Tag: gravity

How the climb rate of an airplane debunks the globe

I have watched and read FlatEarth (“FE”) content that talks about the issue with flight paths, times and other anomalies that are difficult or impossible on a GlobeEarth (“GE”).  However, many GE folks tend to write-off these anomalies by evoking gravity.  Though the gravity excuse can be rebuked, it gives the GE folks a free pass.  This pass is about to be taken away.

My previous work on curvature calculations has given me the idea to track the flight path of a 747-200 class aircraft during the ascent phase of the journey.  On a recent trip to France I recorded the angle of ascent using a bubble level and screen capture software.  This allowed me to compare (in real-time) the ascent angle over time with the flight data from Flight Aware.  The majority of the ascent was between 1-3°.  It lost about 1° of angle every minute.  During the initial take off the angle was much higher at around 12° but it quickly reduced to 2° in about 12 minutes. The final 13 minutes was at a nominal angle of 1°.  A quick calculation shows that TAN(2°) x 185 miles (6.46 miles) is close to the 6.62 miles achieved after 25 minutes.  In other words, the bubble level and the recorded values from flight aware seem to correlate.

If you have an account at flight aware, you can take a look at the actual flight taken.  I will add the data tables here for those who don’t wish to create an account.

The purpose of this post is to show that if there is curvature to the earth, the climb rate over time and the curvature of the earth must place any average international flight at more than double the recorded cruising altitude. This is because the amount of curvature during the ascent phase of the example flight below would be approximately 4.37 miles even without adding any climb at all.

All of the data is publicly available and anyone can do the same calculations I’m going to do.  In fact, any international flight can be used for this experiment.

If you take a look at this graph:

Screen Shot 2017-09-01 at 11.12.22 AM

you can see the ascent is in the first 25 minutes and 25 seconds of the flight (01:59:02 – 02:24:27).  You can compare this to the table data below:

Screen Shot 2017-08-27 at 10.49.42 PM

Screen Shot 2017-08-27 at 10.50.03 PM

Screen Shot 2017-08-27 at 10.50.22 PM

An examination of this data eliminates a potential objection by GE folks that plane is flying with the curvature of the earth since the plane cannot be tracking with the curvature and ascending at the same time.  By it’s very definition, the plane is moving away from the ground (ascending), not tracking with it as it should during cruising altitude.  This might seem obvious to most logically minded folks but we are dealing with hardened attitudes that will use any argument, no matter how illogical, to avoid the conclusions reached here.

Since the plane is ascending, and if we are living on a ball, we must take into account the ascending and curvature rates at the same time.  There is no getting away from this.  This also destroys the idea that gravity keeps the plane moving along the curve, since if this was the case, a plane could never ascend at all.  The ascent angle over time is the key.

For example, if a plane took off and ascended at a nominal climbing rate of 1 meter / hour at an average speed of 450 miles / hour, it would reach a cruising altitude of 4.7 miles in 26 minutes.  If it continued at this rate for 52 minutes (or an additional 26 minutes), it would be at a cruising altitude of 19.8 miles.  The plane can ascend because it’s under it’s own power.  It has already overcome the force of gravity or it would fall to the ground.  But I digress….

Using a real example, if we include curvature into the calculation with the recorded climb rate, the plane should reach a cruising altitude of 35,000 feet (or 6.62 miles) in approximately 13 minutes.  This was calculated using 51 specific data points.  Each point contains time, latitude, longitude, course, direction, kts (knots), km/h, meters, climb rate and reporting facility.  By calculating the speed by the time interval, I can calculate the total distance travelled during each time interval.

The first 12 data points are ignored for distance travelled since the plane took off in a westerly direction and then turned around to a northeast direction.  We still need to include the altitude reach during this time frame which is approximately 1.26 miles.  This altitude was reached at 2:02:17 pm.

To accurately calculate the distance travelled, I had to measure the time difference between each data point (between 15-60 seconds) and multiply by the speed of the plane at that interval.  This eliminated the use of an average speed which would most likely be used as an additional rebuttal against this proposal.

If you examine the table below, you can see each time interval along with the height and curvature reached.

flightawarev2

The 7 columns of the far right show the curvature that would occur over the distance travelled.  The distance travelled was calculated by converting the speed to nautical miles/min (i.e 281 kts / 60 min = 4.68 nm/min).

I then divided the time frame for each data point by 60 seconds to get the fraction of a minute that the plane travelled during that time frame (ie. 25 seconds / 60 seconds = .41 min)

Once I had that value, I was able to multiply that by the plane’s current speed to get the distance travelled (i.e. 4.68 nm/min × .41 min = 1.95 nm).  I then converted the value from nautical miles to miles (i.e. 1.95 nm × 1.15 = 2.24 miles).

Once I had the distance at each interval I could calculate the amount of curvature in feet.  Whew !!!

Since curvature is accumulative, the total arc distance from the starting point to the specific time frame must be included not just the distance travelled in that particular time frame.

I finally converted the last 21 data points into miles of curvature.  Since the plane was already 1.26 miles in altitude when it started heading northeast, I had to add that value to the curvature accumulated from 02:02:17 pm onward.  I calculated both the altitude of the plane without curvature and one with curvature.

As you can see from the table, the cruising altitude should have been reached at approximately the 13 minute mark.  Since the plane does not stop ascending at the 13 minute mark but continues for an additional 12 minutes, we are left with the inescapable conclusion that there is no curvature.

I’ve included a link to a post that calculates the distance covered during take off and the time taken to reach cruising altitude for comparison purposes.

https://aviation.stackexchange.com/questions/14357/how-long-after-takeoff-for-a-boeing-747-400-to-reach-cruise-speed

 

 

 

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Hooke’s Law and the Missing Energy of Gravity.

Recently I’ve been going back and forth with various folks in the twitter realm about the fundamentals of physics.  The biggest stumbling block is the issue of forces and how to define them.  I stated in a previous post that you need 3 specific attributes to calculate any force equation:

  1. Magnitude
  2. Vector (or Direction)
  3. Time

I then present a simple example of a 4000kg car accelerating at 10 m/s² for 10s.  The Impulse of Force of the car at each second is:

Mass x acceleration x time¹ = Impulse of Force¹ (¹ is all reference frames from 1 – 10 seconds).  Multiplying Force · Time is known as impulse and is a valid way of expressing a force.

  • 4000kg · 10m/s² · 1s = 40,000 kg·m/s (velocity = 10 m/s) = 200,000 Joules
  • 4000kg · 10m/s² · 2s = 80,000 kg·m/s (velocity = 20 m/s) = 400,000 Joules
  • 4000kg · 10m/s² · 3s = 120,000 kg·m/s (velocity = 30 m/s) = 600,000 Joules
  • 4000kg · 10m/s² · 4s = 160,000 kg·m/s (velocity = 40 m/s) = 800,000 Joules
  • 4000kg · 10m/s² · 5s = 200,000 kg·m/s (velocity = 50 m/s) = 1,000,000 Joules
  • etc…
  • 4000kg · 10m/s² · 10s = 400,000 kg·m/s (velocity = 100 m/s) = 2,000,000 Joules

However, where I caught the consternation of others in the twitter realm was due to confounding of terms that are very specific in the physics world.  We all experience  the affects of forces all day, everyday in our lives and we generally don’t break them down into units of measurement, types of units, work vs. force, etc.  So my way of describing what I see as a fatal flaw in gravity was met with irate rantings.

