The only digging going on here is you digging a massive hole for yourself, trying to pretend that the reason we haven't made it to step 2 is anything other than the fact you don't understand step 1.
V = R*ω where V = instantaneous tangential velocity (m/s), R = radius of rotation (m), and ω = angular velocity (rad/s).
The tangential velocity at the equator can be calculated using Vequator = Requator*ωEarth.
Substituting the appropriate values for Requator (the Earth’s equatorial radius) and ωEarth (the Earth’s sidereal angular rate, as explained and calculated on the next page) yields: Vequator = (6378.1 km)*(7.292124 x 10 -5 rad/s) = 0.46510 km/s = 1674.4 km/hr (1040.4 mph)
Since the tangential velocity of a point on Earth’s surface is a function of its latitude, the equation V = R*ω can be rewritten as V = R*ω *cosL, where cosL is the cosine of the latitude for a point on Earth. (CosL is 1.0 at the equator and decreases with increasing latitude to a value of 0 at the poles).
A mean solar day is the average time it takes the Sun to make successive apparent passages over a given Earth meridian (longitude) and is the familiar 24-hour period measured by our clocks. However, the Sun appears to move an average of (360° per year / 365.25 days per year) or 0.98563° among the stars during a mean solar day as Earth also revolves around the Sun. Because the Earth's rotation and the Sun's apparent daily motion among the stars are both eastward, the time required for a star to pass over a given meridian as Earth rotates under it can be solved using the following proportion: 24 hrs = one sidereal day 360.98563º 360º Solving for one sidereal day gives (24 * 360º / 360.98563º) = 23 hours 56 minutes and 4 seconds.
Based on the sidereal day, Earth’s true angular velocity, ωEarth, is equal to 15.04108°/mean solar hour (360°/23 hours 56 minutes 4 seconds). ωEarth can also be expressed in radians/second (rad/s) using the relationship ωEarth = 2*π /T, where T is Earth’s sidereal period (23 hours 56 minutes 4 seconds). This method produces a result of ωEarth = 7.292124 x 10 -5 rad/s.
For uniform rotation with constant angular velocity ω, the acceleration is radial and given by rω2 at the distance r from the axis. The radius of the earth at the equator is 6.371 million metres, therefore the outward acceleration at the equator is 6371000 x (0.00007292124) x (0.00007292124) = 0.03387783864 m/s squared.
F=MA
So, let's say you weigh 150kg which I think is probably a reasonable estimate.
F= 150 * 0.03387783864
therefore F = 5.0816757969 N
That would be five newtons of force pushing you outwards from the planet because of the spinning.
That's less than half the gravitational force holding you to the earth, which you also can't feel because it's too small a force.
So there we go, with knowledge of maths, we've simultaneously answered the questions "why can't we feel the spin" and "why don't we fly off the earth because of the spin".
I love maths, me.
Any chance on telling me why the Earth supposedly bulges at the equator based on what you're now saying about how slow it's going?
Laws of motion don't require a medium.
All motion requires a medium. All.
Earth itself requires a medium.
They're called LAWS because they're universal.
Yep meaning it applies to everything. two laws of motion equal using a medium. Both laws are just an extension of one main law. The other so called law does not exist as a law.
They would come with a caveat if they were dependent on a medium, but they have no such caveat, because none is necessary.
A medium is required for everything.
The pressure of the air inside the balloon is what makes the skin of the balloon stretch, yes, and the pressure of the skin of the balloon against the air inside the balloon is what pushes the air out of the hole, yes, but what actually propels the balloon is that the pressurised air is trying to escape in every direction because it's under that pressure, but the only opening into a lower pressure zone (the outside world) from which it can escape is at one side of the balloon, so the motion of the balloon is equal and opposite to the force of the air being pushed out of the hole. Air goes one way, balloon goes the other. Equal and opposite. It's Newton's third law of motion, which you stated is THE ONLY LAW THAT MATTERS a few pages ago.
The air escapes in one direction only. Out of the nozzle against atmospheric pressure.
Medium is irrelevant as long as the pressure forcing the air out of the balloon is greater than the pressure exerted back by the medium it is escaping into.
What do you think a medium is?
You're saying it's irrelevant whilst making it relevant.
As the medium is an almost vacuum, there's next to no pressure trying to force the air back into the balloon, therefore acceleration will occur.
No.
In extreme low pressure there is absolutely little reactionary force against what is expelled which means you get extremely tiny movement.
In so called space which is said to be a vacuum and said to not even be low pressure, your rocket and craft are an impossibility to even get there never mind stay in that set up.
The best any rocket could do is to get to a certain height before arcing back down to the ground. All done in atmosphere.
It's an experiment that would actually produce a more spectacular effect in a vacuum than in the regular air.
Yeah so we're told.
The reality is, it wouldn't work at all.
If you ever get the opportunity to try blowing up a balloon in a vacuum and then releasing it, you should try it. It'll blow your mind.
If you try and see what happens to a near fully deflated balloon in a low pressure chamber you'll see that the air inside it expands as the air around it is evacuated.
Why?
Because the air inside that balloon has to fill the gap left by evacuated air from the chamber, which is why you see the balloon expand.
It's all about clearing the mind and seeing past the stories.
The craft isn't working against itself. The propellant force is acting against the craft.
All the propellant is doing is pushing against the atmosphere. Burning propellant expands massively against the atmosphere and causes that atmosphere to be heavily compressed which creates an immediate crush back reaction to create the very foundation of denser air for the rocket to consistently sit atop of with every second of thrust to weight ratio, ensuring the rocket stays balances and vertical until the fuel becomes spent....in short order.
It actually works much better in a vacuum than it does in air, because there's no air pressure causing friction against the motion, and the propellant is escaping into a medium with much lower density. Have a think about that.
There's none to enable motion.
There's zero reactionary gases to enable a push against. The expelled gases would simply expand massively into the so called space vacuum.
If you don't like the balloon analogy because ti's too complicated for you, try this.
Get a cardboard box, draw a line around its base in crayon, then set off some dynamite next to it.
Same thing.
The dynamite reacts with atmosphere to push the box away.
It creates a massive air blast.
Your space rocket creates nothing. Even by some kind of fantasy of a rocket being placed into this so called vacuum, it's an immediate dead stick.
1000mph in a straight line pretty much is no different to feeling like you're not moving on an aeroplane
We're not talking about a straight line.
In fact there you go that's appx how curved the road would be if it was 12,000 mile radius circle, in fact its actually more curved, you really think you world feel that as a huge spinning curve if you were driving down country on that road?
I think not.
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Not straight, is it?
No, I argued it spins at half the speed of the hour hand on the clock. Keep up at the back.
You can demonstrate this for yourself, by using a 24 hour analogue clock like
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, just point the hour hand at the sun in the morning, and watch it remain in line with it until sunset, and then still be in line with it at sunrise the next day. A simple experiment.
Does your Earth spin at 1000 mph at your equator?
If you say yes then forget using a clock as any analogy to the spin unless you want to also alter how the Earth supposedly bulges at the equator.
Can you explain this?