How do you determine Comet Encke's period?
143259
One obvious way is to keep a history of previous appearances. With Encke this history is particularly rich since its period is only 3.3 years, the shortest of the known comets.  There have been 59 observed appearances of Encke.\nIn the case of another comet which is being sighted for the first time, you must get enough observation data of its positions and velocities during its passage to allow you to fit the data to an ellipse; then relatively straightforward math will yield the period.
234287
Easy.  Use a newer astronomy text (the newer ones have more pictures than the older ones) or web page and check out the section of seasons.  www.astrnomynotes.com has some good stuff.
149918
Rotation period 1.025 957 d
1106342
Halley's Comet returns about every 75-76 years.  The orbital period is not precise but to perturbations from the gravity of the other planets in the solar system.
1063458
Shoemaker 1, tonight!\n\nVery comprehensive calender of comets below:\n\nhttp://www.aerith.net/comet/future-n.html\n\nBut for a simple answer, look at this website, it seems that today (7th June) we will see  the brilliantly named 102P/Shoemaker 1, which you can read about at the second of the two below links, and the third link is wikipedia's little info page on it (the important info is in the table at the right-hand side of the page)\n\nhttp://cometography.com/current_comets.html\nhttp://cometography.com/pcomets/102P.html\n\nSo get that telescope out! :-) Hope that helps
106990
Period is proportional to R^3/2.\nThis is an answer I posted to bombay2pac2000 under Astronomy & Space 8 hours previously, and it applies here as well.\nKepler's 3rd law says: Orbital period T = 2*pi*R^1.5/SQRT(GM)\nwhere R is the radius (in the case of a circular orbit), G is the universal gravitational constant and M is the mass of the central body.  In orbit, centrifugal acceleration CA balances gravitational acceleration GA.  CA=w^2R (where w is omega, the orbital angular rate which = 2pi/T). So CA = (2pi/T)^2*R. From Newton we know GA = GM/R^2. Put 'em together, \nCA = GA\nR*(2pi)^2/T^2 = GM/R^2 \nT^2 = R^3*(2pi)^2/GM\nT = 2*pi*(R^3/2)/SQRT(GM)
853689
Kepler's Third Law states that the ratio of the squares of the periods of any two  planets is equal to the ratio of the cubes of their average distances from the sun.  Here, the "period" simply means the length of time it takes for a planet to orbit the sun once.\n\nIf R1 and R1 are the orbital radii (i.e., the distance from the sun) of planets 1 and 2, and T1 and T2 are the orbital periods, then the Kepler's Thrd Law has the mathematical form:\n\n\n(T1/T2)^2 = (R1/R2)^3\n\nHere, we know that for the Earth, T1 = 1 year, and R1 = 1 AU (1 AU is one Astronomical Unit, which is defined as the average distance from the Earth to the Sun).  We want to find R2 if T2 = 5 years.  So:\n\n(1 yr /5 yr)^2 = (1 AU/R2)^3\n1/25 = (1 AU)^3/(R2)^3\nR2 = (25 AU^3)^(1/3)\nR2 = 2.924 AU
733204
period =t=1/f=1/256=.00391 seconds\n\nv=lf                            l=lambda=wavelength\n\nl=v/f=340/256=1.328meters
1090033
There is actually a comet passing by Earth this month.  To calculate where the comet will be visible go to\nhttp://ssd.jpl.nasa.gov/horizons.cgi#top\n\nType in information for comet 73P
882266
Using Kepler's laws (third law):\n\nThe ratio between the T*T and a*a*a is a constant for our Sun and it applies to all objects rotating around it. where T is the period and a is the average distance. \nSo:\nT2=T1*(a2/a1)^(3/2) ;  ^ is used for power \nT2=1year *[  (4.5*10^9)  /  (1.5*10^9)  ]^(3/2)\nT2=164.3 years     \nTo make sure...see the links
561003
Try this website. You can adjust the date time for just about any period -->\nhttp://www.fourmilab.ch/yoursky/
948805
Hmm....well, since Cepheids are variables it is important to know which part of its 5 day cycle that brightness measurement was taken during.  This is rather important since a cepheids brightness can vary by a factor of 4, so off the bat your estimate could be a factor of 2 off (inverse square law, right?)\n\n\nAssume for now that is the peak brightness.  First convert this energy flux value to the apparent magnitude. (You should have an equation for this someplace if this problem was assigned to you, or you can look it up online or extrapolate from this plot: http://physics.lakeheadu.ca/courses/Astro/2330/Propweb/apmagdef.gif)\n\nOnce you get that go to your handy "cepheid period to absolute magnitude" conversion chart (there is a plot here http://www.astro.lsa.umich.edu/Course/MMSS/Interactive/Ex1.4/ , but again you should have something like this at your disposal already; otherwise, extrapolate from the plot near the bottom of that page).\n\nNow you use that old chestnut of the magnitude distance relationship,\n\nd=10^((m-M)/5+1)\n\nwhere m is the apparent, and M is absolute magnitude.