For one, I was pointing out a potential problem with something they hold in an almost sacred sense (gravity).  And secondly, I was not using terminology correctly for their taste.  I don’t necessarily blame them for wanting exact language, but I would argue that they most likely understood the point I was trying to make but did not want point me in the right direction.  But, for the most part however, they were decent.  In a way, by them resisting what I was presenting forced me to fix and clarify my position.

Force, rate of change, work and energy

What I saw was that over time the acceleration due to gravity should equate to an increase in force on the mass involved.  However, the concept of force is important since in the case of gravity it is, indeed, constant.  I wasn’t arguing that gravity is increasing but the net forces are increasing.  I kept arguing that time must be included in the equation or it doesn’t represent reality (though it was valid see impulse of the force).  But again, my use of the words net force with time was incorrect with respect to physics as we are taught today and net forces has a specific meaning.  It wasn’t until I recalled kinetic and potential energy (work) and how it relates back to force that I was able to present my case in a language that would be acceptable.

From the car example above, if F = ma, then the rate of change is 40,000 N which means the force is constant.  But this is counter-intuitive to most people since they know the car is traveling faster with each second.  However, for acceleration to exist at all,  energy must be constantly applied.  If we accept the conservation of energy law, then that energy must be going somewhere.  In this case it is being expressed as kinetic energy.

In the case of gravity acting on an object or a person on the earth’s surface, gravity is that constant force or energy that is supplying the acceleration.  Either the object or person needs to move or the energy needs to be converted into something else like heat or sound or whatever.  It is the amount of energy that is increasing not the force itself.  We can conclude then, that time multiplies how much energy is in a system with respect to the force being applied.

We can all bear witness to the fact that we are not heating up or emitting noise or expressing some other form of energy release under the stress if gravity.  The counter argument is that since there is no movement there can be no acceleration.  But this is a fallacy since the height of the person is equal to the “x” value in Hooke’s law and the motion is expressed as the compression of the human frame.

There is also a distinction between kinetic energy and potential energy.  A car driving at 100 km/h has a specific amount of kinetic energy but a spring under increasing compression has potential energy that is being stored in the spring itself.  The spring has a maximum compression it can reach before it will begin to become crushed under an increasing load.

Hooke’s law is only a first-order linear approximation to the real response of springs and other elastic bodies to applied forces. It must eventually fail once the forces exceed some limit, since no material can be compressed beyond a certain minimum size, or stretched beyond a maximum size, without some permanent deformation or change of state. Many materials will noticeably deviate from Hooke’s law well before those elastic limits are reached.

Wikipedia

Even though the spring no longer has motion the amount energy will increase as the load capacity is transferred to less elastic structures like the metal of the spring or the surface the spring is sitting on.

In other words, the increase of potential energy will propagate to the surrounding environment and begin to compress the weakest structures first.  In the case of a human being standing on the surface of the earth, the acceleration due to gravity would continue to propagate through the human frame as per Hooke’s law.  I have had many discussions with folks who insist that the forces between the earth and the human being cancel each other out.  But they are confusing force with work.

Any equal and opposite reaction is within the frame of the human being (like a spring) and the ground they are standing on.  The opposing forces balance out, but this would not stop the acceleration; the balanced forces would only resist the downward pull of gravity and subsequently increase the potential energy. Since gravity is supposed to be an acceleration and is supposed to be pulling at a force proportional to the mass of the earth, the human being wouldn’t stand a chance.  It is precisely because of the equal and opposite reaction and the balancing of forces that potential energy is possible.  So they are confusing the forces involved with the energy (work) being applied to the mass.

Potential Energy

assuming conservation of energy: if F=ma and a = 9.8 m/s² and m > 0 then ΔPE > 0. where’s all that energy going? We should all be crushed by now.

Spring energy

The potential energy Uel(x) stored in a spring is given by

{\displaystyle U_{\mathrm {el} }(x)={\tfrac {1}{2}}kx^{2}}

which comes from adding up the energy it takes to incrementally compress the spring. That is, the integral of force over displacement. Since the external force has the same general direction as the displacement, the potential energy of a spring is always non-negative.

This potential Uel can be visualized as a parabola on the Ux-plane such that Uel(x) = 1/2kx2. As the spring is stretched in the positive x-direction, the potential energy increases parabolically (the same thing happens as the spring is compressed). Since the change in potential energy changes at a constant rate:

{\displaystyle {\frac {d^{2}U_{\mathrm {el} }}{dx^{2}}}=k\,.}

Note that the change in the change in U is constant even when the displacement and acceleration are zero.

What does this mean?  It means, a constant and catastrophic amount of potential energy should be building up in every object on the earth’s surface.  As mentioned above, this does not violate F=ma since we are not talking about an increase in the force of gravity but an increase in potential energy due to that constant force.

If we calculate the “elastic” capacity of the human frame and equate it to “k” and multiplied that by “x” which would be the height of the person (since they should be getting compressed by gravity) and then multiply by 1/2 we should get the potential energy stored or expressed within the human frame.  Also, since the value for “k” of the human being is less than “k” for ground, the human being would be crushed into the ground.  However, as we all know, we aren’t springs, and won’t bounce back from such an event.  Most of the energy will be released in the form of noise and heat.  Not a pretty picture.

The modern theory of elasticity generalizes Hooke’s law to say that the strain (deformation) of an elastic object or material is proportional to the stress applied to it. However, since general stresses and strains may have multiple independent components, the “proportionality factor” may no longer be just a single real number, but rather a linear map (a tensor) that can be represented by a matrix of real numbers.

Wikipedia

So we can all say with a high level of certainty that we are not being crushed by gravity, therefore, gravity must be falsified.  We exist in a non-gravitational realm.

Falsification of the Universal Law of Gravity

Note: Just as my previous post, I had to fix some errors in my formulas.  However, the conclusions are the same.

As a follow up to my previous post, I wanted to take a closer look at the gravitational constant of “Big G”.  Upon examination, I found that though Newton used Kepler’s laws of planetary motion, the relationship between them is rather ambiguous.  For example, Kepler’s laws deal with the velocity of a planet around the sun with respect to time – meaning: The time it takes to complete an orbit and the area covered is proportional.

There is no force involved since he was not concerned with the mass of the objects.  The time it takes for a planet to revolve around the sun increases with distance, hence his proportional law of 1/r².  But, again, this has nothing to do with a force acting on the object, rather just the proportions of the ellipse and the time taken to traverse it.

Newton, however, was completely concerned about the mass of an object.  He took the proportions with respect to time and converted it into a force: Gravity.  By replacing the proportions with mass he removed time from the equation.

