26th May 2011 Coral and Day Length

Coral lays down a layer of calcium carbonate every day. These layers can be counted, and so a thicker layer of 365 of them represent a year. Without knowing the length of a year, could we tell it's 365 days? Yes. The thickness of the layers varies with the seasonal variation in day length so there is a cyclic pattern to the layers. [2]

What about coral fossils? What do they tell us? Looking at fossils from 400 million years ago, there are about 410 layers per year. [1] What does this mean? Since the time it takes Earth to orbit the Sun is constant, it means that the length of a day was shorter, about 21 hours long [1,2]. Hence the earth is now spinning more slowly than it was 400 million years ago. Extrapolating back to 4.5 billion years ago, when a massive collision caused the formation of the moon [3], each day would only last about 6 hours.[1]

1. http://www.channel4.com/programmes/catastrophe/episode-guide/series-1/episode-1 16:00-19:30 minutes
2. http://www.don-lindsay-archive.org/creation/coral-clocks.txt
3. http://en.wikipedia.org/wiki/Moon#Formation

16th April 2011 Nibiru and 2012

I decided to look into the calculations I found on the mysterious imaginary planet of Nibiru which is supposed to crash into Earth on 21st December 2012, ending the world as we know it.

How big would Nibiru's orbit be?

Its orbit takes 3600 years and is an ellipse with the sun as one of its two foci. [Kepler's First Law: "The orbit of every planet is an ellipse with the Sun at one of the two foci."]

Kepler's third law states that

"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."  

The semi-major axis is the maximum distance of the orbit from its centre.

Let P be the orbital period of Nibiru (in sidereal years) and a be the semi-major axis of its orbit (in AU). Then

P ² α  a³

Since the constant is 1 for the units given, that is,  (sidereal year)²(AU)⁻³,

P ² = a³

and thus

a is the cube root of 3600  ²

Since Niribu is supposed to come close to Earth or even hit it, we assume that this is the closest Nibiru gets to the sun, one of its foci. This closest point is called the perihelion. So if 1 AU is the perihelion, what is the distance to the aphelion, the furthest point from the sun?

From the diagram, we see that it is simply twice the semi-major axis minus 1 AU. That is, the cube root of 3600² minus 1 which is 469 AU to the nearest AU.  The diagram below shows the comparative maximum distances in AU from the sun

Max distance from the sun in AU

and the table gives the figure for each of the planets.

Planet	Orbital Period years	Max Distance from Sun AU	Min Distance from Sun AU	Average Orbital Speed km/s
Mercury	0.24	0.47	0.31	47.87
Venus	0.62	0.73	0.72	35.02
Earth	1	1.02	0.98	29.78
Mars	1.88	1.67	1.38	24.08
Jupiter	11.86	5.49	4.95	13.07
Saturn	29.46	10.12	9.05	9.69
Uranus	84.32	20.08	18.38	6.81
Neptune	163.79	30.44	29.77	5.43
Pluto	248.09	49.31	29.66	4.67
Nibiru	3600	469	1	???

What would its orbit look like?

Using Dr. Douglas P. Hamilton's Orbital Calculator, the values semimajor axis 235 AU, eccentricity 0.995744 and eccentric anomaly 1.418 give the min and max values described above. [Source]  This gives a semiminor axis of about 20 AU.  Eccentricity is a measure of how far from a circle an ellipse is and ranges from 0 (for a circle) to 1.  The orbit is cigar shaped.

How fast is it going?

Nibiru's orbit sends it 469 AU from the Sun.  The gravity from the Sun on Nibiru at such a distance, if it has about the same mass as the Earth, would be around 1/4 millionth that of the Sun on the Earth. Since the gravity is so little, Nibiru will have almost escaped from the Sun's gravity. In order to do this, when it is close in to the Sun, it must almost reach the Sun's escape velocity.  The Sun's escape velocity at Earth's orbit is about 42 km/s.  That means Nibiru would be travelling at somewhere between the speed of Mercury and Venus.  If my calculations are anything like correct, it would seem to only be going at about 0.09 km/s when it's at its aphelion. [Using Kepler's Second Law:"A line joining a planet and the Sun sweeps out equal areas during equal intervals of time."]
It seems I'm doing something wrong since apparently, Nibiru's orbital velocity is

3.8 x 10⁸ Kilometres per year (or roughly 2.5 AU a year). [Source]

So, for Nibiru to be here on 21st December 2012, it would have to be around 4.2 AU from Earth now. That is around Jupiter's orbit. Surely it would be visible either through infrared or visible light or by gravitational effects?

From this source, Nibiru would be around 7.2 AU from the Sun now, and of magnitude 4.8, if it were the size of Earth.

How will other planets and objects affect its orbit?

Such an elongated orbit is highly unstable (why? I don't know) and so any interaction with a planet or anything else could easily knock it off its orbit.


http://www.2012hoax.org/nibiru
http://www.sjsu.edu/faculty/watkins/orbital.htm
http://astroblogger.blogspot.com/2011/04/looking-for-nibiru.html
http://www.astrosociety.org/2012/ab2009-32.pdf
circumference of an ellipse calculator 

6th April 2011 The Scale of Things

Size

Earth  --- Basketball
Height of Mt Everest --- a few human hairs
Distance you can drive in a day --- a penny
Our atmosphere --- a few mm
Space shuttle, low earth orbit --- a pea above the earth
Geostationary satellites  --- 75cm
Moon --- tennis ball
Earth to moon --- across a 2 lane road (7m)
Earth to Sun  --- walk for about 30 minutes,
Sun --- size of a basketball court
Proxima Centauri, nearest star --- 2 car garage
Proxima Centauri to Earth --- two the distance to the moon!
Asteroid which wiped out dinosaurs, plant life etc --- grain of salt

Times

Circle the basketball in 90 minutes in low earth orbit (pea height)
Apollo, earth to moon --- 3 days
Light, earth to moon --- 1 second
Car speed --- 6 months

Source: 365 Days of Astronomy

6th April 2011 Horseshoe Orbit

I found this diagram most misleading as it looks like it is from the viewpoint of someone outside the bodies involved yet it is actually showing the relative position of the asteroid with respect to Earth and the Sun.

The asteroid is in almost the same orbit as earth. When it is in a slightly smaller orbit, it moves faster, catching up the earth. When it approaches the earth at A, it is pulled into a larger orbit (A->B->C) and thus orbits the sun more slowly, so that it falls behind earth.  Eventually, Earth catches up at D, pulling it into a smaller orbit (D->E) so that it speeds up and travels away from Earth.  At no point does it ever pass Earth in its orbit.

Universe Today Article

NASA: Animations of a Horseshow Orbit

Wiki: Horseshoe orbit