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.