The gravitational constant specifically is:

6.67 ×10−11 m3⋅kg−1⋅s−2

or

6.67 x 10−11 m3/kg/s²

It is a constant that is applied over time as an acceleration but time is not factored in.  An acceleration over time, by it’s very definition, is an increase in velocity over time.  You can’t have a change in velocity without a change in time.

The equation for the “law of gravity” is:

{\displaystyle F=G{\frac {m_{1}\times m_{2}}{r^{2}}}\,.}

However, all force equations require 3 attributes to comply with reality and no variable can be isolated:

  1. Vector
  2. Magnitude
  3. Time

The classic force equation F=ma violates this requirement since it excludes the time attribute.  Math is a descriptor language so it must describe reality not abstractions.  For example, a 4,000kg car accelerating at 10 m/s² for 10 seconds cannot be described by the equations F=ma since it excludes the change in time while it was accelerating.  All it can describe is an abstracted force not a rational one.

Velocity, acceleration, force and momentum all violate these basic requirements since they only include partial attributes.  For example:

a = Δv / Δt  – must include a magnitude to describe reality – ma = Δv / Δt · m

v = d/t – must include a magnitude to describe reality – mv = d/t · m

F = ma – must include time to describe reality –  ΔtF = maΔt

p = mv – must include time to describe reality – tp = mvt

So we must write the “law of gravity” as:

ΔtF = maΔt = GmM/r² · Δt = Δvm

This translates into a significant problem for gravity since the velocity or force involved must increase with respect to time.  Neither the mass nor the acceleration due to gravity changes, just the velocity and the time the acceleration is applied.  A secondary issue is that the acceleration will increase as the distance between them decreases.  In the case of gravity, the acceleration never stops since it is supposedly inherent to matter.

No object could ever separate from any other object of any mass if time is applied.  We neither observer nor experience what should be happening which is a self-evident falsification of the universal law of gravity.  Simply stated: Gravity is Dead.

The Falsification of Terminal Velocity, Gravity and the Source of Mass

Note on update July 2nd: I had to make an update to this post as I had erred in some of the equations.  The conclusions are the same but now the formulas are correct.  Mainly, I had neglected to balance the force equations on both side which resulted in much to-do in the twitter realm.

Before I get into the rational explanation for the falsification of terminal velocity, I need to make it clear what I’m not saying.  My experience with posting controversial ideas tends to make one focused on the negative response rather than the positive.

I’m not saying that their isn’t a maximum velocity that objects in free fall achieve.  This is experimentally demonstrable and not the point of this post.  What I am saying is that the explanation for this phenomenon is incorrect and the mathematical description is invalid.

Some definitions:

Big G

{\displaystyle F=G{\frac {m_{1}\times m_{2}}{r^{2}}}\,.}

The constant of proportionality, G, is the gravitational constant. Colloquially, the gravitational constant is also called “Big G”, for disambiguation with “small g” (g), which is the local gravitational field of Earth (equivalent to the free-fall acceleration).[2][3] The two quantities are related by g = GME/r2 (where ME is the mass of the Earth and rE is the radius of the Earth).

The supposed local value for gravity on the earth is g = GME/r2 which translates to 9.8 m/s².  For the purposes of this proposal, small ‘g’ is sufficient.

Terminal Velocity

The current explanation for terminal velocity as per wiki:

Terminal velocity is the highest velocity attainable by an object as it falls through a fluid (air is the most common example). It occurs when the sum of the drag force (Fd) and the buoyancy is equal to the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration.[1]

There is a significant issue with this explanation which ties into the invalid structure of the mathematical derivation.

According to the standard model, the only downward force that is acting on the object is the force of gravity and any opposite force will result in the reduction of the downward force of gravity itself.  Objects can only accelerate in free fall; they can never achieve a constant velocity.  Again, this is a rebuttal to the current explanation of terminal velocity not a dismissal of terminal velocity itself.

If the net force acting on the object is zero, then the object will simply stop accelerating and since the only downward force is an acceleration, the object should stop moving (akin to neutral buoyancy). Any objection to this conclusion requires either the addition or subtraction of forces.  For example, some will argue that once the net forces are in balance the final velocity achieved at that point, is terminal velocity.  They will also pull out Newton’s 1st law:

An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force

We need to take apart all the assumptions about a free falling object in order to see why terminal velocity as it is currently understood doesn’t work.  As mentioned before, a free falling object only has the force of gravity acting on it therefore the speed is an acceleration not a constant velocity so once the acceleration is zero due to drag, it should simply slow down to zero.  If there was no unbalanced force of drag, the object would continue to fall at an increasing rate.  A simple analogy: two cars tied together and accelerating in opposite directions would go nowhere but the forces are still active. There are no additional forces present.  The cars wouldn’t just start drifting at a constant velocity towards one car or the other.

So where does the velocity value even come from?  To figure that out we need to see how it is derived (from Wiki):

Derivation for terminal velocity

Using mathematical terms, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the drag equation):

F_{{net}}=ma=mg-{1 \over 2}\rho v^{2}AC_{{\mathrm {d}}}

At equilibrium, the net force is zero (F = 0);

mg-{1 \over 2}\rho v^{2}AC_{{\mathrm {d}}}=0

Solving for v yields

v={\sqrt {\frac {2mg}{\rho AC_{{\mathrm {d}}}}}}

If you examine this formula closely you will see where the error creeps in.  It’s not that the math itself is incorrect, but the conclusions are incorrect relative to the math.

There are two distinctive forces present in this formula – gravity and drag.  The portion of the equation that describes the force of gravity is “mg”.  The second portion of the formula is “1/2 pv²AC” which describes the drag forces.  This is the upward force due to drag which counter-balances the supposed force of gravity.

To solve for “v” we need to isolate it like this:

mg = 1/2 pv²AC (multiply both side by 2)

2mg = pv²AC (divide both sides by pAC)

2mg / pAC = v² (get the square root of each side)

v = √  2mg / pAC

The variable “v” is a function of the upward force of drag so it cannot be isolated out in real life.  It’s part of the accelerating upward force.  From a mathematical point of view, it simply describes the upward velocity of the air at a specific point in time.  But for the equation to balance out we need to also take the square root of the force of gravity and the mass of the object.  That doesn’t even make sense.  What does the square root of mass even describe in real life?

So even though the math works (meaning it balances out) it has nothing to do with the actual events we would observe.  The fact that the resultant force is called a velocity rather than an acceleration is also interesting since the downward force on any object at anytime is gravity (which is an accelerating force) according to the standard model.

Secondly, the article goes on to state that buoyancy effects can be subtracted from the mass of object.

Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using Archimedes’ principle: the mass m has to be reduced by the displaced fluid mass \rho V, with V the volume of the object. So instead of m use the reduced mass {\displaystyle m_{r}=m-\rho V} in this and subsequent formulas.

