Ison Rebooted (what for?) [Archive] - The Project Avalon Community Forum

Bill Ryan

29th September 2015, 23:18

.
The first half of this is about Comet ISON... it's a pretty old video, as can be seen by the entire content. Ison, which was just a regular comet, disintegrated a year and a half ago.

(See http://science.nasa.gov/science-news/science-at-nasa/2013/04dec_isonrecap, and many other reports.)

The apparent anomalies in the Ison overlay images were caused by parallax. It's all explained in this post:

http://projectavalon.net/forum4/showthread.php?62457-Is-Comet-Ison-Really-A-Comet&p=729147&viewfull=1#post729147

If this really was made by an anonymous NASA employee (which I profoundly doubt), he was certainly not a scientist.

(copied from that post to here)

~~~~~~~

From http://archive.stsci.edu/hla/ison :

Some bloggers have noted that the Comet C/2012 S1 (ISON) Hubble images (http://archive.stsci.edu/prepds/ison/) have some surprising features. Here we briefly explain the origin of the image structures.

Quick summary: The image is the result of combining 3 exposures that produce the 3 components, and the shapes are produced by the combined motion of the Hubble telescope and the comet. The images look exactly as expected.

http://archive.stsci.edu/hla/ison/combined.jpg
Comet C/2012 S1 (ISON)
Hubble, 2013 Apr 30
Combined 3-exposure image

The image from April 30, 2013 (released on July 16, 2013) using the F606W filter has attracted the most attention. That image (shown above) is the average of 3 separate Hubble exposures. The 3 exposures are available individually for display or download from the web page, as is the combined image. The table below shows the image of the comet in each of the 3 exposures along with information about the exposures.

Comet C/2012 S1 (ISON) Hubble observations on 2013 April 30 using F606W

http://archive.stsci.edu/hla/ison/exposure1.jpg
Exposure 1, 440 seconds

http://archive.stsci.edu/hla/ison/exposure2.jpg
Exposure 2, 440 seconds

http://archive.stsci.edu/hla/ison/exposure3.jpg
Exposure 3, 440 seconds

You can see that the 3 parts of the structure in the combined image result from the 3 different exposures. The comet itself does not have 3 pieces. They are an artifact that results from adding up the separate exposures.

The comet does not look the same in each exposure because both the comet and the Hubble telescope are moving during the exposure. The comet is blurred, just as a picture taken out the window of a moving car will be blurred. For this image the Hubble telescope was pointing very steadily and accurately at the background stars and galaxies rather than tracking the comet.

In this case, the largest part of the blurring is due to the motion of Hubble itself rather than the motion of the comet. Hubble is orbiting the Earth every 95 minutes. The third exposure ended 46 minutes after the start of the first exposure. During that time the telescope moved from one side of its orbit to the other, a distance of about 8000 miles (~13,000 km). Because the comet is relatively nearby (compared with the distant stars and galaxies), its apparent position among the stars changes due to the different viewing position. This effect is known as parallax. It is expected when observing Solar System targets, and can produce very obvious motion or blurring in images as Hubble whips around the Earth.

astronomylive

16th December 2015, 17:10

As Bill mentioned, this was just a natural consequence of parallax due to Hubble's orbit around the earth combined with ISON's orbit around the sun relative to the background stars (which Hubble was tracking on during three separate exposures). Here is all the math, detailed line by line.
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.

http://dropcanvas.com/6eyoa

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:

http://i.imgur.com/RrSi5MF.gif