Dead astronomers...Isons Moons...NASA coverup | |
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Jude User ID: 57596631 United States 11/10/2015 07:52 PM Report Abusive Post Report Copyright Violation | Further update on the vaccine action, another doctor was on the radio saying that five to ten percent of the population carry a retrovirus which is passed from mother to baby at birth. In some of these infants, the intense scheduling of the early immunization sends the immune system into overdrive, the immune system attacks the child's own nerve cells, causing the group of brain changes known as autism. The doctor suggested testing for the retrovirus be done at birth, and for those children affected, a modified immunization schedule be adopted. |
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Dr. Astro
Senior Forum Moderator User ID: 4211721 United States 11/13/2015 02:58 PM Report Abusive Post Report Copyright Violation | Interesting video on Isons moons....Perhaps this is why she appeared to have wings Quoting: Guitarjohn [link to www.youtube.com] No. |
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Anonymous Coward User ID: 13391804 Canada 11/13/2015 03:21 PM Report Abusive Post Report Copyright Violation | Well you Americans better get your shit together so you can handle the truth! Quoting: truth handler 13391804 Pretty dumb statement. tell us all what the truth is. now. Truth is your former President just told you, so lap it up. That's what you Americans do, lap it up. It will never change till enough of you dimwits stand shoulder to shoulder and say, no more. Why are you so afraid of change? |
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Dr. Astro
Senior Forum Moderator User ID: 4211721 United States 11/19/2015 12:06 PM Report Abusive Post Report Copyright Violation | |
Jude User ID: 57596631 United States 12/02/2015 04:29 PM Report Abusive Post Report Copyright Violation | On Friday December 30,1983 Dr. Gerry Neugebauer of the California Jet Propulsion Laboratory confirmed that the IRAS (Infra Red Astronomy Scope) had detected a Jupiter sized or larger body in the area of our solar system's Oort cloud, about 50 billion miles away. This story ran in the Washington Post and is available online or in their archives. The IRAS "stopped working" shortly thereafter, and Dr. Neugebauer had no further comment. Dr. Neugebauer's final silence was secured September 26 2014 by a very rare neurologic disorder which caused progressive paralysis and then death. This post in memory of Dr. Neugebauer. |
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Dr. Astro
Senior Forum Moderator User ID: 4211721 United States 01/26/2016 03:08 PM Report Abusive Post Report Copyright Violation | On Friday December 30,1983 Dr. Gerry Neugebauer of the California Jet Propulsion Laboratory confirmed that the IRAS (Infra Red Astronomy Scope) had detected a Jupiter sized or larger body in the area of our solar system's Oort cloud, about 50 billion miles away. This story ran in the Washington Post and is available online or in their archives. The IRAS "stopped working" shortly thereafter, and Dr. Neugebauer had no further comment. Dr. Neugebauer's final silence was secured September 26 2014 by a very rare neurologic disorder which caused progressive paralysis and then death. This post in memory of Dr. Neugebauer. Quoting: Jude 57596631 No he didn't. They said they didn't know what it was. They later figured it out. Dr. Neugebauer was one of the authors in the papers I cite below revealing the identity of the unidentified point sources. So yes, he did have further comments. You lied. ... Quoting: Esoterica Why this then? [link to planet-x.150m.com] Dr. Astro, I have always admired your expertise, but I have seen the article, with the photos. How do you explain that? How dare you question the word of Astrojesus! L M A O I'm not going to ban anyone for asking a question, but I do ask that you read the answer. Why this then? [link to planet-x.150m.com] Dr. Astro, I have always admired your expertise, but I have seen the article, with the photos. How do you explain that? Please quote from the article where it says "NASAs IRAS infrared telescope discovered an unbelievably huge, massive, what looked like a slowly burning star, a big object, headed right toward our solar system from the region of Orion." That is not what the article says. "So mysterious is the object that astronomers do not know if it is a planet, a giant comet, a nearby "protostar" that never got hot enough to become a star, a distant galaxy so young that it is still in the process of forming its first stars or a galaxy so shrouded in dust that none of the light cast by its stars ever gets through. " It says they didn't know what it was. In fact there were multiple unidentified point-like sources. I've been over this many times before IRAS did not discover any planets. It did find some initially