This is in direct conflict with the concept of mass since the mass of an object is always constant; it is the supposed force of gravity that gives the object weight so it must be the weight that is displaced NOT the mass of the object.  From Wiki:

Assuming Archimedes’ principle to be reformulated as follows,

\text{apparent immersed weight} = \text{weight} - \text{weight of displaced fluid}\,

then inserted into the quotient of weights, which has been expanded by the mutual volume

\frac { \text{density}} { \text{density of fluid} } = \frac { \text{weight}} { \text{weight of displaced fluid} }, \,

yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volumes:

\frac { \text {density of object}} { \text{density of fluid} } = \frac { \text{weight}} { \text{weight} - \text{apparent immersed weight}}\,

(This formula is used for example in describing the measuring principle of a dasymeter and of hydrostatic weighing.)

According to the standard model, the weight of an object is equal to the mass times the force of gravity:

weightf

There are two important factors that need to be considered when looking at drag and buoyancy on a falling object.

In fluid dynamics, the drag equation is a formula used to calculate the force of drag experienced by an object due to movement through a fully enclosing fluid. The formula is accurate only under certain conditions: the objects must have a blunt form factor and the system must have a large enough Reynolds number to produce turbulence behind the object.

So the key to drag is turbulence.  This is one of the effects that slows the object down as it falls.  Turbulence also increases over time.  There is zero turbulence when object is suspended in air and motionless and an increasing turbulence over time once the object is released.

Of particular importance is the u^{2} dependence on flow velocity, meaning that fluid drag increases with the square of flow velocity. When flow velocity is doubled, for example, not only does the fluid strike with twice the flow velocity, but twice the mass of fluid strikes per second. Therefore the change of momentum per second is multiplied by four. Force is equivalent to the change of momentum divided by time.

As the object falls, the flow velocity (meaning the total amount of air pushing back on the object) and subsequently the total mass of air will increase.  The result is an increasing increase in turbulence behind the object.  The total turbulence is directly proportional to the maximum velocity an object can sustain in free fall minus any buoyancy effects.

So the proper formula for buoyancy effects would be mg – pV or W – pV.  Therefore, the proper formula for terminal velocity would be the weight of an object minus buoyancy effects minus drag or v = [W – (pV)] – (1/2 u²pAC).  We can see that this is simply a momentum calculation that subtracts two opposing effects – buoyancy and drag.

The reason for an increasing increase of velocity (acceleration) during the initial release of the object is due to the turbulence requiring time to increase.  But as the turbulence increasingly increases to a maximum value, the object begins to slow down from it’s initial acceleration in direct proportion to the value of turbulence.

We could postulate that the acceleration of an object in free fall would be unlimited if turbulence was zero.  Therefore, the acceleration of an object due to the force of gravity does NOT have to be evoked to explain an object in free fall but only the difference between the density of air and the falling object.  It is the turbulence that is slowing down the object.  Once the density of the falling object meets the density of the ground, the falling stops.  Thus, there is no downward force acting on the object just a difference in density.

If there was indeed an accelerating force due to gravity, the object on the ground should continue to move towards the center of the earth.  Why should the acceleration due to gravity stop just because air gave way to ground?  The fact that we do not continue to accelerate towards the center of a ball when, by all theoretical and mathematical requirements, we should be, is a clear falsification of gravity.

Gravity – an imaginary force

I further postulate that the mass and the weight of an object are the same thing and gravity is an imaginary force that is only applied to imaginary celestial objects (planets and other imaginary celestial objects must all be in free fall around some other object).  By direct observation, an object at rest on the surface of the earth is no longer subject to the imaginary force of gravity since the weight of an object must continually increase due to the constant acceleration.

If Newton’s third law is to be followed, the object “at rest” (but not really) on the surface of the Earth must be continually pushing back with an “equal and opposite reaction” over time since an acceleration is the change of velocity over time.  So with each second the net force being applied to the object must increase.  Let’s take an example from Khan Academy:

(https://www.khanacademy.org/science/physics/one-dimensional-motion/acceleration-tutorial/a/what-are-acceleration-vs-time-graphs)

What does the area represent on an acceleration graph?

The area under an acceleration graph represents the change in velocity.  In other words, the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval.

area = Δv

It might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 m/s² for 9s.

Screen Shot 2017-06-19 at 9.17.25 AM

If we multiply both sides of the definition of acceleration, a = Δv / Δt , by the change in time, Δt, we get Δv = aΔt.

Plugging in the acceleration 4 m/s² and the time interval 9s we can find the change in velocity:

Δv = aΔt = (4m/s²)(9s) = 36 m/s

Multiplying the acceleration by the time interval is equivalent to finding the area under the curve. The area under the curve is a rectangle, as seen in the diagram below.

Screen Shot 2017-06-19 at 9.26.42 AM

The area can be found by multiplying height times width. The height of this rectangle is 4 m/s², and the width is 9s.  So, to find the area also gives you the change in velocity.

area = (4m/s²)(9s) = 36 m/s

The area under any acceleration graph for a certain time interval gives the change in velocity for that time interval.

We can now take the acceleration due to gravity (9.8 m/s²) and apply the same model.  A person standing on the surface of the Earth must experience a constant change in velocity:

Δv = aΔt = (9.8m/s²)(1s) = 9.8 m/s

Δv = aΔt = (9.8m/s²)(2s) = 19.6 m/s

Δv = aΔt = (9.8m/s²)(3s) = 29.4 m/s

Δv = aΔt = (9.8m/s²)(∞s) =  ∞ m/s

According to the standard model of gravity, a person in free fall is weightless, therefore their mass is not subject to measurement.  Mass as a function of weight is only applicable when there is resistance to counteract free fall.  As the mass of the person begins to impact the surface of the Earth, they begin to weigh something and that mass is subject to an increasing velocity which translates into an increasing weight.

How do we measure weight?  This can be measured on a standard scale.  A scale has springs that can compress and translate the downward pressure into a weight.  As long as the springs can handle the increasing weight, so will the person increase velocity and we can measure the increase.  But the force of the mass of a person is dependent upon the velocity at a particular time since:

tΔp = tmΔv

Of course this is not a common way to express momentum since it is a change in momentum due to a change in velocity.  However, in this case, the change in momentum is being translated into a change in weight:

If “Δvm = maΔt”

Then “ΔtW = mΔv” or “ΔtW = maΔt”

So what is a kilogram or a pound?  It is an arbitrary unit of measurement based upon various assumptions.  The primary assumption is that weight is a function of gravity and mass.  However, we can clearly see that if velocity is constantly changing due to the force of gravity then the resultant weight must increase.  It is required by the standard model of gravity that the mass and acceleration of the object are fixed and do not change.  It’s the velocity that must be changing over time. 

The objection to this proposal is that time cannot be a function of Weight or of Force.  The standard way to express the equation is F = ma.  However, this equation contains within it a time function as  ma = Δv / Δt · m.  This results in:

Δvm = ΔtF

ΔtF = maΔt = Δvm

Therefore, if there is no change in velocity, as anyone can observe, then we are left with:

F = W = m

So what is mass?  How does something weigh more than something else?  That is our next topic.

