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.
Terra Firma 