Do not read this if you believe in the globe

First, let me perfectly clear about a few things: There is no Outer Space;  There are no planets; stars are not burning thermonuclear giants; There are no nebulas; comets and asteroids are not chucks of rock flying through the void of space.  In fact all of modern astronomy is a farce. It’s all Science Fiction.  You must be saying I’m totally daft.  How can I possibly say or even think such absurd thoughts?  You must also be asking why in the world would the venerated space agencies present space as a reality if it is, in fact, a fiction?  What would possibly motivate them to sustain such a farce and how could they fool seemingly intelligent people in the reverent halls of modern academia.  This is impossible you say.

Another thing that needs to be clarified: Intelligence or the ability to recite information in a logical manner does not equate to the truth.  If I can describe in minute detail the people, places and events that occur in the Lord of the Rings or Harry Potter, it does not mean they exist, have ever exited or will exist.  I could take a course in “The Physics of Wands in Harry Potter”, read hundreds of esteemed books, write papers, submit thesis, get peer reviewed, and ultimately receive a doctorate, but this will tell me nothing about the real world regardless how much I believed it to be true.  Knowledge is not knowing.

The preponderance of evidence (a concept used by Bob from Globebusters) is so overwhelming that any rational and independent human being would have to take notice:

  • Total lack of curvature
  • Flight paths on a globe
  • Eclipses on a globe
  • Absence of spin
  • Gravity
  • Corpuscular rays
  • Real pictures of Earth from space
  • Bubbles in space
  • Antarctic Treaty
  • Etc, etc, etc, etc

If you are completely unwilling to examine the evidence presented and just accept the word of so-called experts, then I would suggest that you should be a possession of the managing powers of the world.  By willing handing over your will, thoughts and feelings to an outside agency, you have demonstrated your inability to manage your own person and require intervention by such powers.  Discernment is a requirement to understand the truth.  In fact, you should not continue to read this post or any other piece of evidence if that is the case.  You are not ready for the truth.  If that is the case then you must believe what you are being told and continue in your blind ignorance.  Stop reading now.


However, if you are capable of examining ideas with an open and independent mind, then please continue reading.  I would expect that any ideas put forward here are examined and properly verified using your own means.  I’m simply putting forward ideas.

Weightlessness, Free Fall and Outer Space: more proof of a Flat Earth

I wanted to revisit the whole concept of weightlessness as it pertains to so-called “outer space”.  Let’s examine the key points that define how weightlessness is supposed to arise when an object or person is in “space”.

Assumption #1: All Objects in Orbit are in Free Fall

It all starts with a hypothesis by Newton himself (or at least attributed to him) that describes how a cannon ball fired with sufficient velocity in “outer space” would continue to move as per his 1st law of motion and would be kept indefinitely in orbit due to the gravitational pull of the earth and would subsequently “free fall” around the planet.

If we can get past this untested hypothesis and allow for the suspension of disbelief, we can try to drill down into the misconceptions related to it.

First we need to examine what the cannon ball is doing while revolving around the earth.  If fired at a sufficient speed and unimpeded by any other object or force, the cannon ball will move tangentially to the surface of the earth while being “turned” around the globe.

In typical free-fall, the acceleration of gravity acts along the direction of an object’s velocity, linearly increasing its speed as it falls toward the Earth, or slowing it down if it is moving away from the Earth. In the case of an orbiting spacecraft, which has a velocity vector largely perpendicular to the force of gravity, gravitational acceleration does not produce a net change in the object’s speed, but instead acts centripetally, to constantly “turn” the spacecraft’s velocity as it moves around the Earth. Because the acceleration vector turns along with the velocity vector, they remain perpendicular to each other. Without this change in the direction of its velocity vector, the spacecraft would move in a straight line, leaving the Earth altogether


If we apply the same logic to the cannon ball, we can see that it should “turn” around the earth but not “fall” to the earth.  So here is a major misconception: the cannon ball is “turning” around the earth not “falling” around the earth.  In other words, there is no free fall of the cannon ball.  Why is this important?  Simply because floating “astronauts” are impossible if there is no free fall.