The Foundation of Mass

What we are left with are two attributes of mass – Density and Buoyancy.  Gravity, in this instance, is no longer an accelerating force inherent to the mass of an object but a synonym for mass itself.  There is no intrinsic force attributable to mass.

Let’s start with a quote from Newton himself:

“It is inconceivable that inanimate Matter should, without the mediation of something else, which is not material, operate upon, and affect other matter without mutual contact. Gravity should be innate, inherent and essential to matter, so that one body may act upon another at a distance thru a vacuum, without the mediation of any thing else, by and through which their action and course may be conveyed from one to another, is to me so great an absurdity that I believe no man who has in philosophical matters a competent faculty of thinking can ever fall into (for) it. Gravity [mass] must be caused by an agent acting constantly according to certain laws; but whether this agent be material or immaterial, I have left to the consideration of my readers.” – Sir Isaac Newton, Letters to Bentley, 1692

We are indeed considering his words.  What we can extract from this is that gravity or mass is the result of something not the cause of something and a “mediating” entity must be involved.

Here is a quote from Ken Wheeler:

Before embarking on explaining magnetism, first the all important word field must be rigorously and scientifically defined with ambiguity and misconceptions fully removed. A field (Greek: χώρα) is a conjugate nonspatial attribute between the subject, Ether, and the object causing the perturbation (or EM induction), either as gravity [weight/mass], electricity, magnetism, or electromagnetism. A field has relevance only as a relational perturbation between the subject and the object causing its appearance; a field is specified only as regards to: electricity, magnetism, gravity, matter, and dielectricity (and another unmentioned for another article); however as it must be necessitated electricity, dielectricity, gravity [weight/mass], and magnetism are by their very principle not different from the Ether itself. The very term ‘magnet’ merely denotes the electrified mass formerly ‘not-a-magnet’; as such magnetism and magnet are abstractions or distinctions without a difference except as relates to the coherent field charges of the before and after mass. Conceptual disambiguation between a magnet and magnetism itself cannot be enjoined, and is merely a fallacious reification of connotative abstractions.

Ken Wheeler: Uncovering the Missing Secrets of Magnetism p.27

Everything seems to come down to the historically dismissed Ether.  It must be noted that even the high priest of gravity, Newton, proposed an Ether.  The persistent objection to the dismissal of gravity is always, “what is the alternative”?  That an empirically based science that includes the Ether is a valid alternative and has always been a valid alternative seems to elude those so certain of gravity.  It’s the implications of an Ether that frightens them so.

I will not put words into Mr. Wheeler’s mouth so it is important to note that he does not dismiss Gravity per say but the origins of it.  Based upon Mr. Wheelers writings, I can safely say that he is NOT a proponent of the Flat Earth so I can only assume that he believes in planets, comets, gravity, etc.  So in no way am I linking him to the Flat Earth theory.  But I digress…

Current scientific thought requires gravity to be an inherent property of mass itself rather than an affect of a mass within the Ether (like the waves on an ocean is the ocean but not inherent to the ocean; something else causes waves).  If there is NOT an intrinsic force to mass then much of modern astronomy falls apart.

Density and Matter

So what causes mass?  In mathematical terms it would be p * V = m (density * Volume = mass).  In this sense, mass is a function of density and volume.  We know that volume is a function of geometry so what is density?

Magnetism is purely radiative, is the termination of electrification and the end-of-road byproduct of dielectricity. Dielectricity comes before everything else in the four-part schema of Force Unification. Dielectricity and magnetism are the two co-principles of the universe. So how do you get magnetism out of dielectricity, since magnetism requires a subject to emanate from or itself is the termination point of either mass in movement or electricity as it terminates? The answer is that dielectricity terminates into the creation of matter, which itself then has in this conjugate relationship, magnetism as its radiative principle (the proton as found in hydrogen, the most abundant element is magnetically dominant, is the polarized charging dynamo for its discharge plane of interatomic magneto-dielectric volume). Creation (dielectricity) and radiation (magnetism), and their two byproducts, electricity and mass, or gravity [density] as centripetal attributes choate to mass / matter.

Ken Wheeler: Uncovering the Missing Secrets of Magnetism p.38

What Ken Wheeler is calling “gravity” in this instance would simply be the density attribute of mass.  A dielectric object (not-a-magnet mass) has a specific density due to the “centripetal attributes choate” to it.  It is the centripetal/centrifugal attributes of the object that ultimately give objects their density and subsequently makes objects heavier or lighter than others.

A complete thesis developed and proven by Ken Wheeler is available for anyone to read.  It is incumbent upon the read to disprove his empirically based science before dismissing the concept of an Ether.  This is true science and if you fancy yourself a true scientist or a rational human being, you will need to apply the proper scientific rigor rather than falling into Scientism consensus.

 

Gravity: The Baron Munchausen of Forces

 BaronMunch

The Missing force in gravity

Two bodies that are interacting require “something” to push against. For example, in a classic tug-o-war there are two teams pulling against each other using a rope. However, for it to work, they must push against the ground not air or space in order to pull. The classic F=ma equation usually cited does not give the fulcrum from which each body is pushing against. They can’t push against air or space while pulling at the rope. Hence, there is no “equal and opposite” reaction since there is nothing to react against. There is no push to the pull.

Essentially we have an equation that has the earmarks of Baron Munchausen

Tug-O-War2

Centripetal force is not gravity.  This is a common misconception.  Gravity is an independent force due to the mass of the object.  People usually link things together which are not related. For example, angular momentum gives rise to the centripetal and centrifugal forces. The centripetal force is “pulling” at the the centrifugal force and the centrifugal force is “pulling” against the centripetal force and gravity at a rate of ω²r, where ω is the angular speed (=linear speed/r) but it does not cause gravity.

In other words, they have to subtract the emergent centripetal and centrifugal forces from the force of gravity to get the resultant force.  For example, a 72kg person will exert 2.36N of centrifugal force at the equator while gravity exerts (pulls) 705N of force giving a resultant force of 702.64N. But for gravity to pull it must push against something at that’s what’s missing. The earth must be fixed in space to pull.

The only thing that gravity can push against is space and space must have zero mass. For space would need sufficient mass to resist the motion of any and all celestial bodies (galaxies, suns, planets, etc). Therefore, space would no longer be space but mass. So the conundrum for an astrophysicist is the necessity of space and the simultaneous necessity for space to have mass.  They are mutually exclusive.

For this reason, gravity violates Newton’s 3rd law.

Spacetime, curvature and Relativity nonsense

Since gravity is a function of mass, any supposed curvature of space is the result of the mass of the object not the other way around.  Also, curvature of space around an object is impossible. This would require the supposed “fabric” of space is pulled inward towards the object. What exactly is it pulling on? Space by it’s very definition is empty.  How can a force pull on emptiness? Even if gravity was granted such magical powers, it would then, by necessity, require space to have mass and if it has mass then it is not space.  As well, time is an abstraction and a force can’t pull on an abstraction.

