New in cosmology

So what. Are you now comparing a rail gun to an ICBM? 

Not explicitly. But your point is an obfuscation designed to protect your ego.

A better response would be to improve your competence, but that ship sank a long time ago. You have implicitly acknowledged that there is no way for you to become competent enough to do the simpler calculation of the path of a cannonball.

For clarity, note that there are two issues here. The first is the correct understanding of gravity and how it allows the calculation of a ballistic path. The second is that the Earth is spherical, so the gravitational field is non-uniform on a larger scale.

You need to understand the simplest stuff first before you move on to the more complex stuff. Unfortunately, you can't.

Please note that I am not suggesting you use any particular method for calculating the path of a cannonball. While only a correct method will get the answer reflected in reality, you are welcome to use whatever crap you like so we can see what your prediction looks like.

So you believe Werner Von Braun's V1/V2 rockets journeyed into outer space before crashing down into the earth? 

I am surprised you are not able to read the historical facts.

The V-2 was the first object to reach what is defined as near space (its maximum altitude was 175 miles and much of its flight was ballistic) but the V-1 flew at low altitudes - a mere 5 miles - which is not as high as Everest, and well within the atmosphere.

Alright, take a cannon and shoot the cannon ball straight up. 

Typically it will remain aloft for more than 10 seconds. Often closer to 14/15 seconds. 

During that time the earth is alleged to have rotated some 5000 plus meters(480 meters per second). So where will the cannon ball land? 

If you are standing in the aisle of a 737 moving at 500mph, i.e. 223 meters per second, and you jump into the air, hitting the ground 1/5 of a second later, where do you land?

So, now you claim the earth is surrounded by a protective steel/aluminum layer?

Say you were on top of a train traveling 100 mph and standing on the very end. Now jump as high as you can......will you come back down on the train or the tracks? 

The only role of the aircraft shell is to give you something to stand on and take the wind out of the calculation!

On the Earth, you should be aware the airspeed is not high relative to the ground, so it is quite similar.

I understand that this is going to be too difficult for you, but the key thing to the cannonball calculation is that the speed of the Earth's surface changes very little in the time a cannonball is in the air. The change in the velocity of a point on the surface is quadratic in time, which keeps it small for times that are very short.

Lets take your own example of the plane again. 

You would be correct with that assumption, however, incorrect when comparing it to the earth. The only way to truly compare it with the earth would be to make the floor of the aircraft move/rotate, like a conveyor belt within the moving plane. 

Then what would transpire should on remain aloft for a second or two inside the plane if the floor(earth) was rotating underneath them while the plane itself traveled 500 mph? 

You can see how it works ironically enough at many airports that have the moving walkways inside them. Those walking on/with the moving walkway "fly" right by those who are not walking on it reaching the opposite side much faster.. Those walking on/against the moving walkway, move much slower and would necessarily need to walk/run much faster in order to reach the opposite side. Ever try and jump up on one of these moving walkways?

Okay. So, according to YOUR own calculations. 

A moving walkway travels at approximately 1.5mph. For sake of argument lets pretend this walkway is the equator moving west to east approximately 1000mph. 

2 planes sit in the center of this moving walkway. One takes off east and one west. Both traveling 500 mph to each end of the walkway. The walkway is 1000 miles long. Which plane will reach its destination first?

Here is another heliocentric conundrum for you. 

A plane takes off westward from the 1000 mph eastward moving walkway. The planes reaches top speed of 500mph then makes a 180 degree turn after 1 hour flying west and is now heading eastward along with the 1000 mph walkway. If the walkway is on a sphere of 25,000 miles in circumference then how does it reach its eastward destination that has been moving eastward at 1000mph? 

I should point out I am getting a bit bored. It's not that you don't understand these things, it's that your arrogance stops you ever wanting to fix this.

I have told you numerous times that when it comes to theoretical physics/formulas you have me beat.

The problem is that you put far too much faith in them. 

It takes a great leap of faith to believe in the religion of Heliocentrism!

It makes no sense if a plane heads west at 500mph while the earth moves east at 1000mph and that plane can simply turn around and catch up. 

I think it's so simple it goes right over your head!

No, all competent people have you beat in APPLICATION of physics. That's because of two things: they KNOW SOME VALID PHYSICS and they are CAPABLE OF APPLYING IT.

Predicting the path of a projectile (eg a cannonball) given its starting velocity is APPLIED PHYSICS. Historically, while approximate rules were used for a long time, it was Galileo who first arrived at correct way to calculate the path of projectiles and Newton who explained this as part of his theory of gravitation.  Later developments included accurately allowing for air resistance when this matters. 

The key thing is IT WORKS.

Your lack of understanding of the motion of planes on the equator is just the same to me as someone who can't do arithmetic. Your error is lack of clarity about the frame of reference. It's not difficult, but I understand that it is sufficiently non-trivial to confuse some people.

It is not your lack of insight that is unforgiveable, it is your arrogant refusal to make it possible to fix this. I understand that this would be too costly to your ego and you need to psychologically protect yourself.

Let me ask you a question to check that. Would you like you understand the answer to your question about planes flying on the equator? No prevarication - yes or no?

Yes -  please explain how a plane heading west at 500mph while the earth beneath moves east at 1000mph the plane can turn around and catch up to where it took off. If you want to use the equator then by all means do so. 

Then tell me why ALL aeronautical calculations/navigations are based upon a nonrotating stationary flat earth. 

Sure. It's all about frames of reference. It's a little bit more complex (enough to confuse you) because it's a rotating frame, but it's not so difficult on the equator.

The way to make it simple is to stick to a frame that is stationary on the ground. You could define the locations in terms of the number of miles East from a chosen location on the equator. Say the one where the planes start.

**Let's call x the number of miles East of the starting point.**

Then when a plane travels East at velocity v (relative to the ground, of course), x increases at a rate v.  By contrast when a plane travels West at velocity v relative to the ground, x decreases at a rate v (or increases at a rate -v, the same thing).

You can see that in the rotating frame (i.e. everything relative to a point on the equator) the motion is very simple and not confusing.

[There is one physical difference between travelling in the two directions - the centripetal force (or the centrifugal force in the rotating frame) is a little different. This makes the weight of objects flying East a little less than those flying West. But since the effect is of the order of 0.01%, it is perfectly reasonable to ignore it].

Elroch -"The way to make it simple is to stick to a frame that is stationary on the ground."

There is your answer. 

It will only work on a stationary nonrotating flat earth. 

Just like all aeronautical calculations/navigations are based upon. Everything else is theoretical. 

As I said, these are concepts you have trouble with. A frame is just a way of associating numbers with locations (in the rotating frame, time is independent). It is a construct that allows a numerate person to analyse the physical situation.

Just to check, do you accept that locations on the equator do stay at (very close to) constant distances from each other?  If so, you should be able to understand the definition of the frame I gave. Here it is again in big letters:

**Let's call x the number of miles East of the starting point.**

[x is, of course, negative for points to the West of the starting poin. We only deal with distances up to say 10,000 miles in either direction, for simplicity]

Of course, you may have cultured personal dysfunction to such an extent that even such a definition is beyond you. Is this so?

[EDIT: this responded to a deleted post]

Why is your scam better than all the other scams out there that gullible people can be fooled by?