Let’s break this down a little:

The important word to focus on is “perpendicular”.  Since the velocity of the orbiting object (i.e. cannon ball, space craft, etc) is perpendicular to the force of gravity and is hypothetically kept in orbit due to a sufficient velocity along with the centripetal force of gravity, no free fall is possible.

Compare this to an object in free fall where the direction of velocity is in the direction of gravity or parallel to the force of gravity.  For something to free fall it has to actually fall.  When other explanations use elevators or thrust, they are assuming the objects in question are parallel to gravity or to the thrust of the rocket.

Assumption #2: Gravity keeps an airplane from flying off into space

The same action takes place closer to home when a commercial jetliner “turns” around the earth.  The jetliner has constant thrust which overcomes the drag coefficient of air and the “downward” force of gravity.  But don’t take my word for it, even NASA agrees:

Weight is a force that is always directed toward the center of the earth. The magnitude of the weight depends on the mass of all the airplane parts, plus the amount of fuel, plus any payload on board (people, baggage, freight, etc.). The weight is distributed throughout the airplane. But we can often think of it as collected and acting through a single point called the center of gravity. In flight, the airplane rotates about the center of gravity.

Flying encompasses two major problems; overcoming the weight of an object by some opposing force, and controlling the object in flight. Both of these problems are related to the object’s weight and the location of the center of gravity. During a flight, an airplane’s weight constantly changes as the aircraft consumes fuel. The distribution of the weight and the center of gravity also changes. So the pilot must constantly adjust the controls to keep the airplane balanced, or trimmed.

To overcome the weight force, airplanes generate an opposing force called lift. Lift is generated by the motion of the airplane through the air and is an aerodynamic force.Aero” stands for the air, and “dynamic” denotes motion. Lift is directed perpendicular to the flight direction. The magnitude of the lift depends on several factors including the shape, size, and velocity of the aircraft. As with weight, each part of the aircraft contributes to the aircraft lift force. Most of the lift is generated by the wings. Aircraft lift acts through a single point called the center of pressure. The center of pressure is defined just like the center of gravity, but using the pressure distribution around the body instead of the weight distribution.

The distribution of lift around the aircraft is important for solving the control problem. Aerodynamic surfaces are used to control the aircraft in roll, pitch, and yaw.

Weight force and gravity force are the same thing.  Therefore, if the weight has been overcome then so has gravity.  So gravity cannot be used as an excuse for keeping the airplane from flying off into space.  Again, we can consult the folks at NASA:

As the plane’s engines move the wings forward, the air has to flow both over and under them. The wings are designed so that net effect is that the air pushes them upward, countering the downward pull of gravity.

So the upward forces of lift have cancelled out the downward pull of gravity.  People don’t experience weightlessness while flying, so why would the astronauts?  We are in a bit of a conundrum.  The aeroplane has overcome gravity yet the passengers aren’t floating around like astronauts nor has the plane flew of into “space”.  How is this possible?

Assumption #3: Cancelling of forces does not mean weightlessness

The force of gravity at 400 km high is about the same as on the surface of the earth, therefore, there is no major difference between an jetliner “turning” around the earth and an object “turning” around in “outer space”.  Let’s assume a 75kg object or person:

La te xi t 1 5

And in orbit:

La te xi t 1 6

The mass of an average jetliner is about 351,534 kg while the mass of the ISS is about 419,600 kg.  To calculate the centrifugal force you need to take the linear speed squared and divide by the radius.  In both case, the radius is equal to radius of the earth + the cruising altitude.  Since the airliner is going at a slower speed (~ 500 mph) it will require a constant thrust to stay aloft since the the buoyancy force of the atmosphere is constantly pushing the airplane downward (the buoyancy force operates in both directions depending on the object and the fluid it is contained in – but this will be discussed later in this post).