Even if we grant gravity the magical powers to pull on emptiness it still needs to push against something to pull on something.  We get into a circular argument where the space around the earth is being pulled on while simultaneously it is resisting the motion of the earth itself. Hence the allusion to Baron Munchausen pulling himself up by his own pigtails.

The Inertial mass required to fix the earth in space would negate space itself. So regardless of which theory of gravity is evoked, it violates Newton’s 3rd law.

Density/Buoyancy and the falsification of Gravity

Density and buoyancy are a function of the differences between the mass of objects. For example, an object more dense than air will drop to the ground due to the displacement of the surrounding air. Air under compression can push against a heavier object since the air is not being displaced. The mass of an object is also distinct from the force of gravity.

Here we get to the crux of the problem? If a 72kg person is standing on the earth’s surface, they should experience a constant acceleration of 9.8 m/s/s. But this acceleration does not stop just because they are standing on the earth’s surface. Their “weight” will only be 72kg (702.64N) for one second. What happens at the 2nd second? Their “weight” would double since in the 1st second it is 702.64N + 702.64N in the 2nd second (the “weight” of a person in free fall is zero). In other words, we should all be compressed piles of mush. All mass should be pulled relentlessly into the centre of the earth. Therefore, since the acceleration due to gravity is not what we observe, it must be shown as having been falsified.

Other celestial observations, like the gravity bending light, are non-sensical since the bending of light would require either space (which is empty and massless) or light to have mass. Neither of these have mass. The motions of celestial bodies (planets, suns, comet, galaxies, etc) are equally non-sensical since they require gravity and since gravity has been shown to be falsified, the motions of the lights in the sky are due to other phenomenon.

And finally, outer space, planets, comets, etc are all constructs of the theory of gravity. Without gravity they fall apart.  Welcome to the flat earth.

There is an alternative to gravity that has been elegantly presented by Ken Wheeler: https://archive.org/details/magnetism1small

Some thoughts on Gravity and Newton’s Laws

Over the past few weeks I’ve been thinking about gravity and the various laws that Newton proposed.  One of the main questions I have pertains to the equation for F (force) which is F=ma and the so-called law of universal gravitation which is:

F1 = F2 = G⋅m1 x m2 / r(squared)

And this is put into normal language as:

The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them

meaning, if the mass increases then the product increases.  For example, as a mass accelerates the value of F will increase.  This can be seen in the extreme example of a mass that approaches the speed of light;  as it approaches, it’s mass approaches an infinite value.  The difficulty with this concept (as it is only a concept without any empirical evidence) is Newton’s 3nd law which states that “for every force there is an equal and opposite force”.

If you take the 2nd law and apply it to any accelerating mass, you increase the total amount of force which also an increase in energy (e=mc2).

Now if you take the law of universal gravitation (as shown above), and apply the 2nd law to m2, the total force between the two objects will increase.  you can ignore the increase of r(radius) since the increase in radius is insignificant relative to the size of m1.  In other words, an object like a rocket could not possibly launch if gravity acts according to the prevailing theory since the acceleration adds to the objects total energy (as shown above).  By increasing energy you increase the force between m1 and m2 with a “…a force that is proportional to the product of the two masses” and with equal force (as per the 2nd law).  Therefore for every pound of thrust an equal amount of force is brought to bear between the rocket and the earth.

This would also apply to anything accelerating away from the center of the earth (ie. rapidly raising my hand above my head).  In other words, gravity should be acting like a brake against the accelerating body.  To “break free” (think of an inverted pendulum flywheel) of the gravitational force, an object would have to accelerate with a force greater than m1 * acceleration.   Therefore, if we take two equal masses (m3 = m4) and accelerate one of them (m4) it would require the second mass (m4) to accelerate at a greater value than m3 * acceleration.  Essentially, it is an application of e=mc2.  This a far better explanation as to why the mass of an object increases as it approaches the speed of light – the mass itself is not increasing but the affect of gravity increases proportionally to the accelerated mass  therefore the effective mass increases.

The great irony here is that this makes gravity all but impossible since any object on the surface of a globe (ie. Earth) would be held fast against the surface.  Anything pushing against gravity would encounter massive (no pun intended) resistance (like blood flow, plant growth, etc) to the point where no life could form.   Nor could objects be buoyant.  An object floating on the ocean surface is essentially accelerating away from the center of the earth (until it finds equilibrium at the ocean surface).  The gravity of the Earth is far greater then the total outward thrust of the buoyant object (ie. air inflated beach ball).  In a nutshell, buoyancy would be completely overwhelmed by gravity.  As well, the lift experience by commercial airplanes would also be insufficient to overcome gravity.

However, no matter how reasonable this line of thinking is, many gravity apologists will simply drag out their favorite solution:

Einstein – The grand-daddy of excuses

If we push aside Newton for the moment a look at what Einstein proposed, we are actually in a less favorable possible (if you believe in gravity).  Firstly, Space-Time needs to bend or be distorted to create this magical affect.  Looking at the area around an object (like a planet), we see it is spherical.  The so-called gravity “well” needs to encompass the entire planet not just a portion underneath.  I talked about this in a previous post.  The point being, the distortion of space time is not like this:

It needs to be an gravity sphere.  So what about the distortion around objects on the earths surface?  Is this not what causes gravity?  Does space-time wrap around a cube or a oddly shaped stone?

The explanation is somewhat specious:

Bodies with spatial extent

If the bodies in question have spatial extent (rather than being theoretical point masses), then the gravitational force between them is calculated by summing the contributions of the notional point masses which constitute the bodies. In the limit, as the component point masses become “infinitely small”, this entails integrating the force (in vector form, see below) over the extents of the two bodies.  

In this way it can be shown that an object with a spherically-symmetric distribution of mass exerts the same gravitational attraction on external bodies as if all the object’s mass were concentrated at a point at its centre.[2] (This is not generally true for non-spherically-symmetrical bodies.)

https://en.wikipedia.org/wiki/Newton’s_law_of_universal_gravitation

Right.  This explanation effectively reduces all objects to single point masses and runs gravitational vectors between two bodies.  It’s really a rather grotesque idea.  Then the gravity within a body is nullified since all the internal objects of a “single mass” are counted as one:

  • The portion of the mass that is located at radii r < r0 causes the same force at r0 as if all of the mass enclosed within a sphere of radius r0 was concentrated at the center of the mass distribution (as noted above).

  • The portion of the mass that is located at radii r > r0 exerts no net gravitational force at the distance r0 from the center. That is, the individual gravitational forces exerted by the elements of the sphere out there, on the point at r0, cancel each other out.

As a consequence, for example, within a shell of uniform thickness and density there is no net gravitational acceleration anywhere within the hollow sphere.

So now we have a hollow uniform body and only the surface itself has gravity.  So if all the gravitational forces are cancelled out as per the explanation above, then all the gravitational forces most somehow come from the surface.  How is that even possible? Geometrically, the math flattens out the sphere by pushing everything to the surface so the concept of a larger mass having greater density and therefore greater gravity is expunged and we are left with flat plane – in essence.

