Debunked: PLANET X Caught on Camera in Marbella

Astro

Senior Member
A video from late last month purports to show "Planet X" in the skies over Marbella, Spain, shortly after sunset.
There are several problems with this claim. First, and most obviously, the angle of the illuminated portion of the object is incorrect given the lighting conditions. The sun is below the horizon and to the south of the object's location in the sky, so the illumination should appear to be coming from down and to the left, not up and to the right. Less obviously, the video suggests an object traveling at an incredible apparent and relative velocity very close to earth.

Since the location is known, the time and date are known, and the bearings of the distant mountain peaks are known, it's actually possible to approximate the coordinates of the object and compute its orbit under the assumption that it's a real celestial object. I ran all these calculations once through and posted the results to the video uploader's personal forum. I was swiftly banned for "ridicule," but not before he posted a rebuttal involving his exact location (which was only about 84 meters from the location I determined based on the setting of the video and google earth).
http://h.dropcanvas.com/c8z0e/12764708_10209011514672082_604340231389468584_o.jpg
I have now re-run all the calculations using his self-reported location and the conclusion is the same as the first time I ran it; if this object were real, it was on an impact course and would have hit the surface of earth in the middle of the Atlantic ocean just minutes after the video ended.

I started by stabilizing three images from his video, taken at 3:00, 3:30, and 4:00 minutes in his video. I aligned the frames relative to the distant mountains.
http://h.dropcanvas.com/s7vpx/marbellanibirustabilized.gif
The mountains themselves can be used to calibrate the scale and orientation of his image. I labeled the headings of two of the peaks along the mountain range relative to his observing location:
http://h.dropcanvas.com/s7vpx/marbellaheadings.jpg
Measuring the spacing between them as being 493 pixels apart I determined that the image scale was about 52 pixels per degree. Knowing the distance and height of these peaks also allowed me to calculate the altitude and azimuth of the 0,0 origin of the image and create a formula for calculating the approximate altitude and azimuth of a given position within the image:
Azimuth = 253.926713995943+(X*0.0192292089249492)
Altitude = 9.81088356997971-(Y*0.0192292089249492)
As for timing information he gives the time as being 7:03 pm local time, which I assume is for the beginning of the clip he shows. This corresponds to 18:03 UT. We can convert this to the Julian day number with the following equations:
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

We also need to know the current sidereal time.
S = JD-2451545.0
T = S/36525.0
T0 = 6.697374558+(2.400.051336*T)+(0.000025862*T^2)
Add or subtract multiples of 24 as necessary to get the answer to a range of 0-24 hours
Convert the current UT to decimal hours (for example, 18:03 UT = 3/60 + 18).
Multiply the UT in decimal hours by 1.002737909
Add the result to T0 and add or subtract multiples of 24 to get it within the range of 0-24 as needed. This now equals the Greenwich Sidereal Time (GST).
Local sidereal time (for western longitudes) is simply the following:
LST = GST - (longitude in degrees/15)
Once again, add or subtract 24 hours to get the result within the range of 0-24 hours, and this is now the Local Sidereal Time.

Now we can convert the altitude and azimuth to equatorial coordinates with the following equations:
declination = arcsin(sin(altitude)*sin(latitude)+cos(altitude)*cos(latitude)*cos(azimuth))
hour angle = arctan((-cos(altitude)*cos(latitude)*sin(azimuth))/(sin(altitude)-sin(latitude)*sin(declination)))
for the hour angle you must evaluate the numerator and denominator of the arctan function separately in order to determine the correct quadrant of the angle. 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.

Right ascension = LST - hour angle

These are the coordinates at the equinox of date, but you have to precess them back to J2000.

zeta(A) = 0.6406161T + 0.0000839T^2 + 0.0000050 T^3
z(A) = 0.6406161T + 0.0003041T^2 + 0.0000051T^3
theta(A) = 0.5567530T - 0.0001185T^2 - 0.0000116T^3
T = (Julian Day of starting epoch - 2451545)/36525
Then you construct the following matrix:
P' =
[cos(zeta(A))*cos(theta(A))*cos(z(a))-sin(zeta(A))*sin(z(a)) cos(zeta(A)) * cos(theta(A)) * sin(z(a)) + sin(zeta(A)) * cos(z(a)) cos(zeta(A)) * sin(theta(A))

-sin(zeta(A))*cos(theta(A))*cos(z(a))-cos(zeta(A))*sin(z(a)) -sin(zeta(A))* cos(theta(A)) * sin(z(a)) + cos(zeta(A)) * cos(z(a)) -sin(zeta(A)) * sin(theta(A))

-sin(theta(A)) * cos(z(a)) -sin(theta(A)) * sin(z(a)) cos(theta(A)) ]
Then you calculate the following column vector v with the decimal coordinates from the epoch:
alpha = right ascension in decimal coordinates
delta = declination in decimal coordinates
v =
[cos(alpha)*cos(delta)
sin(alpha)*cos(delta)
sin(delta)]
Then you multiply P' and v to form column vector s.
Now you convert to the precessed epoch, recalculating zeta(A), z(A), and theta(A) with T = (Julian day of precessed epoch - 2451545)/36525. In this case the Julian day of the precessed epoch is J2000, 2451545, so T = 0.
Now construct the following matrix, which is really just a transposition of the first matrix but with the new values for zeta(A), z(A), and theta(A):
P =
[cos(zeta(A))*cos(theta(A))*cos(z(a))-sin(zeta(A))*sin(z(a) -sin(zeta(A))*cos(theta(A))*cos(z(a))-cos(zeta(A))*sin(z(a)) -sin(theta(A)) * cos(z(a))
cos(zeta(A)) * cos(theta(A)) * sin(z(a)) + sin(zeta(A)) * cos(z(a)) -sin(zeta(A))* cos(theta(A)) * sin(z(a)) + cos(zeta(A)) * cos(z(a)) -sin(theta(A)) * sin(z(a))
cos(zeta(A)) * sin(theta(A)) -sin(zeta(A)) * sin(theta(A)) cos(theta(A)) ]
Then calculate column vector w by multiplying matrix P by the previously calculated column vector s.
w=
[m
n
p]
Then you extract the precessed coordinates by the following formula:
Precessed Right Ascension = arctan(n/m)
Precessed declination = arcsin(p)