Some will argue that the centrifugal and centripetal forces must balance out.  I completely agree with that statement.  Earth’s gravity is a constant force (I keep hearing).  It can’t pull any harder than 9.8 m/s².  It’s kind of like having a weak string attached to a sphere you are swinging around your head.  If the string breaks, the sphere will fly off.   But until that point the sphere is fixed in place (i.e. orbit).  If you could place a marble inside the sphere and begin to swing the sphere around, what would happen to the marble?  It all depends upon the orbital velocity.  If the sphere orbits at the exact velocity to ensure the string doesn’t break but it also doesn’t allow it to go slack, the marble would not float in the middle of the sphere, it would push against the outer edge of the sphere.  In fact, any amount of angular velocity would force the marble to the outside edge.  In the same way, the astronauts should remain fixed to the ceiling (outer edge) of the ISS.

Let’s look at an example:

For a satellite to stay in orbit, this centrifugal force must balance gravity. Thus,


Simplifying, we find that


and hence that


Thus for a satellite 1000 km above the Earth (and hence 7,400 km from the center of the Earth) traveling in a circular orbit,

(the mass of the Earth is 5.967´ 1024 kg).

No complaints about the logic or the math with that example.  If we take the exact same equation and apply it to the ISS, here’s what we get:

Supposedly, the ISS is traveling at a velocity of 27,600 km/h (7.6 km/s).  The centrifugal force and acceleration acting on the ISS at that velocity at an altitude of 400 km is:

Fc = m ac

    = m v2 / r

Mass of ISS * Velocity of ISS ² / (Radius of Earth + Altitude)

Centrifugal Force = 419,000 kg * 7657 m/s ² / 6,771,000 m = 3472 kN

Centrifugal acceleration = 7657 m/s ² / 6,771,000 m = 8.6 m/s²

We can see that the centrifugal and centripetal forces are balanced but the effect would not be a free fall but cancelling of forces – just like a person who is flying in an airplane.  Do we float around? No.  Since there is no change in distance in the direction of gravity, then a = 0 and if that is the case then F = 0.   The balancing of force does not mean weightlessness.

During orbit, the astronauts should be standing on the ceiling of the ISS and have a weight that is 89% of their weight on the surface of the Earth (being forced to the outer edge by angular velocity).  However, if the ISS could maintain its altitude without orbiting, then the astronauts would have to be standing on the floor of the ISS.  THEY WHOULD NOT BE FLOATING.  As I mentioned earlier, the concept of floating is incorrect since the proposed reason for floating is because they should be falling.  Anyone who has been on a plane during bad turbulence will attest that during the sudden loss of altitude, objects will crash into the ceiling of the plane.  This is not because they are weightless but because the ground beneath their feet has suddenly dropped away.  The “vomit comet” plane shows people falling because the plane itself is falling (it is dropping parallel to the so-called force of gravity) not because the people have suddenly become weightless.

In contrast to this, the ISS is not falling since it is NOT losing altitude.  In fact, the only way for the astronauts to be floating (the way we are shown in countless videos) is if the ISS was in a constant free fall (which it is not nor could it be).  For the ISS to maintain the altitude it does while maintaining it’s supposed path around the Earth, it must maintain a constant speed.  Any variation from this would result in either the ISS crashing to Earth or flying off into “space”.  Hypothetically, the ISS would need to be creating an artificial Lagrange Point.

The Lagrange points mark positions where the combined gravitational pull of the two large masses provides precisely the centripetal force required to orbit at the same angular velocity (essentially, the speed of the orbit) and thus remain in the same relative position (or altitude).

This artificial Lagrange Point is reverse engineered from a mathematical model that requires the ISS to orbit at an absurd velocity.  The speed is required to make the math work as opposed to the math describing reality.

The Angular Velocity Problem

Since the ISS is not in a free fall but rather orbiting at a constant altitude, the astronauts should experience the effects of angular velocity.   Hypothetically, the force of gravity is always perpendicular to the floor of the ISS which means the centripetal force is always acting upon the astronauts in the opposite direction – meaning towards the ceiling of he ISS.  In this hypothetical situation, the forces are balanced but what does this mean in reality versus a mathematical model?

