Simple experiment to show Earth is either Spinning or Stationary

How do we know if the world is spinning or stationary?  Do current images from space or the material in text books prove the Earth is spinning?  Only if you believe the source to be valid.  If, however, you are a natural scientist you would want to verify the veracity of the statements by repeating the experiments.

Using these two examples, let’s look at what is experienced when a person walks upon the earth.  If the earth is spinning it is comparable to the conveyor belt example except that the earth’s axis acts as the central wheel that “belt” of the earth’s surface or ground moves around.  It cannot be the train example since the floor of the train is not in motion.  We can see that additional energy is provided by the moving ground whereas no additional energy is provided to the person who is beside the belt.  We could then create an experiment where a person would leap against the direction of the belt (both on the belt and off).   We could place an object 5 feet from the person both on and off the belt and ask them to leap towards the object.  Since the additional energy of the belt is continually compelling the object forward, we would theorize that the total distance between the person and the object would be less than then person leaping beside the moving belt.

Why is leaping important in this experiment?  If the person simply walks on the belt, the forward momentum will be continually added to the individual since there is always contact with the belt.  By disengaging with the belt in a direction against the motion, we are subtracting (for a brief moment) the forward momentum of the belt and allowing the belt to move underneath.  It would be possible to land nearer or even surpass the object whereas the person leaping beside the belt would gain no advantage like that.   The question is how quickly does the forward momentum of the belt dissipate once the person leaps?  If the person can leap (towards the object like a long jump) with an acceleration of 1 m/s/s and can stay airborne for 2 seconds then a total distance of 2m can be achieved (with a final velocity of 2m/s).  Since the belt is moving at 1 m/s then in the first second the momentum has been overcome and the object has moved 1m closer.  In the 2nd second the object moves 2m closer if we add the motion of the belt and the acceleration of the person.  The total distance covered would be 3 meters.  In the case of the person beside the belt, they would only be able to cover 2m if they leaped with the same acceleration.  (you can work out the acceleration equation if you want)

We could increase the acceleration and distance by using something like a canon.  If we use the same arc and acceleration the ball should land at a greater distance from the canon if on a moving belt than if stationary.  If we assume that the acceleration of the canon is 10/m/s/s, the tennis ball can reach 80 m/s velocity and has a total airtime of 10 seconds, we can calculate the total distance travelled (640m).  If the velocity of the belt is 1 m/s then within 1 second the momentum of the belt has been overcome and we can add an additional 9 seconds of belt movement to the distance for a total of 649m.  If we then turn around and fire the canon in the opposite direction then the distance travelled by the canon is subtracted from the total distance of the ball (1m/s x 10 seconds = 10m).  This would mean that the total distance between the canon and the ball would be 630m.

If we take the above example and apply it to a spinning earth then a similar result must take place.  Taking the spin as the same as the conveyor belt one need only launch an object (a tennis ball thrower would suffice) to the west and then to the east.  If the objects land at similar distances from the thrower then we are not spinning.  However, if they land at different distances then we must be spinning.

Advertisements

Response to Spherical Geometry and Empirical Evidence

In response to my latest post, I received a comment that due to refraction, objects at certain distance would be visible even if they are below the horizon.  He also talked about adjusting curvature every 8-10 miles.  I’ve started reading his blog but the reason for the adjustment is not clear.  However, I think he is talking about this. In the end, the height of the person and the object being observed are taken into account.  Here is the original comment:


“You are using an improper formula for curvature that has to be constantly corrected for every 8-10 miles. The correct forumula for curvature is found here. https://chizzlewit.wordpress.com/2015/05/13/working-with-the-curvaure-of-a-spherical-earth/ Also you are over water which tends to make refraction a bit of a problem. You will find however that with the height of you the observer on the deck as you pointed out and the height of the lands you put in that it does indeed fit in with the sphere earth model especially when you account for refraction. If you want to use a crude formula for distances between lighthouses and boats that includes for refraction you can use this one though its only an average correction for refraction. http://www.pajack.com/stories/pitts/viewdistance.html I suggest you try this with surveying equipment though if you are at sea the choppiness of the water will make it tough. Also being over water with refraction conditions makes it tough. You will find though that if you use surveying equipment you can clearly measure landmarks and see the curvature of the earth. https://www.youtube.com/watch?v=IOFmLHHwtic


