Ok since you think I am just repeating what others say, and you do not understand the math, I will copy paste from some of my earlier posts. It was dealing with the force that would "throw you off" the Earth from the 1000 mph speed.

Also the Earth isn't starting and stopping suddenly, you are riding on it. Therefore, if you want to simulate that just think about the train example. Your motion while on the train and the trains constant motion have no real bearing on each other. You were already traveling at the speed of the train, you moved and then stopped relative to the train, therefore you would not "feel" the movement of the train in relation to what you do. Yes I understand if the train is turning, or not on absolutely smooth rails, or not actually travelling at a constant speed you would feel those changes in velocities. But we are talking about the Earth rotating at a constant rate (roughly, it is slowing down, and wobbling a bit).

Lets do some fun math

In the formula I posted above for centrifugal force, and Alpha showed earlier also I will use catboy's numbers:

So

** a** is the centrifugal acceleration, aka the force trying to throw you off of the spinning Earth in this case.

And

** r** is the radius of the Earth ( 1/2 of 7,926.3352 miles then multiply by 5280 to get feet = 20925524.928 ft)

And

** v** is the velocity of the spin at the equator (1,037.554 mph multiply by 5280 to get ft per hour = 5478285.12 then divide by 3600 to get ft per second = 1521.746 ft per second)

This gives:

a = (1521.74 ft/s)

^{2}** /** 20925524.928 ft

a = 2315692.6276 ft

^{2}/s

^{2}** / **20925524.928 ft

a = 0.1107 ft/s

^{2}Gravity = 32.174 ft/s

^{2}So what percent is that of gravity?

0.1107 ft/s

^{2}** /**32.174 ft/s

^{2}equals 0.00344

So 1/3 of 1 percent of the force of gravity reduced at the equator.

Amazing how math works.

If

Then we could do a couple of things, first we could figure out what the velocity would have to be to overcome gravity.

So lets figure out the velocity needed to overcome gravity.

So if

currently we showed

a = 0.1107 ft/s

^{2} we need something greater than 32.174 ft/s

^{2}so we will use

a = 33.174 ft/s

^{2}and

r = 20925524.928 ft

then to solve for v we would need to rearrange the formula

v

^{2} = a*r I multiplied both sides by r to move it from one side to the other. I love how algebra works.

v

^{2}= 33.174 ft/s

^{2} * 20925524.928 ft

v

^{2} = 694183363.961472ft

^{2}/s

^{2}take the square root of both sides

v = 26347.3597 ft/s

The diameter of Earth is 7,926.3352 miles according to catboy, I think its pretty close we should use it.

so divide v by 5280 ft to get miles (which would be per second for velocity here) (26347.3597 ft

** /** 5280 ft ) = 4.9902 miles per second (lets say 5 mps)

Multiply the 5 miles per second by 3600 to get mph 18000 miles per hour (funny that close to what the ISS has to travel at to keep in orbit)

18000 miles an hour? The diameter of Earth is only 24,901.295401 miles. So we would be seeing a full day and night in about 1.38 hours. And that's basically barely counteracting the force of gravity at the equator. Still not fast enough to throw us off really. I guess as long as you were traveling at that velocity you could jump really high and as you slowed down you may sink back down to the Earth. But hey lets just say that's where we need to be at.

Ok lets do a second scenario. With the current rotational velocity what would the circumference have to be to overcome gravity?

Back to the formula above, we would rearrange it to be

r = v

^{2} ** /**a If my algebra is correct.

So from the previous fun

a = 33.174 ft/s

^{2}and

v = (1521.74 ft/s)

^{2}v = 2315692.6276 ft

^{2}/s

^{2}so

r = 2315692.6276 ft

^{2}/s

^{2}** /** 33.174 ft/s

^{2}This gives us a radius of 69804.4441 ft. Multiply that times 2 for the diameter

Diameter of 139608.8881 ft divde this by 5280 for miles

In miles we have a diameter of 24 miles. Wait???

Diameter times pi for the circumference gives 75.40 miles?

?

So to keep the 1000 mph figure to have enough force to overcome (barely) gravity we have to be living on a world that is 75.4 miles around???