If we compare this to a car travelling with a constant velocity in a straight line along a road, no effects of that motion would be experienced by the driver.  If, however, the driver turns a corner, the effects will be felt as a pull in the opposite direction of the turn.  Assuming a constant turn rate and velocity, an object sitting on the dashboard will only begin to move once the forces of friction and inertia are overcome by the angular velocity forces – the object breaks free of the forces of friction and inertia and moves away from it’s static location or in the case of the ISS, it flies of into “Space”.  In this example, the forces of friction and inertia are equivalent to “gravity” and the angular velocity is the motion of the ISS.  Too much velocity and ISS will fly off into “Space”; too little and it will fall to the ground.  But until then, the object stays put on the dashboard surface and the ISS stays in “orbit” even if the opposing forces are equal.

In reality, the ISS would function much like a fast moving aeroplane.   In the hypothetical environment of “Outer Space”, there is no atmosphere to slow it down but there also isn’t an atmosphere to create lift.  So how does it stay aloft?  From the artificial Lagrange point I wrote about earlier.  If an aeroplane was able to travel at the same velocity and constant altitude (it doesn’t matter what altitude since we have artificially removed the atmosphere) as the ISS, it would not cause the passengers to become weightless.  It would allow the plane to stay aloft without additional thrust just like the ISS is supposedly doing – Another artificial Lagrange point.  The only difference would be a slight change in speed due to the increased “gravity” at a lower altitude.

The G-Force problem

There is a serious problem with the description of g-force as written in wikipedia:

The gravitational force, or more commonly, g-force, is a measurement of the type of acceleration that causes a perception of weight.

In the case of an object on the earth’s surface, weight is produced by the supposed force of gravity acting towards the centre and the opposing force of the earth itself pushing back on the object. (how the earth can pull an object towards the centre and simultaneously push back with the same amount of force is equally baffling)

When the g-force acceleration is produced by the surface of one object being pushed by the surface of another object, the reaction force to this push produces an equal and opposite weight for every unit of an object’s mass. The types of forces involved are transmitted through objects by interior mechanical stresses. The g-force acceleration (except certain electromagnetic force influences) is the cause of an object’s acceleration in relation to free fall.[1][2]

However, the g-force is not apparent on the surface of the earth:

Gravitation acting alone does not produce a g-force, even though g-forces are expressed in multiples of the acceleration of a standard gravity. Thus, the standard gravitational acceleration at the Earth’s surface produces g-force only indirectly, as a result of resistance to it by mechanical forces. These mechanical forces actually produce the g-force acceleration on a mass. For example, the 1 g force on an object sitting on the Earth’s surface is caused by mechanical force exerted in the upward direction by the ground, keeping the object from going into free fall. The upward contact force from the ground ensures that an object at rest on the Earth’s surface is accelerating relative to the free-fall condition. (Free fall is the path that the object would follow when falling freely toward the Earth’s center). Stress inside the object is ensured from the fact that the ground contact forces are transmitted only from the point of contact with the ground.

How can an object at rest still be accelerating?  This paragraph is full of contradictory concepts.  For the ground to be pushing up, there must be a continuous force pulling down.  You can’t be accelerating if you are at rest.  And this has NOTHING to do with the supposed spin of the earth.

Some would argue that if they throw an object up in the air, the point where the object comes at rest in mid-air (i.e. maximum height before falling back down), the acceleration due to gravity is acting upon the object even though it is at rest.  This example is making liberal use of the word “rest”.  If an object was thrown with a slight arc (even a tiny one), the object would never be “at rest” since the object would follow an arc pattern and be in continuous motion.  That the arc of an object throw directly upward is zero is simply a mathematical description of one possible scenario where the object transitions from upward to downward in a single moment.  For any rational person, an object thrown into the air is always in motion until it is actually at rest on the surface of the earth.  To say it is “at rest” while in motion is sophistry exemplified.