To be fair to the individual, I wanted to address the objections empirically and see if he had valid points.  You can read my response here:


“I appreciate your comment. The main issue with the video is that it was taken with a crappy Samsung smart phone. It is possible to take a stable view from land (from either side) which is what I plan to do so any shakiness could be taken out. This will require a telescope with a camera to bring a sharper focus at a greater length. There are distances of 70 miles or more that can be observed from Vancouver. Specifically from Kits beach to Denman Island.

I would like to address your blog and the concept within them. If we wish to proceed using a scientific process (which must involve empirical evidence not just math), we must apply your math to observations.

If I misrepresent your ideas or math in anyway please forgive as I’m simply trying to apply them the best I can. If there are errors, please point them out and we can move forward from that. If we disagree, then the empirical evidence must be the final judge, not our opinion. Do you agree with that?

If you can apply your formula to the distances given in my post (29 and 33 miles) that would provide me with an actual representation of how your formula represents curvature. From there we can compare the two formulas and see the differences. Are you ok with doing that?

Refraction is defined as “the bending of a wave when it enters a medium where its speed is different. The refraction of light when it passes from a fast medium to a slow medium bends the light ray toward the normal to the boundary between the two media.”

As I’m looking towards Vancouver from the deck of the ferry, I would not be looking into water which would be a slower medium. As well the refraction requires the object be within the slower medium (or at least in between my eye and the object in question). When looking at this wiki article, the superior mirage is the only explanation that you are referring to that could bend light (if you have a different one, please explain). It would require that the object in question is within an area of higher temperature than the observer. As well, if this object (like a building) occupies various temperature strata then the object would appear quite distorted since different temperature regions would overlap and the image would not make much sense to the eye (https://en.wikipedia.org/wiki/Mirage#/media/File:Superior_and_inferior_mirage.svg). The video that I shot shows objects that are consistent in shape and size without any distortion. With both superior and inferior mirages the objects are inverted and/or distorted. The other issue is that the temperature differential required for a superior mirage is significant:

“The “resting” state of the Earth’s atmosphere has a vertical gradient of about -1° Celsius per 100 metres of altitude. (The value is negative because it gets colder as altitude increases.) For a mirage to happen, the temperature gradient has to be much greater than that. According to Minnaert,[1] the magnitude of the gradient needs to be at least 2°C per metre, and the mirage does not get strong until the magnitude reaches 4° or 5°C per metre. These conditions do occur with strong heating at ground level, for example when the sun has been shining on sand or asphalt, commonly generating an inferior image”

The temperature differential from the bottom of UBC to the peak would require a consistent temperature change over 81 meters. According to the above formula, the temperature from the bottom of UBC (let’s say 12 degrees C) to the peak would be impossible. It makes sense at small distances (like asphalt) since the temperature difference can be significant. However, the temperature from sea level to 81 meters will not be 4-5 degrees per meter. So unless you see an error in that formula, we can safely rule out a superior mirage. So once I can see your math using your equations then we can move forward with the next discussion. Hopefully I will be able to get a sharper image next time that will show more detail.

Thanks again for your comment.”


