Is the world a heliocentric globe or a geocentric plane.
Why not also consider a geocentric globe or a heliocentric plane?
there are plenty of explanations inside of Youtube.
Many things you find on youtube, especially allegedly supporting FE, is wrong.
However, after seeing plenty of evidence to support the case that there was no lunar landing and it is questionable whether any NASA/SpaceX rocket has even left our upper atmospheres, I do put a big question mark on the Globe theory.
I am yet to find any evidence to support that case, just nonsense. Perhaps you can provide some?
Regardless, even if the moon landing was faked there is still plenty of evidence that rockets have left the atmosphere, such as GPS and various other technologies provided by satellites which depend upon them to work.
Even without that, Earth was known to be a globe thousands of years ago, long before NASA, so that isn't a reason to question it.
1) Can you provide an accurate mathematical formula for calculating the curvature of a sphere? This same formula must be consistent with any sphere no matter what it’s size (Tennis ball, wrecking ball, Moon, Earth or Jupiter).
That depends upon what you mean by curvature, and to what extent.
I will assume you mean how much it drops for a given distance.
For small distances you can use a right angle triangle.
You have a line tangent to Earth with a length of d.
You have a line at right angles connecting to the centre with a length of R.
You have a line going from the centre to a point along the first line with a length of R+h.
This gives:
(R+h)^2=R^2+d^2
R^2+2*R*h+h^2=R^2+d^2
2*R*h+h^2=d^2
h*(2*R+h)=d^2
Then with the approximation that R is much greater than h (again, this is for small distances) you can take 2*R+h~=2*R.
This gives:
h*2*R=d^2.
thus h=d^2/(2*R).
If you stick in some values and convert it you get approximately the 8 inches per mile squared.
However this doesn't include refraction which depends upon the atmosphere and conditions.
This formula can be done the other way around to find the distance to the horizon (d) based upon eye height (h).
Also, when calculating how much should be hidden you need to consider both sides of the horizon, the section between you and the horizon and the section between the object and the horizon.
2) How does a person visually experience the curvature of the Earth? My understanding of basic logic shows that as a person is closer to ground level, his viewing distance is shorter. As a person rises in altitude the is more available to see. In saying this, how high does a person need to be in order to see the ground drop away? This can be proven, in my opinion, if an object with a height of several hundred metres can only be seen as being a few metres above the horizon from a distance.
The horizon is how.
You can measure the angle of dip again using a right angle triangle. (the same one as before).
The angle of dip will be the angle subtended at the centre (if you need this explained I can).
This is theta in the following formula:
cos(theta)=R/(R+h)
So how high you need to go depends upon what you are using to try and measure it.
Surveyors with accurate theodolites can easily measure it at ~1.5 m above ground level.
Without a tool to measure it, you don't stand much chance of easily seeing it until quite high up.
However, the mere existence of the horizon (other than at mountains) is the curvature.
The horizon can only be the edge of Earth. Otherwise it would just fade to a blur.
The fact that the horizon moves around as you move shows the edge of Earth is everywhere, which only happens on round objects (like a sphere).
We can also clearly understand the horizon is the result of curvature rather than perspective due to how objects act near the horizon. If it was the result of perspective, as objects approach the horizon they would just shrink and disappear as a point, some still quite high in the sky.
If it was due to a curve, assuming the object was large enough, it would go over the horizon and start to disappear from the bottom up.
This is an image with a bunch of possibilities shown:
The distance I used was d2, and the height h1, and the angle was a.
3) The August 21, 2017 Solar eclipse was only visible in totality from the Earth in a small band of 71miles wide. Google states that the moon’s diameter is 2159 miles. Using a single light source, how can this shadow area be smaller than the object itself?
Because the light source is larger.
This shadow is where the entirety of the light from the object is blocked.
You can try it with a circle on your computer screen and a coin.
For a generic shadow, there are three regions, the penumbra, the umbra and the antumbra.
The umbra is the region of totality, that is the region where all light from the source is blocked.
That is any region where the coin above completely hides the circle on your screen.
The antumbra only exists for objects which are smaller than the light source.
That is where the obscuring object appears entirely inside the light source, i.e. where the coin appears inside the circle on the screen, with parts of the circle all around.
This is equivalent to an annular solar eclipse. At this distance, the shadow (the region of totality) has no size at all (or could be extrapolated to a negative size).
The penumbra is then basically any other religion where some light is blocked, i.e. regions where the obscuring object partially but not completely covers the light source, without being entirely inside the light source.
Again, a diagram showing this (not to scale):
The red circle is the sun, the blue circle is the moon.
The umbra, penumbra and antumbra are labelled U, P and A respectively.
The surface of Earth sits right around the intersection, where U,P and A all meet.