I have written about this absurdity in a previous post so I won’t go too much detail.  Suffice to say that if a person is to stand on a scale, the acceleration due to gravity MUST continue to act until the object is at rest.  Meaning the spring should continue to compress regardless of the mass of the object.  The only object, according to the current gravitational hypothesis that can be “at rest” is the atom/particle at the very center of the earth (gravity is zero).  It is this magical object that is able to pull everything into the center while pushing back with an equal amount of force.  All objects on the earth’s surface should be in a continual downward pull towards the center.  What stops the downward compression at the very center of the earth?  The explanation is that the gravitational forces cancel each other out.   This doesn’t even work mathematically.  If we take the formula to calculate the volume of a sphere (V = (4/3)πr) and simply subtract the total volume of the earth from the volume of the earth at half the radius, we can see that the total volume on the periphery is many times greater than the center:

V = (4/3)π(3959 mi) = 2.6×1011 = Volume of earth

V = (4/3)π(1975.9 mi) = 3.23×1010 = Volume of earth at half distance

Volume of earth at periphery = 2.27×1011

There is an order of magnitude more volume at the periphery than the center.  In other words, the center would be pulled towards the periphery due to the greater mass.  This would make planetary formation through gravitational attraction impossible.  For as more mass accumulated on the periphery, the less attractive the force in the center.

The Atmospheric Pressure Problem

The amount of air above a persons head pushes down with a force of approximately 7 pounds / square inch.  The average human head is about 21 inches in diameter or π x r2 = 3.14 x 10.52 = 346 square inches. At 7 pounds per inch, there should be 2,424 lbs / square inch of additional weight on the head of each human being.

If I stand on a scale and weigh 72kg and then pick up 2 weights of 10kg each, the scale would show an increase of 20 kg to 92kg.  For the same reasons the weight of every human being on earth should be their weight + the weight of the atmosphere above their head.  This is what the theory of gravity must predict to be true.   However, no one experiences an additional 2,424 lbs of weight on their head.  This is another falsification of gravity.

We also need to differentiate between air pressure and air mass (or weight).  Pressure is caused by air being forced into an enclosed space which causes more air molecules of air to be present in the same space.  Air weight is the downward force caused by gravity.  What causes air pressure to be perpendicular to the force of gravity?  According to the theory of gravity, once an object is in contact with the surface (in this case the human head) it will experience resistance from the skull which pushes back with an equal amount of force thereby stopping the downward motion.  The earth is not a pressurized container since the only thing holding the atmosphere in is gravity and gravity is in one direction only – downward.  Therefore, there is no known force perpendicular to the force of gravity which can explain air pressure.

Since air cannot be lighter or denser than itself, density and buoyancy cannot be be used to explain this problem away.

The Net-Force Argument

That the net force between an object a rest on the surface of the earth is zero, does not mean there is no force between them.  It’s mathematical slight of hand to make it seem like no force is present.  If that was the case then all objects would weigh zero.

The key word is NET.  As I wrote earlier, if I pickup an additional 20kg of weight the NET force between myself and the ground is still zero even though the scale has registered an increase in weight.

If I place my hand on the ground and place 2000kg on it, my hand would be crushed even though the NET force is equal to zero.

Another way to think about this is to examine how a balancing scale works.  If I place a 1kg weight on both scales the forces are balanced – or equal to zero.  However, there is a force of 1kg still pushing down on each scale.  That they are balanced does not negate the forces in question here which is implied with the word ZERO.  We can test this by removing one of the weights.  The scale will become imbalanced and tip in the direction of the remaining weight.

Buoyancy – The separating force

Many of the globe folks comment that if buoyancy is responsible for the downward force on objects, what is it that makes them “fall” downward rather than “upward”.  The answer is that objects do “fall” upward – Helium balloons for example.

Objects of different densities separate at a constant rate plus or minus the addition of external forces.  What we experience as “down” is really the separation of lighter material from more dense material.  Objects go in opposite directions which creates “up” and “down”.

So why is “up” above our heads and “down” everything below our feet? Because of the density of the average human is greater then air.  We, as physical humans, will tend towards objects of equal density – the ground beneath our feet vs. atmosphere above our heads.  If we became lighter then the density of the atmosphere, “down” would become what we experience as “up”.

Just as “down” for a helium balloon is “up” for humans.  Similarly, if a beach ball filled with air is forced under water to any depth, it will rise at a specific rate.  But it is “falling” “down” towards a location of equal density or equilibrium.

Just as it takes energy to submerge the beach ball, it also takes energy to place a human above the surface of the ground.  The atmosphere is constantly pushing the human towards the ground and simultaneously pushes a helium balloon towards the sky.

That’s it for now.



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