It occurred to me that the image of Chicago across lake Michigan

was described as a superior mirage.  As I wrote in my post above, the temperature differential between the lake surface and the highest object would require at least 2 degrees per meter.  If you think about it, an object 30 meters high (~100 feet) would require 60 degrees difference for the affect to occur.  And that is not a very strong affect.  According to the above article, you would need 4-5 degrees for it really be visible.  There simply is not that kind of temperature differential in such short distances.  As well, the object would need to be inverted or distorted which is obviously not the case in this video.  Near the end of the video, they show a “duct” between the hot air and the cold air.  For the affect to work, the temperature difference must be so great as to bend the light.  The distance between the hot air and cold air is greater than 1 meter.  The important thing to remember is that the “temperature gradient” must change not just one portion but the entire length of the object.  In the case of a small object on the horizon at a great distance (70 miles), atmospheric distortion would occur and you would see inverted images.  But if the object(s) in question are visible at 70 miles at a greater distance above the horizon (like a tall building) then temperature must be that much greater.  Finally, if the image in question was taken with a camera with a zoom lens it completely removes the possibility of a mirage since the focal point of the lens would bypass the affect as seen in this video:

As the lens sharpens the image, any distortion would be removed and any mirage would disappear.

You can read about Fata Morgana to see how these distance object are explained within the commonly accept ideas about the curvature of the earth.  However, mirages are distinctive by their distorted image and changing conditions.  In the above video of Chicago, there is no distortion.

Spherical Geometry and Empirical Evidence

I took a trip recently to Vancouver Island via the Horseshoe Bay ferry to Nanaimo (https://en.wikipedia.org/wiki/Nanaimo).   It is approximately 36 miles (58km) west of Vancouver.  I took a few videos of the beautiful scene from the deck of the ferry.   On the way back to Vancouver (just as the ferry left the harbour) I took video of Vancouver and the UBC endowment lands.

As you can see in this video, the land mass to the east is clearly visible on the horizon, The ferry itself is has 7 decks (excluding the bridge).  The bottom deck is quite high to allow for over-height vehicles.  But if you allow for double the height of the bottom car deck (5.4m) which are decks 2-3 and every other deck is 2.7m then the total height to the highest passenger deck is 4×2.7m + 5.4m = 16.2m or 51 feet.   I wanted to ensure that we include the height of the ferry in the final calculation.

The video shows a small island called snake island which is approximately 33 miles from Vancouver and 29 miles to the UBC endowment lands.  Using spherical geometry, the endowment lands should be 8x29x29/12 or 560 feet *below* the horizon.  If you go to the UBC endowment wiki page (https://en.wikipedia.org/wiki/University_Endowment_Lands) you will see that it is only 260 feet above sea level.  If we add the height of the ferry we get a total of 311 feet which would require the endowment lands be 249 feet below the horizon.  And since the endowment lands are the highest peak along the coat between Richmond and Vancouver, no lands mass should be visible at all.

I’m going to work on getting better video with a zoom lens to show Vancouver from Nanaimo.  Anyone riding the ferry will be able to do this experiment by themselves while using their own eyes (which have greater definition than the camera).  You don’t need to take my world for it.

In conclusion, it would seem that either the concept of a spherical earth is incorrect or the size of the earth is larger than we are being told.  In either case, it seems that the knowledge being given to us is incorrect.

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 https://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 https://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.

Does Gravity make sense?

If you actually think about how gravity is supposed to work, you might find that there are significant issues with the theory.  I’m just pointing out some factors that might be furiously refuted since without gravity (as it is proposed by the majority of scientists), the universe would be a completely different place.  So I’m moving into a territory that is highly guarded and sanctified.

If we grant that there is a force “pulling” towards the center with a force of 9.8 Newtons/kg at all points on the ‘globe’ simultaneously *and* a centrifugal force that pushes away from the center at such insignificant values, then magical traits would have to be applied to the centripetal force. To put this into perspective, if you are standing on a disk (r=10m) that spins at 10 m/s and a person with a mass of 72kg stands on the edge of this disk, they would feel 720N of outward force (centrifugal). The only way that the centrifugal force can be ‘balanced’ is by having the person hold onto the disk with a greater force than 720N. In other words, the centripetal force. If this person stands close to the center of this disk [cos(89)X10] then then would only have to hold on with 0.2N.

http://www.calctool.org/CALC/phys/newtonian/centrifugal

So what is the difference between gravity and centripetal force? Of course centripetal force is valid if the object is attached to the center of a spinning globe by a rope or some kind of real tensor (like a tree rooted in the ground). However, centripetal force can not be applied to an object that is not tethered to the center of a spinning globe unless you are applying some kind of magical grappling force to it (I hope not). So for objects like people or cars or dogs, you can only apply gravity and the centrifugal force (and no the dog is not tied to a tree by a leash). So we are left with a tiny centrifugal force and gravity.

