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#1
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Mxsmanic wrote:
El Maximo writes: It's called orbital velocity http://electronics.howstuffworks.com/satellite3.htm Any object following a ballistic trajectory (disregarding aerodynamic effects) is in orbit. Anthony, that statement makes a road-kill 'possum smarter than you. Try again, and get back to us when you've figured out your error. Rip P.S.: Still waiting for answers to the other issues you ran away from. Yawn... |
#2
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Rip writes:
Anthony, that statement makes a road-kill 'possum smarter than you. Try again, and get back to us when you've figured out your error. Get back to me when you're ready to explain the alleged error. |
#3
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Get back to us when you get a life.
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#4
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"Mxsmanic" wrote in message
... Rip writes: Anthony, that statement makes a road-kill 'possum smarter than you. Try again, and get back to us when you've figured out your error. Get back to me when you're ready to explain the alleged error. I'm sure there are enough definitions of orbit for you to choose from that you wouldn't bother learning, only arguing. In any event, you have exposed another hole in your knowledge. |
#5
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El Maximo writes:
I'm sure there are enough definitions of orbit for you to choose from that you wouldn't bother learning, only arguing. In other words, I am not in error. |
#6
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![]() Well, MX, consider the case of a ballistic track that starts off with an initial vertical velocity that is less than escape velocity. Up, then down, until it hits what someone so nicely described as 'the hard edge of the sky". Is your assertion that is orbital? |
#7
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Well, MX, consider the case of a ballistic track that starts off with
an initial vertical velocity that is less than escape velocity. Up, then down, until it hits what someone so nicely described as 'the hard edge of the sky". Is your assertion that is orbital? Suppose that, as the rock was passing through the soft part of the sky (we neglect air friction because this is powered freefall flight), the earth shrunk to the size of a walnut. The flight path would stay the same, and orbit the walnut. Escape velocity is the velocity required to escape the earth's gravitational field (go up and not come down at all). What we call "orbital velocity" is the velocity required to remain in a circular orbit around the earth, at an altitude of about four thousand miles above the earth's center. At the apogee of a sufficiently elliptical orbit, the actual velocity can be arbitrarily small. Jose -- You can choose whom to befriend, but you cannot choose whom to love. for Email, make the obvious change in the address. |
#8
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Jose writes:
Suppose that, as the rock was passing through the soft part of the sky (we neglect air friction because this is powered freefall flight), the earth shrunk to the size of a walnut. The flight path would stay the same, and orbit the walnut. Exactly. At least someone here is capable of abstract thinking. |
#9
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I used the escape velocity limitation to disallow the obvious
argument. The intitial velocity limitation says 'vertical'. I don't care what size the 'walnut' is, it will be impacted given there was no horizontal velocity component. Even if there was, by the way, the Newtonian physics demand the path will cross the starting point, and the implication to most readers would be that was on a surface. Jose, I don't think from an initial impluse, which is the model I described, you can avoid impact unless the mass of the projectile is such that conservation of momentum would demand the launching surface move enough out of the way of the way to avoid the impact. You really can't go orbital with a initial impulse launch from a realizable surface unless the object gets redirected as it gains altitude. I did not allow that redirection in my model. As for abstract thinking mentioned in another post? I might use a word other than abstract. |
#10
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