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**Flat Earth Theory / Re: Why do objects fall at dofferent speeds?**

« **on:**July 27, 2021, 01:38:16 PM »

This is the difference between arithmetic and science. You are putting data into a calculator and expecting it to work it out for you. You can't do that if you haven't got your head around the science first. If we actually wanted to know the terminal velocity of the skydiver, or how long it takes him to impact, the calculator would be great. But we don't. If we had 2

Lets see where we disagree on the science:

Round Earth.

The skydiver has mass, and he will be accelerated downward by a force of 9.81 m/s/s multiplied by his mass.

At the instant of release, he will feel weightlessness, as the force of gravity is not being opposed.

As soon as he develops airspeed he will feel an opposing force upward due to aerodynamic drag.

The drag force is proportional to his size and shape, and to the square of his airspeed.

As soon as he develops airspeed, and hence drag, he will begin to feel some of the effect of gravity.

Because he is feeling a downward force (Gravity) opposed by an upward force (drag), his airspeed will increase at <9.81m/s/s.

As airspeed increases, the force of drag will increase.

At a particular airspeed, the increasing force due to drag will equal his mass multiplied by 9.81 due to gravity. He will stop accelerating. His airspeed will be constant thereafter, until impact. This is the Terminal Velocity (TV).

He is no longer weightless, as he is being supported by the force of drag so feels the effect of gravity.

The force of drag is not a constant; it increases with airspeed until stabilizing at terminal velocity.

If he was a styrofoam mannequin, but with the same aerodynamic size and shape, the upward drag-force would equal his gravitational down-force at a reduced airspeed, so his TV would be lower, so time-to-impact would be longer.

Flat Earth.

Prior to jumping, the skydiver is at identical altitude to the RE case.

The human skydiver has identical mass and aerodynamics.

Prior to jumping, skydiver has the same upward velocity and (UA) acceleration as Earth. His airspeed is zero. (His speed relative to the universe is the same as Earth, a gazillion mph, but we are only interested in speed relative to the Earth/Atmoplane).

He jumps....

For a few milliseconds his airspeed remains zero, but he is no longer accelerating upward and his universal speed is constant. He experiences weightlessness.

At no point is he subjected to acceleration downward.

The Earth continues to accelerate upwards at 9.81 m/s/s. As the Atmoplane is static with respect to Earth, it also accelerates at 9.81m/s/s, unmitigated by any aerodynamic forces.

As the skydiver has constant Universal Velocity, but the atmoplane is accelerating upward, then he will begin to experience an upward airspeed.

As the airspeed increases at 9.81m/s/s, the skydiver will become subject to the force of drag, which will increase identically to the RE case, as his aerodynamic properties are identical.

As soon as he develops airspeed, and hence drag, he will begin to feel some of the effects of UA; no longer weightless.

This upward force will be opposed by the inertia of the skydiver due to his mass, which is identical to the RE case.

As he continues to accelerate upwards with respect to the universe, the rate of acceleration will increase, due to increasing airspeed, until the upward force of drag equals his mass multiplied by 9.81 (due to UA). This is TV.

At TV, his acceleration is now identical to the Earth/Atmoplane, but his Universal Velocity is lower, so the Earth will be catching up.

He is no longer weightless, as he is being accelerated at 9.81m/s/s by the atmosphere.

If he was a styrofoam mannequin , the upward drag force would equal his mass multiplied by UA at a lower airspeed, so he would have a lower TV. Because his TV is lower, his UV is higher, hence closer to that of Earth. Because he is now accelerating at the same rate as Earth, but spent less time un-accelerated, earth will still impact him, but time to impact would be longer.

In summary; same mass, same drag, same acceleration. Different pos/neg signs.

Please let us know if you (or anyone) have a problem any these statements.

**identical**spheres of pure gold, and wanted to know the weight one of them we could weigh it using some kind of scales to balance against a known mass. Or we could determine the volume of water it displaces, and multiply by the density of pure gold. But we don't, we only want to know if they are identical, so we can just balance them together on scales. No calculations. No numbers.Lets see where we disagree on the science:

Round Earth.

The skydiver has mass, and he will be accelerated downward by a force of 9.81 m/s/s multiplied by his mass.

At the instant of release, he will feel weightlessness, as the force of gravity is not being opposed.

As soon as he develops airspeed he will feel an opposing force upward due to aerodynamic drag.

The drag force is proportional to his size and shape, and to the square of his airspeed.

As soon as he develops airspeed, and hence drag, he will begin to feel some of the effect of gravity.

Because he is feeling a downward force (Gravity) opposed by an upward force (drag), his airspeed will increase at <9.81m/s/s.

As airspeed increases, the force of drag will increase.

At a particular airspeed, the increasing force due to drag will equal his mass multiplied by 9.81 due to gravity. He will stop accelerating. His airspeed will be constant thereafter, until impact. This is the Terminal Velocity (TV).

He is no longer weightless, as he is being supported by the force of drag so feels the effect of gravity.

The force of drag is not a constant; it increases with airspeed until stabilizing at terminal velocity.

If he was a styrofoam mannequin, but with the same aerodynamic size and shape, the upward drag-force would equal his gravitational down-force at a reduced airspeed, so his TV would be lower, so time-to-impact would be longer.

Flat Earth.

Prior to jumping, the skydiver is at identical altitude to the RE case.

The human skydiver has identical mass and aerodynamics.

Prior to jumping, skydiver has the same upward velocity and (UA) acceleration as Earth. His airspeed is zero. (His speed relative to the universe is the same as Earth, a gazillion mph, but we are only interested in speed relative to the Earth/Atmoplane).

He jumps....

For a few milliseconds his airspeed remains zero, but he is no longer accelerating upward and his universal speed is constant. He experiences weightlessness.

At no point is he subjected to acceleration downward.

The Earth continues to accelerate upwards at 9.81 m/s/s. As the Atmoplane is static with respect to Earth, it also accelerates at 9.81m/s/s, unmitigated by any aerodynamic forces.

As the skydiver has constant Universal Velocity, but the atmoplane is accelerating upward, then he will begin to experience an upward airspeed.

As the airspeed increases at 9.81m/s/s, the skydiver will become subject to the force of drag, which will increase identically to the RE case, as his aerodynamic properties are identical.

As soon as he develops airspeed, and hence drag, he will begin to feel some of the effects of UA; no longer weightless.

This upward force will be opposed by the inertia of the skydiver due to his mass, which is identical to the RE case.

As he continues to accelerate upwards with respect to the universe, the rate of acceleration will increase, due to increasing airspeed, until the upward force of drag equals his mass multiplied by 9.81 (due to UA). This is TV.

At TV, his acceleration is now identical to the Earth/Atmoplane, but his Universal Velocity is lower, so the Earth will be catching up.

He is no longer weightless, as he is being accelerated at 9.81m/s/s by the atmosphere.

If he was a styrofoam mannequin , the upward drag force would equal his mass multiplied by UA at a lower airspeed, so he would have a lower TV. Because his TV is lower, his UV is higher, hence closer to that of Earth. Because he is now accelerating at the same rate as Earth, but spent less time un-accelerated, earth will still impact him, but time to impact would be longer.

In summary; same mass, same drag, same acceleration. Different pos/neg signs.

Please let us know if you (or anyone) have a problem any these statements.