354664
go to this website: http://tycho.usno.navy.mil/vphase.html and type in the information, they'll tell you what the shape is for that day.  Then go to this website to look up the name: http://www.calculatorcat.com/moon_phases/moon_phases.phtml
647693
lots :-)  Let me figure it out...\n\nFirst the number is spelled 'googol' and is equal to 10^100 (1 followed by 100 zeroes).\n\nA year nominally has 31,556,926 seconds (365.242199 days x 24 hours x 60 minutes x 60 seconds).  I used the number 365.242199 because that is how long it takes for the earth to go once around the sun... we round this to 365 and add an additional day every 4 years (with some exceptions 3 out of every 4 centuries, blah, blah, blah)...\n\nAnyway, I digress.  Back to the problem at hand.\n\n10 x 10^99 / 3.1556926 x 10^7 = (10/3.1556926) x 10^92\n\n3.16887646 × 10^92 years.  There is no English name for this word but it is *lots* like I said at the beginning.  :-)\n\nThe closest I can get to a name would be to use vigintillion (10^63) as a base and go from there...\n\n3.16887646 x 10^29 x 10^63 or\n\n316 x 10^27 x 10^63 +\n887 x 10^24 x 10^63 +\n646 x 10^21 x 10^63\n\nSo that would be:\n316 octillion-vigintillion, 887 septillion-vigintillion, 646 sextillion-vigintillion years\n\nOkay, a little more research and I found that my names were close but not exactly correct.  This number should be called:\n  \n316 octovigintillion 887 septenvigintillion 646 sexvigintillion years!
819223
If I understand correctly, one revolution of the disk is equal to one period.  You need to know the time (T) taken for one period.\n\nFrequency (f):\n f = 1/ T\n T = 1/ f\n T = 1/ 2200\n T = 0∙000 454 545....\n T ≈ 454∙454 µs\n\nAngular speed (w):\n w = Θ/ t\n w = 2π/ 454∙454 x 10^-6\n w = 6∙283 185 307.../ 454∙454 x 10^-6\n w = 13823∙007 68...\n w ≈ 13,823 rad/s\n\nIf the period is 20 for one cycle then:\n w ≈ 13,823 / 20 rad/s\n w ≈ 691∙15 rad/s
572283
stardate...\n\nor the Aztec calender they were only off 1 day in 6000 years...
761664
Pulsar's period, T = 0.033 sec and is increasing at a rate of 1.26 x 10^-5 sec per year\n\n(a) Angular acceleration, α = ∆ω/∆t\n\nwhere ω = 2π/T = 6.283/0.033 = 190.39 rad/sec\n\nSince the period increases by 1.26 x 10^-5 sec per year, then\n\n∆T = 1.26 x 10^-5 sec => T' = T + ∆T = 0.0330126 sec\n\nω' = 2π/T' = 6.283/0.0330126 = 190.32 rad/sec\n\nHence = ∆ω = ω' - ω = -0.07267\n\nNow, α = ∆ω/∆t = -0.07267/31536000 = 2.3 x 10^-9 rad/sec^2\n\n(b) ω' = ω + αt, set  ω' = 0, solve for t:\n\nt = -ω/α = (190.39)/(2.3 x 10^-9) = 82,624,112,123.83 sec\n\nor 2,620 years\n\n(c) ω = ω(0) + αt, set t = (2006-1054)x365x24x60x60 sec\n\nω(0) = ω - αt = 190.39 - (-(2.3 x 10^-9)(30022272000))\n\nω(0) = 190.39 + 69.05 = 259.44 rad/sec or its initial period\n\nT(0) = 0.024 sec