DYNAMISM by GEORGE MARTIN WILLIAMS - HTML preview

VOLUME

DENSITY = 7.372497349961709 x 10^-51

1.410784666635855 x 10^-71

DENSITY = 522581335023166798298 kg/m^3

THIS IS DENSER THAN A NEUTRON STAR.

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Eight extremes: The densest thing in the universe

At the modest temperatures and pressures of Earth’s surface, the densest known material is the metallic element osmium, which packs 22 grams into 1 cubic centimetre, or more than 100 grams into a teaspoonful. Even osmium is full of fluff, however, in the form of electron clouds that separate the dense atomic nuclei. Although rarefied, these clouds are robust, and even the immense pressures deep within the planet can only compress solid matter to a modest degree.

Far greater pressure is found within the collapsed core of a giant star, a remnant we know as a neutron star. There, matter is in some exotic and ultra-dense form – most probably neutrons, and possibly a few protons and electrons, packed cheek-by-jowl. One cubic metre of “neutronium” matter from the centre of a neutron star could have a mass of up to 10^18 kilograms, or a million billion tonnes.

https://www.newscientist.com/article/mg20928026-900-eight-extremes-the-densest-thing-in-the-universe/

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THE NEUTRON STAR'S CORE IS ESTIMATED TO HAVE A DENSITY OF 1 x 10^18 kg/m^3.

ESTIMATED NEUTRINO DENSITY = 5.225,813,350,231,667,982,98 kg/m^3

MY ESTIMATE FOR THE NEUTRINO'S DENSITY IS 5.225 x 10^20 kg/m^3.

WHICH IS FIVE HUNDRED TIMES DENSER THAN THE NEUTRON STAR'S CORE.

THE NEUTRINO IS THE DENSEST OBJECT IN THE UNIVERSE. THUS IT STANDS TO REASON THAT ALTHOUGH

ITS MASS IS ALMOST NEGLIGIBLE, THE INFINITESIMAL SIZE OF ITS DIAMETER GIVES IT AN EXTREMELY

SMALL VOLUME AND THEREFORE EXTREMELY HIGH DENSITY. THIS EXTREMELY HIGH DENSITY MEANS

THAT GRAVITY PLAYS AN IMPORTANT PART IN KEEPING THE ELECTRONS, POSITRONS, AND QUARKS

TOGETHER, DESPITE THE FACT THAT GRAVITY IS THE WEAKEST OF THE FOUR FUNDAMENTAL FORCES.

THE PRESUMED HIGHEST POSSIBLE FREQUENCY FOR ANY ELECTROMAGNETIC WAVE IS LESS THAN 1 x 10^30 Hz.

I PRESUME THIS TO BE RELATED TO THE DIAMETER OF THE NEUTRINO AS WELL AS THE HIGHEST

POSSIBLE FREQUENCY FOR ANY OBJECT WITHIN THE PHYSICAL DIMENSION. THIS MEANS THAT ANY

OBJECT THAT VIBRATES AT A HIGHER FREQUENCY THAN THIS WOULD AUTOMATICALLY BE IN A HIGHER

DIMENSION THAN THE PHYSICAL.

OF COURSE, THIS IS A ROUGH ESTIMATE. ONLY THROUGH OBSERVATION AND EXPERIMENTATION CAN THE

EXACT FREQUENCY BE DETERMINED.

TO CALCULATE THE FORCE OF GRAVITY BETWEEN TWO NEUTRINOS THAT ARE IN CONTACT WITH EACH

OTHER; WE NEED:

1. GRAVITATIONAL CONSTANT = 6.67430 × 10^-11 N·(m/kg)^2

2. N1 = N2 = NEUTRINO MASS = 7.372497349961709 x 10^-51 kg 3. N1 x N2 = NEUTRINO MASS SQUARED = 5.4353717175192421907943766200681 x 10^-101 kg 4. R = DIAMETER OF NEUTRINO = (2.99792458 x 10^-24) m

5. R^2 = DIAMETER OF NEUTRINO SQUARED = (8.9875517873681764 x 10^-48) m THE ACTUAL FORCE OF ATTRACTION BETWEEN ANY TWO GIVEN NEUTRINOS WILL VARY ACCORDING TO

THE INVERSE SQUARE LAW.

