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Use the terminal velocity formula, v = the square root of ((2*m*g)/(ρ*A*C)). Plug the following values into that formula to solve for v, terminal velocity.[1] m = mass of the falling object g = the acceleration due to gravity. On Earth this is approximately 9.8 meters per second per second. ρ = the density of the fluid the object is falling through. A = the projected area of the object. This means the area of the object if you projected it onto a plane that was perpendicular to the direction the object is moving. C = the drag coefficient. This number depends on the shape of the object. The more streamlined the shape, the lower the coefficient. You can look up some approximate drag coefficients here.
The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object IN A VACUUM near the surface of the Earth
originally posted by: DutchMasterChief
a reply to: choos
I'm no good at math but I think you are forgetting terminal velocity here, like I pointed out several times before. But hey, just keep ignoring that while posting your flawed calculations.
You also seem to agree that the plane would not be near topspeed at the top of the parabole.
Did you take this into account?
The fatal flaw, with the 9.8 m/s2 you keep using in your "calculations"
Furthermore, what if the parabola is made larger so that the top section is bigger?
ISS is a international thing (hint: it's name? International Space Station.
originally posted by: ngchunter
No.
Tell me, how is it the ISS always manages to appear exactly where these telescopes expect it to be? That wouldn't be the case for random observers scattered around if it were just a "plane" flying in the atmosphere; topocentric would reveal the hoax.
originally posted by: DutchMasterChief
a reply to: choos
So even after it was pointed out to you that the number you base your calculations is based on falling in a vacuum, and that you are ignoring several other factors, you still insist that your speed calculations are correct?