So does the math coincide with reality?  If the math says all the gravity is on the surface, then is it really on the surface?  I mean really only on the surface.  If yes, then how could a surface, no matter how big, generate a sufficiently potent gravity field as to warp space-time?

How can space-time have a gravity well?

This is re-post of a comment I sent to TheMorgile.  I’ve been follow his work and found some very enlightening ideas.  Though there are some errors in the idea of being squished at the north pole (since the time per rotation is so slow), I have not found any other errors in his work.  The concept of gravity is so counter-intuitive since it takes real and observable forces (centripetal and centrifugal forces) and defines one (centrifugal force) as imaginary and the centripetal force as a falling force (hence orbiting objects are constantly “falling” towards the more massive body but never hit due to linear motion perpendicular to the “falling” direction – ok…right).  The “attractive” nature of gravity is only to change direction but no force is applied.  It’s so bizarre and without any relationship to the real world, you have to wonder how we accepted the concept in the first place.  However it seems that Newtonian mechanics have so many flaws that Relativity was needed to fix the issues.  So is gravity due to Newtonian laws or Relativity or QM or………

Here is a video trying to describe gravity via QM / Relativity by having objects on a flat, elastic surface and the mass of the object causing a depression.  This depression then causes objects to “fall” towards the depression.  Huh?  So space is bent in one direction which happens to be underneath and the linear motion of the smaller object is in perpetual motion since it never loses velocity and therefore stays in orbit.  Are we really believing this stuff?  What about the space *above” the well?  What direction does that go in?  Or to the left and right?  I’ve seen some images trying to deal with this absurdity.

qm_gravity

In this image, gravity sucks in space rather than pushes it away.  Because space has three directions, you would get a distorted cube, not a sphere.  In any event, even this image flies in the face of the scientific consensus.

———- original post ———–

I’ve written an article http://eternalworldorder.com/2015/07/22/action-at-a-distance-as-an-article-of-faith/ that talks to your comments about gravity.  I was debating a couple of folks online about gravity based upon a video that you created https://plus.google.com/u/0/118362508845435062812/posts/UPyqZuDZA4F

I did the equations for centrifugal and centripetal force at the equator and north pole.  In the spinning ball model, the spin of the earth between the two locations only reduces the centrifugal force by a few Newtons per kilogram because the time for one rotation takes so long.  However, the entire basis of gravity *requires* action-at-distance which is absurd.  Anything which is not tethered to the ground will experience lift due to the spin.  The heavier the object the greater the centrifugal force (as well as the faster the rpm).  It’s like spinning on a merry-go-round.  Centripetal force can only take affect if the object is attached to the spinning body (like a hammer throw).  I calculated that anything greater than 330kg at the equator will float off the surface of the earth due to the centrifugal force http://eternalworldorder.com/2015/07/10/a-discrepancy-in-the-use-of-the-centripetal-force/.  This obviously does not happen.

During the debate (and as Newtonian mechanics started to fall apart) the last refuge for gravity became quantum mechanics.  If you look at all images for the warping of space time, you see a sphere pushing down on a flat surface.  The warp is supposed to cause a well underneath which is supposed to cause the object to fall towards the heavier object (but the original velocity of the smaller object keeps it from hitting the larger object).  But this is an incomplete image even if you accepted the model.  Since space is all around the object in 360 degrees it should be a gravity sphere *not* a well.  If fact, the concept of a event horizon for black holes is incomplete since it should be a black sphere.  A massive gravity sphere should surround the object.  This would cause light to never reach any object through space since all objects have a gravity sphere surrounding them.  The Sun should have all its light bent and distorted as it leaves the Sun and as it enters the Earths gravity sphere.   All light would be scattered and undefined.  This also is not the case.

Also, as it *must* be a sphere then in what direction is this warp?  It makes no sense once you take “space-time” out of a flat plane and into spheres.  It’s kind of ironic that the basis of Einstein’s theory sees space as a flat plane and the surface is warped.

To me this invalidates quantum mechanics as having any relationship to space or the actions of bodies in so space.  Let me know your thoughts.  Good work by the way.

Action-at-a-distance as an article of faith

To come full circle…it seems to me that to accept gravity as real you also have to accept “action-at-a-distance”(https://en.wikipedia.org/wiki/Action_at_a_distance).  It is obvious that action-at-a-distance (other than to the most ardent believers) is a tenuous concept that has zero empirical evidence (http://www.newappsblog.com/2012/07/newtonian-gravity-as-as-action-at-a-distance-you-know-a-sympathetic-process.html) and instead relies upon metaphysical arguments and conjecture.

To claim that any object orbits due to the gravitational attraction of bodies is speculating and is attributing special, hidden and unverifiable qualities to matter.  In essence, gravity cannot be proven since it can’t be shown to not exist.  For example, I can solve the buoyancy of a balloon without the need to include gravity.  Volume, differences in density and temperature will give an exact value for lift.  However, someone could simply argue that since gravity is inherent within all objects (“the conspiring nature”), you don’t need to include it; therefore, I can give gravity any arbitrary value and the results will match (ie. the total amount of lift will be the same).  Try it yourself – just solve a buoyancy equation but give gravity a value of 1 or ignore it altogether.  If you think about it, how is terminal velocity, buoyancy and density any different than acceleration due to gravity (ignoring action-at-a-distance)?

In fact, all equations that have a gravity function can be removed without altering the real outcome.  However, the only equations which will not work are objects in orbit that require action-at-a-distance.  An orbiting body like the ISS requires a continuous change in direction.  The change in direction involves a net zero force.  So just like the schwarzschild radius can divide by zero (https://www.thunderbolts.info/thunderblogs/archives/guests08/061108_sjcrothers.htm) a change in direction requiring zero force requires a complete suspension of disbelief (https://www.reddit.com/r/askscience/comments/3be0zs/how_are_orbiting_objects_not_accelerating_due_to/).   Honestly, how can something act on an external body with zero force?  It doesn’t matter if you qualify it as a *net zero force* or not.  The result is zero…meaning no force.  This was the greatest achievement of Newton – to separate the relationship between force and motion.  But can anyone really claim that this has any reality?  To do that, you *have* to believe in action-at-a-distance *and* that a change in direction is possible *without* force.  Other than orbiting bodies (which in themselves are not verifiable), is there any empirical evidence of this?

Many have convinced themselves of the validity of such an action because the implications are too devastating to consider – What if gravity is a false premise?  But it’s impossible to prove a negative.  One would then have to start questioning and challenging the supposed authorities on these matters.  Most of science can still move ahead by removing gravity from their equations.  However, one particular science cannot.  I will let you guess which one that is.  What would happen if an orbiting body is impossible?  What does that say about the information that we are presented with everyday?

