees)". This is the equation for comuting local sidereal time, not RA. $\endgroup$– Greg MillerMar 28 at 15:20
-
$\begingroup$ You didn't mention which epoch your D is measured from. It looks like it's midnight at the start of 2000-Jan-1. The GMST at the J2000 epoch is ~280.46062°. $\endgroup$– PM 2RingMar 28 at 16:58
|
Show 1 more comment