unidentified point-like sources. The 1983 article claims it "could" be a jupiter-like planet OR extra-galactic, they had no idea when it was first observed. IRAS did not find a planet though, it did find a number of infrared-luminous galaxies and intragalactic dust. In fact, it initially found 9 unidentified point-like sources of infrared light, listed here: Quoting: Astromut 0358+223 3h 58m 2.8s 22d 18 0 0404+101 4h 4m 44.7s 10d 11 52 0412+085 4h 12m 32.3s 8d 31 13 0413+122 4h 13m 47.3s 12d 17 16 0422+009 4h 22m 54.0s 0d 56 6 0425-012 4h 25m 12.1s -1d 14 50 1703+049 17h 3m 1.4s 4d 57 50 1712+100 17h 12m 57.8s 10d 4 8 1732+239 17h 32m 51.4s 23d 56 36 Here are the papers that identified these objects: Unidentified point sources in the IRAS minisurvey, Houck et al. Astrophysical Journal, vol. 278, March 1, 1984, p. L63-L66 Unidentified IRAS sources - Ultrahigh-luminosity galaxies, Houck et al., Astrophysical Journal, vol. 290, March 1, 1985, p. L5-L8. Optical counterparts of unidentified IRAS point sources Infrared luminous galaxies, Aaronson and Olszewski, Nature, vol. 309, May 31, 1984, p. 414-417. Oh, and by the way, the above coordinates from the original IRAS paper are given in the B1950 epoch which was still in use in the 80's. Be sure to precess forward before use. You guys all know how to do that, right? |
Dr. Astro
Senior Forum Moderator User ID: 4211721 United States 01/26/2016 03:13 PM Report Abusive Post Report Copyright Violation | Mr. Wiltgen will not have the chance to say what I am certain that he noticed, the changing time of the winter solstice indicates a significant changing of the earth's tilt and rotation.- This post in memory of Nick Wiltgen Quoting: Jude 57596631 Bullshit. If earth's tilt and rotation changed that significantly (enough to change the date of the solstice from what it should be) then polar aligned telescopes would stop working. That would make images like these impossible to take using telescopes aligned based on where the axis of rotation used to be. Why are you bumping an ancient thread about ISON and its so-called "moons?" It's debunked, get over it, it wasn't "Nibiru" nor are they killing people over it. Here is all the math, detailed line by line. Quoting: Dr. Astro The first thing you need are the orbital elements of earth, ISON, and Hubble. Here is a scan of a page containing the orbital elements of earth from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith. [link to dropcanvas.com] Earth Orbit Epoch 1990.0 Orbital period Tp 1.00004 (tropical years) Longitude at epoch E 99.403308 Longitude of perihelion w' 102.768413 eccentricity e 0.016713 semi-major axis a 1 (AU) inclination i 0 Argument of perihelion O 0 Hubble Space Telescope Orbit Orbit Epoch (Julain Day) 2456412.251 Orbital period (years) Tp 0.0001821 Longitude at Epoch 358.4212 Longitude of Perigee 59.6612 eccentricity 0.0002971 semi-major axis (km) 6934.189 inclination 28.4694 Longitude of ascending node 230.3343 Argument of perigee 189.3269 C/2012 S1 (ISON) Orbit Perihelion Date (Julian Day) 2456625.264 Argument of Perihelion w 345.54102 Longitude of Perihelion w' 641.2278526 Longitude of Ascending Node O 295.6868325 Inclination i 62.16095792 Eccentricity e 1 Perihelion Distance (AU) q 0.012466817 The rest is just math. First we need to find the number of days since the epoch of those elements. The first step is to convert the date and time of when we want to know ISON's apparent position as seen from Hubble to a julian day number. y = year m = month d = day (fraction of a day) The procedure to convert this to Julian day number format is as follows: If m = 1 or 2, subtract 1 from y and add 12 to m, otherwise y' = y and m' = m If the date is > or = 1582 October 15 calculate: A = the integer part of y'/100 B = 2-A+integer part of A/4 otherwise B = 0 if y' is negative calculate C = integer of ((365.25 x y')-.75) otherwise C = integer part of (365.25 x y') D = integer of (30.6001 x (m' + 1)) JD = B + C + D + d + 1720994.5 For calculating the position of earth relative to ISON we also want the JD for 1990.0 (which is the epoch of earth's orbital elements), which equals 2447891.5. Therefore, the days since epoch 1990.0 which we will call D We will first calculate the position of earth at this timepoint before we do the calculations for ISON. Mean anomaly = 360/365.242191 * D/Tp + E - W where D = days since epoch 1990.0 Tp = Period E = Longitude at epoch (1990.0) W = Longitude of perihelion Refer to the above scanned page for all of these values for Earth. In most cases that will give you a number greater than 360 for any date past 1991 because earth will have gone around the sun more than once, so you need to reduce it down to a value within 360 degrees using the following formula: d = starting value (8568.4260132099 in this case) (d/360 - integer of (d/360))*360 = d' if d' < 0, then 360 + ((d/360 - integer of (d/360))*360) = d' d' is the new value. This formula will be reused later on, I will simply refer to it as the "360 degree routine." Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. This calls for an iterative process to solve Kepler's equation (E - e*sin(E) = M). We start by making an approximation of E = E0 = M Find the value of sigma = E-e*sin(E-M) where M = mean anomaly e = eccentricity if sigma < 10^-6 radians then take the present value of E as the correct solution, otherwise continue for another iteration For the next iteration find delta E = sigma/(1-e*cos(E0)) E1 = E0 - delta E Substitute E1 back in the equation above to find a new value for sigma and repeat this routine as necessary until convering on a solution where sigma < 10^-6 radians. tan (v/2) = Sqrt((1+e)/(1-e))*tan(E) where v = true anomaly e = eccentricity E = solution to kepler's equation from the iterative process above find tan(v/2), take the arctangent and multiply the result by 2. You need to evaluate the numerator and denominator within the arctan function separately in order to do a quadrant disambiguation routine. If the numerator is positive and the denominator negative, add 180 to the result of the arctan function (after converting the result to degrees of course). If the numerator is negative and the denominator positive, then add 360 to the arctan result unless that makes it greater than 360 degrees, in which case add 180 degrees. Finally, if both the numerator and denominator are negative, then add 180 degrees to the result of the arctan function. Otherwise if you have made no other additions and the result is negative, add 180 degrees, or else keep the result of the arctan function without adding or subtracting anything. Now you have v, the true anomaly. Now we need to calculate heliocentric longitude, l. l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W) where D = days since epoch Tp = Period e = eccentricity E = longitude at epoch W = longitude of the perihelion Use the 360 degrees routine to get l within 360 degrees. Now we need the radius vector, that is to say, distance from the sun, r. r = (a*(1-e^2))/(1+e*cos(v)) where a = semi-major axis of the orbit e = eccentricity v = true anomaly Now we repeat these calculations for Hubble given the values for Hubble's orbit above. For Hubble we're dealing with an inclined orbit, so we also need to find the latitude. This is given by U = Arcsin(sin(L - Omega)*sin(i)) where L = true longitude of Hubble (same formula as heliocentric longitude for earth) omega = longitude of the ascending node i = inclination Now we need to find the geocentric longitude; true longitude consists partly of values which are inclined to the equator by Hubble's inclination. L' = v+w'-O where v = true anomaly w' = longitude of perigee O = argument of perigee Perform the 360 degree routine on L' Now we re-orient L' back into the plane of the equator so that we can find the geocentric longitude with respect to the vernal equinox, effectively geocentric right ascension. geocentric right ascension = arctan((sin(L')*cos(e))/cos(L'))+O where L' equals the result from above e = eccentricity O = argument of perigee Remember to evaluate the arctan function using the quadrant disambiguation routine described above Now we need to find the geocentric longitude. We now know the angle of Hubble east of the vernal equinox, but we need to know the number of degrees the vernal equinox is of the 0 line of longitude, the prime meridian, in other words, we need to know Greenwhich mean sidereal time. GMST = 18.697374558+24.0657098244191*(JD-2451545) GMST*15 = GMST in degrees Perform the 360 degree routine on GMST in degrees to get the value within 360 degrees. Now take the geocentric right ascension and subtract GMST in degrees. This is the eastern longitude of Hubble. If the value is negative, it's west longitude, if the value is positive it is east longitude. Save this value and the latitude calculated above for later. Now we need to calculate the position of ISON relative to earth. The orbital elements are listed above. We will approximate ISON's orbit as a parabolic orbit to aid in the ease of calculation. First find W W = (0.0364911624/(q*sqrt(q))) * d where q = perihelion distance d = days since perihelion (perihelion date in Julian Days - JD) Now we need to solve s^3 + 3s - W = 0 First approximate s = s0 = W/3 calculate sigma sigma = s0^3 +3*s0 - W if sigma < 10^-6 degrees then take s as the correct value otherwise continue with the iterations and proceed below s1= (2*s0^3 + W)/(3(s0^2 + 1)) substitute s1 back into the formula above to find sigma and repeat until the value is within the accuracy of sigma <10^-6 degrees. Then take that value of s and proceed Find the true anomaly v = 2*arctan (s) Find the distance from the sun r = q*(1+s^2) where q = perihelion distance in AU find heliocentric ecliptic longitude l = v+w' where v = true anomaly w'= longitude of perihelion find the heliocentric