## Orbits and Faith The Universe as viewed by some.

I was browsing through the Internet one day. In the merry, merry month of July. I was taken by surprise, because right before my eyes, I saw a lot of really stupid things out there.

Browsing a while back, I came across a website that proposed that the Sun revolved around the earth. If that wasn’t bizarre enough, they were offering \$1000 to anyone who could prove them wrong. So I said to myself, “Self, this has to be easy; I bet it can be done with basic high school physics and math.” Then myself challenged me with, “Bet you can’t.” The argument ensued, “Can too!” “Can not!” “Can too!” “Can not!” “Can too!” “Can not!” “Can too!” “Prove it!” “Damn, how do I keep losing arguments with myself?” Well, I did have to prove it. So, that is what I did. I set out to prove that the Earth actually does revolve around the Sun and not vice versa.

I submit my humble, high school, physics, attempt at a proof that the Earth revolves around the sun for your consideration and enjoyment.

First, we must agree upon these as facts:

Let Newton’s three laws of motion and Law of Gravity be agreed upon as correct and relevant (1):

• Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed on it.
• Force is equal to the change in momentum (mV) per change in time. (For a constant mass, force equals mass times acceleration. F=ma)
• Every action has an equal and opposite re-action.
• Force = Gravitational constant * Mass-a * Mass-b / distance^2 (2)

We also agree that:

• The Solar Mass is 1.989 x 10^30 Kg. (3)
• The Mass of the Earth is 5.976 x 10^24 Kg. (4)
• The distance between the Earth and the Sun is 149,597,892 Km. (5)
• The Circumference of a circle “C” is calculated as C=2*pi*r. (6)
• The velocity “v” of an object is calculated as v=distance/time. (7)
• The acceleration of an object in a circular path is a=v^2/r. (8)
• G is a natural constant which has been measured in a laboratory and equals approximately 6.6742×10^-11 m^3 s^-2 kg^-1. (9)(10)

We finally agree for the sake of this proof:

• The motion of the earth, or sun, would be in a curve around the other.
• We will accept that the orbits of the celestial bodies are circular instead of elliptical to simplify the math and we agree that this does not affect the results of this proof.
• We agree that each body would complete a single orbit around the other once per year, and in the case of a geocentric model, the earth spins to create day and night.

If you can’t, or don’t, agree to the above, you’re not living in the same universe as I am. Go back home on your transcendental, rainbow, space ship.

Now to the proof.

Let m1 = mass of the sun
Let m2 = mass of the earth
Let t = 1 year or 365 1/4 days or 31,557,600 seconds
Let r = 149,597,892,000 meters

v=d/t
c=2*pi*r
a=(v^2)/r
F=m*a

Note on Mathematical Notation: * denotes multiplication, ^ denotes exponents, ()’s are used to group operations, / denotes division, pi denotes the value 3.14159265358979323846… (11)

Simple physics which was all agreed upon above. Now, we will create a hypothetical Geocentric model where we assume the Sun revolves around the earth.

The sun will travel the circumference of the circular orbit (also agreed upon above for simplicity) around the earth. So, the distance is the Sun travels is the circumference ‘c’ of the circular orbit c=2*pi*r. d=c=2*pi*r, where ‘r’ is the distance between the sun and the earth, r=149,597,892,000 meters.

We can then solve for the velocity of the sun in this scenario by substituting (2*pi*r) for ‘d’ in the equation v=d/t, v=c/t, or v=(2*pi*r)/t, where t=31,557,600 seconds.

Then we solve for the acceleration of the sun around the earth by substituting the sun’s velocity, above, into a=(v^2)/r to get a=(((2*pi*r)/t)^2)/r.

This allows us to solve for the Force of the sun traveling in this scenario around the earth. F=m*a, we substitute F=m1*((((2*pi*r)/t)^2)/r), where m1=1.989×10^30 kg.

