Beautifully put Alan.
Beautifully put Alan.
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I think it's a science. It's logic, and that is what we use.Science is supposed to be the investigation of the world around us by observation and experiment.
Not sure logic fits that, pure maths doesn’t.
Maths has lots of proofs, which science doesn’t.
You’re not ignoring the laws of conservation of momentum are you, what’s your coefficients of restitution for these collisions, gasses and solids ?I carefully did not violate momentum conservation. I simply computed the desired momentum and gave that to my small 'meteor'. It's small enough to have no gravitational effect on the experiment before the collision.
I think coefficient of restitution has more to do with collisions between two objects that remain reasonably distinct. So we have two objects before, and a reasonably uniform moving blob of plasma afterwards. Coefficient of e=0 in the ideal case. It doesn't bounce back (e > 0) if that's what you're asking. More like a e<0 (like a human getting hit by a bullet that passes through), which is why I break it up just before collision. I want to move Earth, not shoot a hole through it.
Heck, the whole problem can be solved by having the major mass (sun say) be a black hole. That can't explode when you smack it with something. e=0 necessarily. Fire a small but very high momentum object at it and the black hole will acquire all that momentum without bits flying all over the place. Perfect for our experiment. How long before Earth deviates from its path when the sun abruptly changes velocity by .9c in some direction? GR can answer that because there's no violation of any laws in it, and we don't need to worry about the dynamics of a supernova-scale explosion.
Would you care to put a number to the mass of a photon?
ii. If the Sun suddenly disappeared, it would take about eight minutes for Earth to become dark (due to the speed of light). How long would it take to feel the absence of the Sun's gravity?, Astronomy magazine, 2012
And the force propagates through the spring. Gravity does that only indirectly. It pulls the weight which changes the stress on the spring. If you were put in place of the weight you would feel the acceleration as the stress on the spring slowed you down. You would not feel this acceleration in free fall. It's force Jim, but not as we know it.
So JefferyWhile you can state F = ma this is not the same as saying F = mg. No force is felt in the second equation. So F ≠ mg.
If what you state is correct, how do my bathroom scales work ?
Your bathroom scales work by preventing gravity from moving an object. The tension in a spring will do it. The tension propagates through the spring until it overcomes gravity. Gravity does not propagate through objects, otherwise you would be able to measure your acceleration during free fall.
Look at the slinky drop video in slow motion and see the tension acting against gravity. It is not gravity propagating through the slinky but the release of tension.
This is why the two Fs in the equation about are not the same. Also why I believe you will never be able to properly measure a difference between mi and mg.
While you can state F = ma this is not the same as saying F = mg. No force is felt in the second equation. So F ≠ mg.
But if we delivery a 10 ton payload in a 10 ton transporter parked alongside will the ISS slow down and by how much?? We can assume the ISS weight 400 tons so it increases in combined weight by 5%
Is there any method that you can think of that could achieve this? Have a stab at it.Similar to the sodium lamp, I believe it would be possible to derive the speed of light using the red/blue shift due to interaction of light with a gravitational field, for example
'Come to rest' is dang undefined without frame reference. If the ship is constantly accelerating, then it is gaining speed and not coming to rest. It can cease accelerating and therefore be at rest in its own frame, but the clocks will be very out of sync after that.
So if I follow that logic correctly if the clocks were in sync at departure they would be in sync once they/the rocket came to rest in reality ?.
But as far as I know, Einstein never addressed the question of simultaneity-at-a-distance, according to the person who sometimes accelerates.
I'm now thinking that Einstein DID address the question of simultaneity-at-a-distance, according to a person who sometimes accelerates. I remember that Einstein predicted that for two clocks stationary in a gravitational field, the clock higher in the field (farther from the source of the field) will run faster than the clock lower in the field. He got this result long before he published his paper on his GR theory (and long before he arrived at his GR theory). He got it by solving an SR problem, and then invoking his "principle of equivalence" between acceleration and gravitation.
The SR problem he solved was for a rocket, not in the presence of any gravitation field, with a clock at the front and a clock at the back, that is undergoing a constant acceleration. He determined that the clock in front will run faster than the clock at the back. So he DID determine what an accelerating observer at the rear of the rocket will conclude about the current time at the front of the rocket.
I've heard that Einstein wrote a paper in 1907 (2 years after his SR paper, and 8 years before his GR paper) that may have discussed the above result. I'm currently trying to find that 1907 paper (translated into English, of course). I'd like to know if he used CMIF (co-moving inertial frame) simultaneity in getting his result, and whether he made any assumptions in justifying that choice.