F = G X N1 X N2

R^2

F = FORCE G = GRAVITATIONAL CONSTANT N1 = MASS OF NEUTRINO ONE N2 = MASS OF

NEUTRINO TWO R = DISTANCE

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F = (6.67430 × 10^-11) * (5.4353717175192421907943766200681 x 10^-101) / (8.9875517873681764 x 10^-48) F = (4.036394149653267871291822628385 x 10^-64)

THAT ANSWER OF 'F = (4.036394149653267871291822628385 x 10^-64)' DOESN'T SEEM LIKE MUCH. BUT TO

PUT IT IN ITS PROPER PERSPECTIVE, WE HAVE TO USE THE DENSITY FIGURE FOR THE NEUTRINO, WHICH

IS: NEUTRINO'S DENSITY = 5.22581335023166798298 x 10^20 kg/m^3.

TO DO THIS WE HAVE TO EQUATE THE VOLUME TO ONE CUBIC METER.

Volume of a sphere = 4/3 π r^3

VOLUME = 4 π r^3

3

1 = 4 π r^3

3

3 = 4 π r^3

r^3 = 3

4 π

r^3 = 0.2387324146378272916

r = 0.62035049089940001666800681204777816735078620018600

d = 1.24070098179880003333601362409555633470157240037200

r = radius d = diameter π = 3.14159265358979323846264338327950288419716939937510

WHEN THE NEUTRINO IS SUPERSIZED SO THAT IT HAS A VOLUME OF EXACTLY ONE CUBIC METER, THEN

ITS RADIUS IS 'R = 0.62035049' AND ITS DIAMETER IS 'D = 1.240700981798'. THE DIAMETER FIGURE IS

EXACTLY THE DISTANCE BETWEEN THE CENTER OF MASS OF TWO NEUTRINOS THAT ARE IN CONTACT

WITH EACH OTHER.

TO CALCULATE THE FORCE OF GRAVITY BETWEEN TWO NEUTRINOS THAT BOTH HAVE A VOLUME OF ONE

CUBIC METER AND ARE IN CONTACT WITH EACH OTHER; WE NEED:

1. GRAVITATIONAL CONSTANT = 6.67430 × 10^-11 N·(m/kg)^2

2. N1 = N2 = NEUTRINO MASS = 522581335023166798298 kg

3. N1 x N2 = NEUTRINO MASS SQUARED = 273091251714595297765025836203380215696804 kg 4. R = DIAMETER OF NEUTRINO = 1.24070098179880003333601362409555633470157240037 m 5. R^2 = DIAMETER OF NEUTRINO SQUARED =

1.5393389262365063316037311053470604415548755849719 m

THE ACTUAL FORCE OF ATTRACTION BETWEEN ANY TWO GIVEN NEUTRINOS WILL VARY ACCORDING TO

THE INVERSE SQUARE LAW.

F = G X N1 X N2

R^2

F = FORCE G = GRAVITATIONAL CONSTANT N1 = MASS OF NEUTRINO ONE N2 = MASS OF

NEUTRINO TWO R = DISTANCE

F = (6.67430 × 10^-11) * (2.730912517 × 10^41) / (1.5393389262365063316037) F = (1.1840751313779757754321226889035 × 10^31)

TO PUT THIS FIGURE INTO ITS PROPER PERSPECTIVE, IF THESE TWO NEUTRINOS IN CONTACT WITH EACH

OTHER, AND EACH WITH A VOLUME OF ONE CUBIC METER, WERE IN THE VACUUM OF SPACE WITH

NOTHING AROUND FOR MILLIONS OF KILOMETERS, THEN:

'F = 1.18 × 10^31'

WOULD BE THE FORCE OF GRAVITATIONAL ATTRACTION THAT THEY WOULD HAVE FOR ONE ANOTHER.

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How much does Earth weigh?

By Katherine Irving published March 31, 2024

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Earth's mass took hundreds of years to estimate, and even now, experts don't agree on the exact number.

Our planet holds everything from hard rocks and minerals to millions of species of living things, and is covered in countless natural and human-made structures.

So how much does all of that weigh? There's no single answer to that question. Just like humans weigh much less on the moon than we do at home, Earth doesn't have just one weight. Earth's weight depends on the gravitational force pulling on it, which means it could weigh trillions of pounds or nothing at all.