Then we can go back to Newton’s Law of Gravity which states F=G*(m1*m2)/r^2.  So we replace F with our calculation above to get m1*((((2*pi*r)/t)^2)/r)=(G*(m1*m2)/r^2).

Now we substitute our variable above and we can solve for G, or in this case G1, the gravitational constant in our geocentric model.  Our equation is pretty straightforward:

G1=(r*(((2*pi*r)/t)^2))/m2, where m2=5.976 x 10^24 Kg.

Now we just have to repeat the calculation above, but instead of calculating the force of the sun revolving around the earth we will calculate the force of the earth revolving around the sun. Same equation, just different masses.  G2 is the gravitational constant in our heliocentric model.

G2=(r*(((2*pi*r)/t)^2))/m1, where m1= 1.989 x 10^30 Kg

Now we compare G1 and G2 to the accepted value of G above taking into account the cocktail napkin mathematics employed here will not produce an exact value, but should be within a reasonable margin of error.

G1 = 2.2208 x 10^-5 m^3 s^-2 kg^-1
G2 = 6.6726 x 10^-11 m^3 s^-2 kg^-1

Not only is the difference between G1 and G2 astronomical (pun intended) but my approximation of G is much closer than I would have even expected. This proves for the Sun to maintain an orbit around the earth the gravitational constant would have to be one million times stronger than the observed, accepted, laboratory measured, value we agreed upon at the beginning of this proof. This isn’t a trivial difference nor can one argue against the rough calculations I employ here.

Hence, I put forth that my rough calculation of G in these two models proves a heliocentric orbit beyond all rational doubt.

I asked the website to please feel free to publish my email address and my Internet nickname on their site. I would provide my name and my work address in another email from this same account so they could send the \$1000 check. The funny thing was, I never heard back from them and they quickly and quietly took the website down.

It’s also known that some people feel the Earth is completely stationary in the universe. They actually believe that the Earth doesn’t even spin. For this to be true, the Sun would have to complete a full orbit around the Earth every 24 hours. So, in our calculations above, the time value ‘t’ would be 24 hours * 60 minutes/hour * 60 seconds/minute = 86,400 seconds instead of 31,557,600 seconds. In that case, a Geocentric Gravitational Constant calculates out to

G3=(r*(((2*pi*r)/t)^2))/m2, where m2=5.976 x 10^24 Kg. and t=86,400 s.

In this case

G3=6.1543×10^+17 m^3 s^-2 kg^-1

The difference between that number and the scientifically measured and accepted value of the Gravitational Constant is Ten Billion, Billion, Billion (28 zero’s to be more precise) times greater.

Now let’s think about how silly this is. If gravity was stronger, we’d feel heavier. Fighter pilots will black out under 9Gs of force. That’s only nine times (9x) the force of gravity. In both of our geocentric models gravity would have to be millions (10^6) or ten-billion-billion-billions (10^28) times stronger. I’m glad the Earth revolves around the sun, or we’d all be paste.

Geocentricity is bunk!

End Notes:
(1) Newton’s Laws of Motion = http://www.grc.nasa.gov/WWW/K-12/airplane/newton.html
(2) Newton’s Law of Gravity – http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
(3) Solar Mass – http://scienceworld.wolfram.com/astronomy/SolarMass.html
(4) Earth Mass – http://scienceworld.wolfram.com/astronomy/EarthMass.html
(5) Earth Sun Distance – http://www.astro-tom.com/getting_started/earth-sun_distance.htm
(6) Circumference – http://www.mathgoodies.com/lessons/vol2/circumference.html
(7) velocity – http://ceres.hsc.edu/homepages/classes/astronomy/fall97/Mathematics/sec5.html
(8) acceleration of a object in a circular orbit – http://www.mcasco.com/p1cmot.html
(9) Gravitational Constant – http://en.wikipedia.org/wiki/Gravitational_constant
(10) Gravitational Constant – http://scienceworld.wolfram.com/physics/GravitationalConstant.html
(11) The Value of Pi – http://en.wikipedia.org/wiki/Pi