Be sure to use the quadrant disambiguation routine on the above arctan numerator and denominator as described earlier. This now gives us the coordinates, precessed back to the J2000 standard epoch. These coordinates, along with the location and the time of the observations, can be loaded into a number of astronomy programs to calculate the orbit. I calculated the coordinates and times from 3:00, 3:30, and 4:00 in the video:
Time in UT: 18:03:48
RA: 20h 09m 37.69s Dec: -06d 17' 04.6"

Time in UT: 18:04:18
RA: 20h 08m 59.23s Dec: -06d 17' 00.3"

Time in UT: 18:04:48
RA: 20h 08m 17.89s Dec: -06d 16' 01.0"

I loaded this information into FindOrb, along with his precise location. Here's the station information for his location in standard minor planet center format:

W00 355.110230.804742+0.591635Marbella, Spain

I used the Simplex routine in FindOrb which is well-optimized for short arcs of observations. Simplex quickly converged on a solution which fits the observational data quite well given the limited resolution of the observations. Unfortunately for the video's creator, the solution indicates that the object was just about to hit earth when it was filmed, and impacted in the middle of the Atlantic ocean just minutes after the end of the video, long before it was even uploaded to YouTube.

Marbella Nibiru

Perigee 2016 Jan 18.766468 TT = 18:23:42 (JD 2457406.266468)

Epoch 2016 Jan 18.0 TT = JDT 2457405.5

q 3897.19938km (2000.0) P Q

H 13.7 G 0.15 Peri. 108.23621 0.25216793 0.81305282

Node 313.89226 0.34589529 -0.58217588

e 2.0547487 Incl. 46.73230 0.90375205 -0.00404309

From 3 observations 2016 Jan. 18 (59.6 sec); mean residual 10".550.

IMPACT at 18 Jan 2016 18:16:38.01 lat +30.16948 lon W44.87728
http://dropcanvas.com/vd5g3

The above link also contains a link to the spreadsheet I created to convert from X and Y coordinates to J2000 coordinates. I apologize for any "malicious site" warnings your browser gives you; unscrupulous characters use dropcanvas to host malicious files but I promise I am not one of them.

We can now use the orbital solution to calculate the approximate distance and therefore the size of the impactor. The distance of the object at the start of his clip from his location is predicted to be 13721 km according to FindOrb. Given the scale of his image and the size of his object (0.0192 degrees per pixel, object diameter roughly 43.4 pixels) and the distance, his object should have been about 199 km wide. Even a porous asteroid that large would have devastated our planet in an instant, in stark contrast to his claim that it only created a mild earthquake and weird weather.

In fact, an object that large hitting earth at that velocity would generate an earthquake larger than any ever seen in history. It would be felt around the world and it would create a gigantic complex crater more than a 1000 km in diameter, even if it were made of porous rock. At his location 3754 km from the point of impact, the fireball would have caused his clothing to spontaneously ignite. He would have been killed by his object long before he even had the chance to download his video from his camera. His impactor would have been an order of magnitude larger than the Chucxulub impactor that is thought to have killed the dinosaurs.
http://impact.ese.ic.ac.uk/cgi-bin/...theta=45&tdens=1000&wdepth=8000&wdepthUnits=1

Last I checked, humanity did not come to an abrupt end on January 18, 2016.
 
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Chew

Senior Member.
Can you calculate the heliocentric orbital elements? I would be curious to see if this "Nibiru of the week" had a 3600 year orbital period.
 

Astro

Senior Member
Can you calculate the heliocentric orbital elements? I would be curious to see if this "Nibiru of the week" had a 3600 year orbital period.
No problem. Here are the Keplerian orbital elements relative to the sun, I just set FindOrb to interpolate them for one day prior to the observations (doing so when it was still far from earth causes FindOrb to automatically switch to the sun as the most appropriate central body).

Orbital elements:
Marbella Nibiru
Perihelion 2015 Nov 29.338424 TT = 8:07:19 (JD 2457355.838424)
Epoch 2016 Jan 17.0 TT = JDT 2457404.5 Earth MOID: 0.0000 Ma: 0.0114
M 32.26533 (2000.0) P Q
n 0.66305574 Peri. 109.66169 0.67341614 -0.73721074
a 1.30247487 Node 297.88164 0.65279193 0.62794642
e 0.3836143 Incl. 3.57097 0.34694868 0.24940649
P 1.49/542.93d q 0.80282678 Q 1.80212297
From 3 observations 2016 Jan. 18 (59.6 sec); mean residual 10".550.
 

Hevach

Senior Member.
0.0192 degrees per pixel, object diameter roughly 43.4 pixels
So, 50 arcminutes? The full moon is only 31. This would have been quite visible - no amount of cover up or cutting the live feed will work on the eyes of the hemisphere facing such a large object in the sky that is quite obviously not the moon.
 

Astro

Senior Member
So, 50 arcminutes? The full moon is only 31. This would have been quite visible - no amount of cover up or cutting the live feed will work on the eyes of the hemisphere facing such a large object in the sky that is quite obviously not the moon.
Quite. Yes, it was about 0.83 degrees wide, significantly larger than either the sun or the moon in the sky. There's no way it would have gone undetected.
 
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