It absolutely does, and once you've grasped and acknowledged those mathematical fundamentals, we can move onto step 2.
Showing how we work out the moon size and distance relies on acknowledging some mathematical fundamentals.
If I just start throwing equations at you for the moon size you're just going to say "no, this makes no sense, prove it from the absolute ground up, the fundamentals", just like you have for everything else we've shown you, so this time that's where I'm starting.
Once we get past basic geometry we can move onto the next step: trigonometry.
Trying to deflect away from being called out on deflection. I'm loving this new meta. What shall we call it, deflectception?
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.
No it doesn't, it remains curved. We showed you the numbers earlier, and you can test it yourself if you obtain sufficiently accurate measuring equipment. It doesn't matter how many times you parrot your lies about it being flat, it still isn't flat.
Laws of motion don't require a medium. They're called LAWS because they're universal. They would come with a caveat if they were dependent on a medium, but they have no such caveat, because none is necessary.
Just like I described with the balloon. That's how.
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.
Yes, you've mastered this observation. Now follow the rest of what's happening logically. I've spelled it out for you above.
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. 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.
It's an experiment that would actually produce a more spectacular effect in a vacuum than in the regular air. 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.
The craft isn't working against itself. The propellant force is acting against the craft. 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.
It really isn't, you daft sod.
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.