A 72kg person at the equator will have 705N of force due to gravity on them at all times (9.8 N/kg * 72kg). The centrifugal force at the equator is only 2.4N or an effective force of 702.6N. At the 89th parallel it is effectively null. So the centripetal and centrifugal forces are inconsequential to our experience on this ‘globe’.

What we are now dealing with is the 702.6N-705N of force that is pulling down on the average human. So 9.8N of force is being applied to every atom (and supposedly a lot of empty space) in that body so that the person in question ‘weighs’ 72kg. However, the force of gravity on an object reduces at the square of the distance from the source. So the hydrogen atoms at the top of my head weigh less than the hydrogen atoms in my feet – unless I’m lying down or standing on my head.

https://en.wikipedia.org/?title=Inverse-square_law

To get a real measurement of gravity we need to get the source value of gravity on Earth at its core. As far as I can tell, there is no clear answer as to how gravity functions below the surface. Some theorize that it decrease as you leave the surface and head towards the core. But how can that be? This means that the surface of the Earth is the center point of mass. So how can we get the actual source value for gravity? Without that, no one can give a real answer as to what the value for gravity on the earth’s surface is. The only answer I have is to reverse the formula when heading towards the core of the earth. At 4000 miles from the earth’s core gravity is 9.8 N/kg. The surface area is A = 4πr^2 @ 4000 miles = 200960000. So the energy or intensity decreases (divided by 4) as the distance r is doubled (or increases and inverted going in the other direction). At 2000 miles then the intensity of gravity would be 9.8 N/kg * 4 = 39.2 N/kg. At 1000 miles it would be 156.8 N/kg. 500 miles would be 624 N/kg; 250 miles 2496 N/kg…etc..etc.

Therefore, 8000 miles from the core of the earth would be 2.45 N/kg which makes the idea of weightlessness at 200 miles above the surface a little suspect. I don’t see how an object like the ISS (370,131kg) is able to ‘float’ at a orbit of only 4200 miles from the earth’s core. It would still be have an effective ‘weight’ very similar to what it would have on the earth’s surface (about 2.5% less ‘weight’). Even at 8000 miles the ISS would ‘weigh’ 92,532kg and would still fall back to earth. Until the object approaches the Moon at approximately 110,000 miles, the gravity of earth will always be greater. However, gravity due to the sun will begin to pull at a greater force than both the moon and the earth at similar distance. With this in mind, I don’t see how space travel is possible without massive amounts of fuel and propulsion.

As an aside, you can’t really have an effective gravity ‘at sea level’ since there can’t be a ‘sea level’ on a globe due to the roundness of the water (which really makes no sense at all). And if we are on a spinning globe, there seems to be a magical grappling effect on every water molecule via the centripetal force that keeps the water in a spherical shape. This then puts the concept of orbiting objects into question since the centripetal force is the only thing keeping it in a circular orbit (or is it pear shaped…I can’t keep up) and the centripetal/centrifugal force is a fraction of the ‘pull’ of gravity, then all objects must fall. The counter argument that since the object is in the ‘vacuum’ of space and a force perpendicular to the pull of gravity (ie. Newton’s first law) which keeps it in orbit is nonsensical since the force of gravity is greater than the centrifugal force and weightlessness can’t be within such proximity to the earth’s surface. A constant pull is being applied perpendicular to all orbiting objects therefore they cannot ‘float’ in ‘space’ without the addition of acceleration (ie rockets or other forms of propulsion). And finally, the question of *how* a planet stays in an orbit without additional forces is beyond me. What is supplying the additional acceleration? The gravitational forces of the sun would have long ago exhausted any capacity to stay in an orbit.