ecliptic latitude U = Arcsin(sin(l - O)*sin(i)) where l = heliocentric ecliptic longitude O = argument of perihelion i = inclination Now we need l' which is the heliocentric longitude projected onto the plane of the ecliptic. l' = arctan((sin(l-O)*cos(i))/cos(l-O))+O where l = heliocentric ecliptic longitude O = argument of perihelion i = inclination remember to perform the quadrant disambiguation routine on the arctan function Now we calculate the geocentric ecliptic longitude (lam) and latitude (beta) of comet ISON. If ISON's radius vector from the sun is less than earth's at the time you are evaluating you can regard it as an inner planet in which case the formula is this: lam = 180 + Le + arctan ((ri*sin(Le-l'))/(re-ri*sin(Le-l'))) where Le = heliocentric longitude of earth re = radius vector of earth rI = radius vector of ISON l' = heliocentric longitude of ISON projected onto ecliptic Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. If ISON's radius vector from the sun is greater than earth's at the time you are evaluating you can regard it as an outer planet in which case the formula is this: lam = arctan((re*sin(l'-Le)/(rI-re*cos(l'-Le)))+l' where Le = heliocentric longitude of earth re = radius vector of earth rI = radius vector of ISON l' = heliocentric longitude of ISON projected onto ecliptic Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. beta is calculated with the following formula: beta = arctan((r'*tan(U)*sin(lam-l'))/(re*sin(l'-Le)) where Le = heliocentric longitude of earth re = radius vector of earth l' = heliocentric longitude of ISON projected onto ecliptic U = heliocentric latitude for ISON r'= radius vector of ISON Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. Distance of ISON from earth reI=sqrt(rI^2+re^2-2*re*ri*cos(l'-Le)*cos(U)) where Le = heliocentric longitude of earth re = radius vector of earth rI = radius vector of ISON l' = heliocentric longitude of ISON projected onto ecliptic U = heliocentric latitude for ISON For the next step we will need the obliquity of the ecliptic. This can be calculated with the following formula, which will need time "t" centuries from 2000.0. We can calculate t thusly: t = ((JD-2451545)/36525) where JD = julian day number Obliquity of the ecliptic (Obl) can then be calculated with this formula: Obl = 23.43928-0.013*t+0.555*(10^-6)*(t^3)-0.0141*(10^-8)*(t^4) Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of ISON at this timepoint. Right ascension = arctan((sin(lam)*cos(Obl)-tan(beta)*sin(Obl))/cos(lam)) where lam = geocentric ecliptic longitude beta = geocentric ecliptic latitude obl = obliquity of the ecliptic Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lam)) where lam = geocentric ecliptic longitude beta = geocentric ecliptic latitude obl = obliquity of the ecliptic Now we have the geocentric coordinates for ISON, but we need to account for the parallax induced by Hubble's orbit. First find u u = atan(0.996647*tan(U)) where U = geocentric latitude of Hubble Next find p*sin(theta') p*sin(theta') = 0.996647*sin(u)+(r/6378140)*sin(theta) where u = u above r = altitude of hubble (convert to meters) theta = geocentric lattitude of hubble find p*cos(theta') p*cos(theta') = cos(u)+r/6378140*cos(theta) where u = u above r = altitude of hubble (convert to meters) theta = geocentric lattitude of hubble pi = (atan((6378140+r)/149597870700)*3600)/reI)/3600 where r = altitude of Hubble (in meters) reI = distance of ISON from earth LST = (GMST*15)+Lh where GMST = Greenwhich mean sidereal time Lh = geocentric longitude of Hubble Perform the 360 degree routine to get LST in degrees for the equation below delta RA = atan((p*cos(theta')*sin(pi)*sin(LST-RA))/(cos(declination)-p*sin(theta')*sin(pi)*cos(LST-RA)) where RA = geocentric right ascension of ISON declination = geocentric declination of ISON. Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. subtract delta RA from the geocentric right ascension of ISON to get the parallax corrected right ascension of ISON which we will call the hubble centric RA gamma = atan(tan(p*sin(theta')/p*cos(theta'))*cos(.5*delta RA)*(1/cos(LST-.5*(geocentric RA + hubble centric RA))) delta declination = (p*sin(theta')*sin(pi)*sin(gamma-geocentric declination))/(sin(gamma)-p*sin(theta')*sin(pi)*cos(gamma-geocentric declination)) subtract delta declination from the geocentric declination to get the parallax corrected declination of ISON which we will call the hubble centric declination. Now we have both the hubble centric RA and hubble centric declination for this point in time. If you repeat these equations over multiple points in time covering the imaging times for the Hubble images you can generate a list of coordinates for ISON over the imaging session and graph the predicted trail of ISON as seen from Hubble over that period of time. Doing so results in the following graph which matches with the images taken by Hubble: [link to i.imgur.com] |