That is why a large section of Earth experiences the penumbra, and we can have total eclipses where we are in the umbra, or annular eclipses where we are in the antumbra.
This also helps show the common FE model with a close, tiny sun and moon is wrong, as it shows the moon and sun must both be larger than the largest ever observed total solar eclipse.
This is because we know the light source (sun) must be larger than the obscuring object (moon), otherwise we would never have annular solar eclipses. This means the shadow (umbra, region of totality) must be smaller than the moon.
4) Can you provide the mathematical equation for measuring the tilt of the earth’s axis? How/Where can this tilt be proven correct, not a religious statement of ‘believe with no evidence’
This is (IMO) most easily done by measuring the angle of elevation of the sun at solar noon over the course of a year, in regions outside the Arctic and antarctic circles.
The tilt of Earth is evidenced by the sun not remaining in the same position. It is the angular offset of the axis of Earth's rotation with the axis of Earth's orbit.
If it was 0, there would be no variation and the sun would always appear at the same angle of elevation (at solar noon).
If there was a tilt, then the sun would appear to move back and forth between the 2 extremes. The change in this angle is twice the tilt.
However, this could also be explained in a geocentric globe model with the Earth rotating and the sun moving along an orbit which is tilted with respect to that axis.
(The FE model can't even explain the apparent position of the sun so it can't even begin to approach this question).
5) Based on the commonly known speed equation(speed=distance/time), earth rotates at the equator at a speed of 1037.5 mph. Why can we not feel this incredible speed? If I hover a quadcopter 30 feet above the ground for 10 minutes, why is it still above my head and not 173miles west of me?
You don't feel speed, at all, ever.
Humans are completely incapable of detecting speed.
Instead they detect acceleration.
If you are riding in a jumbo jet, do you feel like you are flying really fast? No.
You can get up and walk around just as easily as on the ground. All you feel is the turbulence, the change in motion.
As for your quadcopter, that is likely because it is using a camera to keep track of its position and correct for it, or a very good GPS lock.
Slight breezes or imperfections means they will not stay in the same spot without correcting.
I have one of my own and it will happily stay mostly in place with its GPS lock (it drifts around a bit due to GPS only being accurate to a certain degree).
If I instead switch it to alt-hold mode, so it no longer keeps its position fixed, it will drift with the wind.
However, it doesn't drift the 173 miles, because the air, for the most part, is moving with Earth.
Again, this can be related to planes and trains.
If you jump on a plane do you find yourself slamming against the back of it? No.
That is for 2 reasons:
You have inertia, and thus will keep moving along with the train.
The air is also moving with it and thus wont slow you down.
However, if you change your latitude enough you can have some effect. That is seen with large scale weather systems rotating in one direction in the northern hemisphere and the other in the southern hemisphere. But this is a result of the velocity required changing.
6) How does a passenger airplane land on a North-South runway if the Earth is rotating 1037.5mph in a West-East direction?
Because it is also moving sideways that amount.
7) If I take a plane from Bangkok(Thailand) to Bangalore(India) that are both 13degrees N, the flight time is 3hr 40min going an east to west direction. Distance between them is 1539miles. Knowing that the rotational speed at 13degrees N is slower than the equator, but assuming it’s still pretty fast, let’s say 800mph – Why does this flight not take less than 2 hours? All the plane needs to do is reach a height and simply ‘wait’ for Bangalore to come to it.
It would need to leave the atmosphere, and ditch all its speed, then speed back up to land on the runway without overshooting.
Instead planes fly in the atmosphere, keeping their speed.
Also note that they can do some pretty extreme cross wind landings.
Is there a speed at which point an object is moving faster than the gravitational pull? Is it possible for an object with no self-propulsion method to overcome gravity and leave the atmosphere into zero-gravity space?
It would need a propulsion method to do so, such as a rocket.
It would also need to continue propelling itself to oppose air resistance.
However, once in space with negligible air resistance, it can continue going without further propulsion and without crashing back down.
Satellites in orbit will do so. (but the slight air resistance does mean they slowly fall).
If you could ignore air resistance you just need to be at the escape velocity, which is around 11.2 km/s.
So if you had a magic object which had no air resistance and you fired it at around 11.2 km/s or faster, it would not fall back down.
Also, space isn't 0 g. 0 g is an object in free fall such that all parts experience roughly equal acceleration from gravity with no outside forces acting upon it.
7) Can you explain seasonal changes? Why is it winter in the central part (northern hemisphere) while it is summer in the outer part (southern hemisphere)?
Typically they explain this by having the sun change position.
During the northern summer the sun circles north of the equator, making it hotter there and colder in the south. During the southern summer it circles south of the equator, making it warmer there and colder in the north.