What if you started testing and thinking with your own eyes and mind?  Ideas that were previously blocked from consideration might become viable.  But this requires a person to recognize that they believe something as a matter of faith and not because it is self-evident or a testable hypothesis.  We are all subject to articles of faith (even atheists).  If a person doesn’t think that they are subject to those articles then woe to them.  And to argue that the presence of the moon is proof of gravity merely reinforces the idea of that article of faith.  The moon and its motions are not fully understood and anyone claiming they do understand it are being intellectually dishonest at best or deliberately misleading at worst.

A Discrepancy in the use of the Centripetal Force

This is a re-post of an article I wrote in reply to JimSmithInChiapas.  He has been kind enough to respond to some of my ideas on gravity, centrifugal and centripetal force.  Though he does not agree with my assessment of these forces, we continue to communicate in a respectful manner.  Below is a recent reply to a comment he had on centripetal forces.  Ultimately, I feel that the centripetal force has been fundamentally misapplied in spinning frames of reference.  It is my contention that a single frame of reference for spinning body requires that for any object to be apart of that frame, it must be attached directly to that spinning body.  This has massive (no pun intended) implications.  You can read my argument below:

Jim…thank you for your reply.  I initially wanted to clarify a slight misrepresentation in your article.  My article is called “Does Gravity Make Sense?” not “does gravity exist?”.  They are different concepts.  I’m questioning how gravity is being represented by mainstream science today not that objects have mass or that they fall from the sky.

In any event, the example that I used provided a single frame of reference with respect to a spinning disk.  I used a disk with a 10m radius that spins at 10 m/s with a person of 72kg on the outer rim.  The person would experience 720N/kg of centrifugal force.  I don’t think that is in question.  I then go onto saying that unless the person holds onto the disk (via an attached handle of some kind), they would be flung from the disk at 720N/kg.  Again, I don’t think that is in question either.  The centripetal force is an inward force that requires the handle and the person to be attached to the disk at all times.

A more illustrative example would be the Olympic hammer throw.  The athlete is holding onto a tether which is attached to heavy weight.  The spinning motion of the athlete creates the centrifugal force which flings the weight outwards.  The strength of the athlete keeps the hammer from leaving a circular orbit via a centripetal force.  The inward force (centripetal) is provided by the athlete which is ‘balance’ by the centrifugal force.  But all the spinning objects in that frame of reference are and must be attached together. 

There is no empirical evidence of a centripetal force acting on a body that is not attached to the spinning body.  How could it?

There are no real world examples of a free floating object being acted on by a centripetal force.  You can mathematically present a centripetal force acting on an object but it is missing the real world necessity of being attached to or apart of the spinning object.  The centripetal force is a function of a spinning object; it is not a separate force that can be applied to an object outside that frame of reference.  To be part of that frame of reference, an object would, by necessity, need to be attached to the spinning object. 

For example, the person in the disk example above, is not part of the frame of reference unless they hold onto the disk with sufficient force.  They are literally removed from the frame via the centrifugal force.

Therefore, I would conclude that any object that is rotating around the earth must, by necessity, be attached to the earth to be part of that frame of reference for any object not attached is subject to the centrifugal force and will be removed from the frame. 

If we take a real world example of a person of 72kg standing on the surface of the earth and if they are standing at the equator and if the earth is spinning at 1000miles/hr then they are subject to a centrifugal force of 2.2N/kg.  If gravity is acting on the person with 9.8N/kg then a total force of 7.6N/kg is present.  The centripetal force is not part of the frame of reference for that person as shown above.

As the mass of the object increases, the centrifugal force increases.  Therefore, an object greater than ~330kg should become “weightless” on the surface of the earth.  This is obviously not happening nor are people 22% lighter at the equator than they are at the north pole.  Additionally, the person would not feel heavier if they grabbed hold of something attached to the earth. 

What I’m showing is that there is a discrepancy between real world situations and the mathematical examples presented by modern science.  At this juncture I can only conclude that the centripetal force is being improperly applied across multiple spinning  frames of reference to account for the discrepancy and if that is the case then we cannot be in a spinning frame of reference (as shown above).

The Impossible Flight of the ISS

I was looking at some additional sites from NASA that try to explain the nature of gravity at certain altitudes.  https://www.grc.nasa.gov/www/K-12/airplane/wteq.html

The final sentence of the explanation is “…But the high orbital speed, tangent to the surface of the earth, causes the fall towards the surface to be exactly matched by the curvature of the earth away from the shuttle. In essence, the shuttle is constantly falling all around the earth.”

As mentioned in my previous posts, the centripetal force only makes sense for something that is tethered to the spinning body  (If you feel that the centripetal force *does* have special powers, please provide a clear empirical example that can be tested). Neither the space shuttle nor the ISS are tethered to the earth unless we grant the centripetal magical grappling abilities (see hammer throw). https://www.youtube.com/watch?v=KnHUAc20WEU

As well, for the shuttle to be constantly “falling” but not actually falling downwards, a constant acceleration would need to be applied (ie. rockets) plus a continual adjustment of direction or the shuttle would fly off into space (see what happens when the hammer is released).  Again, for apparent “weightlessness” in space, it would require objects to be falling at a rate of 9.8N/kg (or m/s/s) which would mean a constant counter-force of equal value would need to be applied or they would rapidly fall to earth.  So the “floating” objects and people in space would need to be in a free fall all the time.  This is obviously not the case since the ISS would have crashed to earth a long time ago.  In essence the ISS is just like a airplane at a higher altitude and would require constant thrust to stay in “orbit”.  If you turn off the engines of an airplane at 30,000 feet will it stay in “orbit” because “…the high orbital speed, tangent to the surface of the earth, causes the fall towards the surface to be exactly matched by the curvature of the earth away from the [airplane]? ”  I don’t think any scientist would want to be in that airplane at 30,000 feet.   It should be noted that the standard equation for centrifugal force for any object at the equator great than ~317kg would have a centrifugal force greater than gravity.  Unless the centripetal force is magically grappling those objects, they should all start floating and since objects like elephants weigh ~4000-7000kg, they should all be floating thousands of miles above the earth.

If we grant the ISS a value of 3217N/kg (centrifugal force) due to its orbit around the earth (@ 17,150 miles/h & 4200 miles & ~331,000kg) – what force was initially used to get it to that speed?), then an equivalent (but opposite direction) for it must be present via the centripetal force.  In order for a centripetal force to be present the object must be tethered to the earth.  However, to obtain 3217N/kg, this would require the object to be traveling at a faster rate than the earth’s rate of spin.  So an object that travels faster than the earth’s rate of spin *must* be under its own propulsion and not tethered to the earth.  Since the ISS is traveling at such a high rate of speed and is not tethered to the earth, then it *must* be under its own propulsion and heading.  This is plainly not the case.  If the centripetal and centrifugal forces are equal but opposite directions, then we are left with 9.8N/kg (the force of gravity) on all objects.

In conclusion, if the centripetal force only applies to objects that are tethered to a spinning object (ie. Earth) then objects above the earth’s surface must be constantly under their own propulsion (like an airplane) to stay above the earth’s surface.  In other words, the ISS should be falling out of the sky.