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It's already September 2012 as I write these words, but where is Nibiru, the great planet that's supposed to be bombarding the inner solar system in 2012? When a comet or an asteroid becomes bright enough to observe with binoculars - or even if it's not - I can always find a reliable charts at Heaven's-Above or Skyhound. But why is it so onerously difficult to find a reliable Nibiru sky chart?

Look on the right for what's known as an ephemeris, giving you the coordinates for the sun, moon and planets and even the dwarf planet Pluto for early September 2012. Noticeably absent, however, is the planet Nibiru! Click on Solar System Live for an up-to-date ephemeris, courtesy of Fourmilab.

Where are all the reputable and easy-to-use finder charts highlighting Nibiru's place in front of the backdrop stars? For instance, I can't find Nibiru on this handy sky chart, this interactive planetarium and elsewhere. Why are Jupiter and Saturn so clearly visible to the unaided eye whereas Nibiru is lost in the ether somewhere?

But, then, perhaps Nibiru can only be seen with binoculars or the telescope? Click here to see the present positions of the eight solar system planets. Or here or here to see the positions of the planets all the way to the dwarf planet Pluto. Alas, I see no Nibiriu anywhere. Certainly, this rogue world should be inside Pluto's orbit by now.

I can find all kinds of sky charts, giving me all kinds of information. Here's a wonderful chart of the four major moons of Jupiter, two of which are about the size of our moon, and two which have diameters some 1.5 times the lunar diameter. Here's another sky chart showing where to find the outer planets Uranus and Neptune. Moreover, the charts atThe Digital Astrolabepinpoint the positions of the fainter and more distant dwarf planets Pluto, Haumea, Makemake and Eris. But where can I find a viable Nibiru chart?

**Is Nibiru an elaborate fiction?**

Let's put it this way. The planet Nibiru appears to be the brain child of a fellow named Zacharia Sitchin. As the story goes, Sitchin translated Sumerian cuneiform tablets and supposedly discovered that the ancients had been aware of the planet Nibiru, which has an orbital period of 3,600 years.

Purportedly, inhabitants of Nibiru (Anunnaki) first arrived on Earth about 450,000 years ago to mine for gold. As I understand it, the technologically advanced Anunnaki created humans (Homo sapiens) to serve as slaves. They performed this feat by the use of female apes and genetic engineering.

Apparently, Sitchin did not predict the return of Nibiru in 2012 but in 2900. A woman named Nancy Lieder predicted the return of Nibiru in May 2003, and when that failed to materialize, doomsday enthusiasts forwarded the arrival date to 2012.

*Figuring Nibiru's orbit with Kepler's Third Law*

Kepler's Third Law of Planetary Motion is so beautifully simple that we'd be remiss not to make use of it to figure out Nibiru's orbit:

a^{3} = p^{2}

In the above equation, p = period of planet's orbit in Earth years and a = the semi-major axis of the planet's distance in astronomical units. Given that Nibiru's orbital period = 3600 Earth years, we can now compute its semi-major axis in astronomical units (AU):

a^{3} = p^{2} |

a^{3} = 3600^{2} |

a^{3} = 12,960,000 |

a = semi-major axis = 234.892 AU |

*Computing Nibiru's orbital speed in inner solar system*

One thing to keep in mind. The orbit of any planet is an ellipse. Two different ellipses could have the same semi-major axis yet varying eccentricities. And a change of a planet's eccentricity necessarily changes the distances of perihelion and aphelion, but not necessarily the lengths of the major and semi-major axes. Play around with this ellipse to change its eccentricity without altering the lengths of the major and semi-major axes.

Because we know Nibiru's semi-major axis, which we'll round off to 235 astronomical units (AU), we can also compute Nibiriu's orbital speed at any distance from the Sun. We'll figure Nibiru's orbital speed at one astronomical unit (AU) - the Earth's distance from the Sun.

If Nibiru gets as close as one AU from the Sun, the outer edge of its orbit must be a whopping 469 AU from the Sun (2 x semi-major axis - 1). That gives an orbital eccentricity of 0.9957447 (234/235 = 0.9957447)!

*The magical Vis-viva equation figures out Nibiru!*

Now that we know Nibiru's semi-major axis, we can compute Nibiru's orbital speed at any distance from the Sun with relative ease. Using the magical Vis-viva equation to perform the feat, we have r = distance from the sun = 1 AU, and a = semi-major axis = 235 AU

We figure the orbital speed in kilometers per second, with r = 1 AU and a = 235 AU:

orbital speed = 29.8 x the square root of (2/r - 1/a) |

orbital speed = 29.8 x the square root of (2/1 - 1/235) |

orbital speed = 29.8 x the square root of(2 - 0.0042553) |

orbital speed = 29.8 x the square root of 1.9957447 |

orbital speed = 29.8 x 1.4127083 = 42.098707 kilometers per second |

*Nibiru's extremely unstable orbit*

At Earth's distance from the Sun - at one astronomical unit - Nibiru would be flying at nearly 42.1 kilometers per second. Quite by coincidence, the escape velocity from the solar system at one astronomical unit from the Sun is 42.1 km/sec. According to this web page, that makes for a highly unstable orbit!

According to Wikipedia (under the subheadingScientific criticism), the astronomer Mike Brown claims such an orbit would only allow a planet to last for a million or so years before being ejected from the solar system.

By the way, the illustration on the left ofEARTH TO COLLIDE WITH NIBIRUwas taken fromWeekly World News(The World's Only Reliable News). If interested, read the accompanying articleEarth to collide with Nibiru on November 21, 2012!

*Planet with greatest known eccentricity in Milky Way galaxy*

After all this investigation of the Nibiru drama, I couldn't help but wonder if any known extrasolar planet exhibits a comparable eccentricity to that of Nibiru. Much to my surprise, I found the planet with the greatest known eccentricity, known as HD 80606 b, with a semi-major axis of 0.453 AU and a periastron point of 0.03 AU. It has an eccentricity = 0.9336 (Nibiru's eccentricity = 0.9957).

Should a planet with these characteristics orbit our Sun, what would its orbital speed be relative to escape velocity at 0.03 AU from the Sun? Given an escape velocity of 42.1 km/sec at the Earth's distance from the Sun, the escape velocity can be found for any other distance by this equation:

escape velocity from solar system = the square root of 42.1^{2}/AU distance |

escape velocity from solar system = the square root of (42.1^{2}/0.03) |

escape velocity from solar system = the square root of (1772.41/0.03) |

escape velocity from solar system = the square root of 59080.333 |

escape velocity from solar system = >243 km/sec |

Now to use the Vis-viva equation to compute the planet's orbital speed at periastron, with r = distance from sun at periastron = 0.03 AU, and a = semi-major axis = 0.453. Hopefully, I got it right!

orbital speed = 29.8 x the square root of (2/r - 1/a) |

orbital speed = 29.8 x the square root of (2/0.03 - 1/0.453) |

orbital speed = 29.8 x the square root of (66.667 - 2.2075055) |

orbital speed = 29.8 x the square root of 64.459161 |

orbital speed = 29.8 x 8.0286463 = >239 km/sec |

copyright 2012 by Bruce McClure