2012 and Nibiru

*This post re-published & paraphrased from cache – August 29th 2009*

I understand that there’s a lot of concern over the 2012 year and the so-called Planet X (Nibiru).  So much that I could calculate a significant statistical average of questions coming across Q&A sites such as Yahoo Answers.  Just to guess, I’d say 1 in 10 questions have something to do with this topic.

While it should be stated that there are very few things that can be stated with complete certainty.  These claims do not have substantiated data to back them up and the greater scientific community denounces it as a hoax.  While this has wrongly happened in the past (just think back to when the Earth was the center of the universe and the world was flat), I hope that we’ve moved past short-sighted, systematic denials and instead, evaluating a claim or situation as bogus because evidence or even Occam’s razor points to the rational that it isn’t true.

Given this, the Nibiru portion of the 2012 scares is a hoax.  The suggestion that the year 2012 as the end of time came about because of the Mayan calendar which ends in 2012; specifically December 21st 2012.  Aside from that, there are no grounds for this case.  Nibiru comes from an ancient Sumerian fable and no evidence of its existence has ever been found and beyond the fable, nothing suggests it existed or exists today.

To everyone who wants more information about this rather lengthy topic NASA has set up a site that gives more specific details about these claims:  Nibiru and Doomsday 2012

Now, there is one thing that should be added to this which is substantial and does have science to back it up.  That is, that the year 2012 will bring about the next peak in the Sun’s solar cycle.  At this time the Sun’s magnetic field will be so twisted up that it will actually snap out of it and as a result reverse its poles.  This time also means elevated activity in the way of sun spots and solar flares.  We know this because it has happened time and time again over the years.  The Sun cycles on an ~10.66 year cycle (new data on this, down from the 11 years that was previously used) and reverses polarity at each change of cycle.  Because of the increased solar activity during peak times we are at risk of such problems as electrical interruptions (Look up information on 1859 solar event, 1998 and 2003 incidents), satellite and communications related disruptions for anything that is unprotected.  This type of situation is a real threat based on what we’ve seen in the past and is something to be aware of.

How to Calculate Weight on Another Planet

*This post re-published from cache – August 28th 2009*

In working on answering a question today, I was rather perturbed by the fact that I was not able to find a definitive answer to a question that I felt should be well defined.  That question was: If I weigh 100lbs on Earth, how much would I weigh on Saturn?

Simple question right?  Not by what I found online.  There are a number of different sites that can be used to get a quick ( and possibly incorrect) answer.  I found the sites that I visited in search of an answer to be at the least, inaccurate and at the most, dead wrong.  The answers ranged from around 55lbs to about 115lbs.  On top of this, the method for calculating weight on another planet was based on the gravitational acceleration at the surface.  This is something that is defined already depending on the site or information you look at, but again, the values varied significantly.

Here’s an easy way to do it and get a reliable answer:

Find the gravitational acceleration, or more specifically the ratio of acceleration in relation to Earth, where Earth is a value of 1.  Now, find a table online with the planet’s ratio that you want to calculate.  For Saturn I used 1.065 (1.07 significant digits) found here.  Simply cross-multiply to get your answer:

\frac{100 lbs.}{1} = \frac{x}{1.07}

100 * 1.07 = x * 1

x = 107 lbs

Now that we have the easy answer I think it is important to understand that without already having the gravitational ratio this is still calculable by obtaining the gravitational ratio on our own, as shown below.

Before we begin, it should be noted that the radius of a planet type body varies from point-to-point.  This means that your answer will also vary depending on where you measure from.  This could account for some of the discrepancies that I found on other sites.  For example, you will be heavier at either of the poles and lighter at the equator.  This is due to the bulge at the equator caused by rotation.  This increases distance from the center of mass and reduces the gravitational forces on you slightly, causing you to weigh less.  For our purposes, we will use the mean radius vs. a specific measurement.

First, we need my mass in kilograms (kg), rather than weight in pounds (lbs) so we use the formula here:  \frac{x}{2.2} = x kg

\frac{100}{2.2} = 45.5 kg

Great, now that we have the mass, we can move forward.  We will need additional information at this point.  The formula that will allow us to make further calculations, Newton’s Law of Universal Gravitation.


It should be noted that the value for r should be in meters, the values of m_1 and m_2 in kilograms and the resultant value of F will be in Newtons.

This formula will allow us to calculate the force exerted between objects of a given mass, at a given distance from the center of gravity.  In turn, allowing us to calculate the respective weight.  For this formula to work, we will need additional statistics.  Since we are calculating for Saturn specifically, I took the values found from the NASA stats page linked earlier.

  • G = The Gravitational Constant (6.67*10^-11 N^2x m^2/kg^2)
  • m_1 = Object 1 Mass, Saturn (5.68*10^26 kg)
  • m_2 = Object 2 Mass, Person (45.5 kg)
  • r = Mean Radius, Saturn (5.82*10^7)

Now that we have all the variables needed, we can use Newton’s formula to get the result:

F_1 = (6.67*10^-11)\frac{(5.68*10^26)(45.5)}{(5.82*10^7)^2}

F_1 = \frac{1.72*10^18}{3.39*10^15}

F_1= 508N

Now we have the gravitational force (F) for Saturn.  In order to make a comparison based on weight, we will also need to do the same calculations for Earth.  Since we already have the gravitational constant and our calculated mass, we need to acquire Earth’s mean radius and it’s mass in kilograms.  These can both be found through the site linked earlier.

  • G = The Gravitational Constant (6.67*10^-11 N^2x m^2/kg^2)
  • m_1 = Object 1 Mass, Earth (5.97*10^24 kg)
  • m_2 = Object 2 Mass, Person (45.5 kg)
  • r = Mean Radius, Earth (6.37*10^6)

Once again we can use Newton’s formula to calculate the value of F:

F_2 = (6.67*10^-11)\frac{(5.97*10^24)(45.5)}{(6.37*10^6)^2}

F_2 = \frac{1.81*10^16}{4.06*10^13}

F_2= 446N

Now that we have both values we can compute the ratio:

R = \frac{F_1}{F_2}where F_1 is the value of Saturn and F_2 is Earth.

R = \frac{508}{446}

R = 1.14

Curiously enough, the ratio arrived at in this equation is higher than the 1.065 cited in the NASA document.  I am assuming this is due to differences in the value used for each planet’s radius.  I will see what I can do about further investigating this discrepancy and amend my post if needed.

Now, to finish our process.  We once again cross-multiply using 100lbs to 1 for the value on Earth.

\frac{100 lbs.}{1} = \frac{x}{1.14}

100 * 1.14 = x * 1

x = 114 lbs

In conclusion, the variations in answer can be explained by taking into account the differences in radius values used in the calculations.  Polar radius, Equatorial radius and mean radius will all give values slightly different from one another in a difference of P_r < m_r < E_r.

With our calculations it can be assumed that we have a possibility for error.  Based on this information and given a discrepancy of a significant 7.4 lbs, this could account for roughly 7-13% variance in answers to this question.  That said, make certain to note which radius measurement is being asked for when calculating these types of problems.  It could mean the difference between a right and wrong answer.