What scientists have spent centuries determining, however, is Earth's mass, which is its resistance to movement against an applied force. According to NASA, Earth's mass is 5.9722×10^24 kilograms, or around 13.1 septillion pounds. This equates to around 13 quadrillion of Egypt's pyramid of Khafre, which itself weighs around 10 billion pounds (4.8 billion kilograms). The Earth's mass fluctuates slightly due to the addition of space dust and gases leaking out of our atmosphere, but these tiny changes won't affect Earth for billions of years.

https://www.livescience.com/planet-earth/how-much-does-earth-weigh ___________________________________________________________________________________________________

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The Sun

The Sun, at the center of our Solar System, is at the beginning of this scale model of the Solar System. In this model, the Sun is represented as a ball 4 inches in diameter. This makes the scale of our model 1 inch = 180,000

miles. Each step that you take (28 inches) is then 5.0 million miles.

Our Sun is a huge, massive, spherically shaped object, containing about 99.8% of all the matter in our Solar System.

(The planet Jupiter contains most of the remaining material.) The sun has a mass of 1.9891x10^30 kg =

4.384x10^30 lb = 2.192x10^27 tons, or a mass 333,000 times that of the Earth. The radius of the Sun is 696,265,000 meters = 696,265 km = 432,639 mi or a radius 109 times that of the Earth. The volume of the Sun is so huge that it could hold over 1 million Earths.

https://www.mccc.edu/~dornemam/Planet_Walk/Sun/the_sun.htm ___________________________________________________________________________________________________

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Mathematical relation between kg And newton

The relation between kg and newton can be mathematically expressed using Newton’s second law of motion as follows-1 N = kg x m/s^2

Where,

• N is the force in newton.

• kg is the mass in kilograms.

• m is the distance travelled in metres.

• s is the time duration in seconds.

kg and newton

In physics, kg is directly proportional to Newton. Which means that-

• When the mass of the object in kg is high, the force required to move it in N is also high.

• When the mass of the object in kg is low, the force required to move it in N is also low.

Refer the table below for kg to newton and newton to kg values.

Values

Kg to newton

1 kg = 9.81 N

newton to kg

1N = 0.10197 kg

/t https://byjus.com/physics/relation-between-kg-and-newton/

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THE ARTICLE ABOVE BASICALLY SAYS THAT AT THE EARTH'S SURFACE, ONE KG IS APPROXIMATELY EQUAL

TO 10 NEWTONS.

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SO THE FORCE OF GRAVITATIONAL ATTRACTION BETWEEN THE TWO NEUTRINOS IS:

'F = 1.18 × 10^31 NEWTONS'

DIVIDING THIS NUMBER BY TEN GIVES US A FIGURE IN KILOGRAMS OF:

'1.18 X 10^30 KG'

THIS IS A SIMPLIFICATION TO SHOW THE GRAVITATIONAL ATTRACTION BETWEEN THESE TWO MASSES IN

RESPECT TO HOW MUCH MASS THAT WOULD REPRESENT IF IT WAS JUST A SINGLE MASSIVE OBJECT

RESTING ON THE EARTH'S SURFACE.

THE MASS OF THE EARTH IS ESTIMATED TO BE:

EARTH'S MASS = '5.9722×10^24 KG'

(1.18 x 10^30)/(5.9722 × 10^24) = 197582

THE MASS OF THE SUN IS ESTIMATED TO BE:

SUN'S MASS = '1.9891x10^30 kg'

(1.18 x 10^30)/(1.9891 x 10^30) = 0.5932331205067618520939

THE GRAVITATIONAL ATTRACTION BETWEEN THE TWO NEUTRINOS REPRESENTS A MASS ON THE SURFACE

OF THE EARTH WHICH IS ALMOST TWO HUNDRED THOUSAND TIMES HEAVIER THAN THE EARTH AND

WHICH IS ALMOST SIXTY PERCENT OF THE MASS OF THE SUN.

'The radius of the Sun is 696,265,000 meters = 696,265 km'

r = 696265000

SUN'S VOLUME = (4 π r^3) / 3

SUN'S VOLUME = (4 π (696265000)^3) / 3 m^3

SUN'S VOLUME = (4 pi (696265000)^3) / 3 = 1.413879191397011864322683272 x 10^27 m^3

DENSITY = MASS