Stationary with respect to what?
You have now.
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DAMN YOU TO HELL!!
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“…the 61-year-old limousine driver…”
I here I was, thinking that limo drivers were only idiots.
PS: Why doesn’t he just launch a camera? Oh. Perhaps he doesn’t want to take the chance of having the camera switched out before he gets to the rocket.
Here’s one:
- Are voter literacy tests (a) legal (b) a good idea.
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Well darn. I didn’t see this in time. Sony for my other reply.
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So many good lines in that article, almost seems like The Oinion and not the WaPo.
“Hughes promised the flat-Earth community that he would expose the conspiracy with his steam-powered rocket, which will launch from a heavily modified mobile home — though he acknowledged that he still had much to learn about rocket science.”
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Donald Trump was elected president, and every second since then has been Poe’s Law in action.
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Actually engaging a flat earther is the problem, the second you do you lose. Never start an argument with an idiot, they will only lower you to their level and then beat you with experience.
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Well, that’s micro-gravity. Macro-gravity works different.
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Whatever you like, whatever makes the most sense to you given the nature of the problem. You get the same answers for each individual observer’s interpretation of relative velocities no matter what point you choose as stationary.
The formula for determining the relative relativistic velocity u’ between any two observers, as a function of their velocities u and v according to a stationary observer is u’ = (u - v)/(1 - uv/c2). If you measure all velocities as fractions of c, that last 1/c2 is just 1 and it drops out. If u and v are both small relative to c, then the last term approaches zero and the relative velocity difference u’ = u - v, which is what we expect at every day slow velocities where two cars approaching each other at 50 km/h have a 100 km/h collision. But that’s not what happens if they have speeds of 0.5 c.
Suppose you have three observers in a line, A in the middle, B on the left, C on the right. A is stationary (velocity = 0), B is approaching A at velocity +0.5 (left to right) and C is approaching A at velocity = -0.5 (right to left). According to A (velocity = 0), B is approaching at u’ = (+0.5 - 0)/(1 - (0.5)(0)) = +0.5/1 = +0.5, as expected. According to A, C is approaching at u’ = (-0.5 - 0)/(1 - (-0.5)(0)) = -0.5/1 = -0.5, as expected. According to C, B is approaching at u’ = (+0.5 - -0.5)/(1 - (+0.5)(-0.5)) = 1/(1 + 0.25) = 1/1.25 = 0.8, not 1, because relativity. So C thinks B is coming at it at +0.8, and B thinks C is coming at it at -0.8.
But what if it’s not A that’s stationary, it’s C? Then velocity C = 0, velocity A = +0.5 (moving left to right toward C), and velocity B = +0.8 (as we just determined). How fast does moving A think moving B is approaching it? u’ = (0.8 - 0.5)(1 - (0.5)(0.8)) = 0.3/(1 - 0.4) = 0.3/0.6 = +0.5. Which is exactly the same thing A thought when we thought A was stationary.
According to B, A approaches at -0.5 and C at -0.8. According to A, B approaches at +0.5 and C at -0.5. According to C, B approaches at +0.8 and A at +0.5. It doesn’t matter which one you choose as the stationary point, or even if the stationary point is something else entirely. Same observations for all three. Which is the whole point of relativity: there is no universal stationary point, the reference frame is relative to the observer.
Dude. Must be!
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