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The HARP looked a bit complicated on investigation, might go back there at some later point, but shoot a ray of light away from M, it keeps on going, surely?
Note that vesc = sqrt(2GM/R), i.e. is independent of the mass of the escaping entity, therefore applies equally to photons of no mass.
So - when velocity of m is not escape velocity - at a height from M, a height that will be dependent on the magnitude of the velocity of m, any value of m will encounter a moment of inertia where velocity = 0.
If p = mv, and v = 0, then does p = m?
So - if not moment of inertia, then what is the moment called when m not travelling at v(escape) stops going upwards when v=0?
Does v reduce by the inverse square law with distance from m, I guess is what I'm interested in......and a quick Google shows me that Jeff asked this question here at Naked Scientist's in 2013...So based on what I just read there, a velocity does not reduce by the inverse square law to distance because of time contraction and spatial dilation, but if these are both accounted for, by what proportion to a straight upward distance (from Earth) does a velocity reduce by?Is it -9.807m/s^2?
And where the distance is spatially and temporally extended, by what proportion is it extended by?Is it 1.5?i.e: In that escape velocity from radius is 1.5 times orbit velocity at that radius...or have I got that wrong?And surely if the distance is spatially and temporally extended on the outbound, it will be the exact equivalent on the inbound?...Hence my thought path of inbound=+9.807m/s^2, and outbound=-9.807m/s^2.
I know that you have told me that p is not really caused by anything, but if a property requires a force to cancel it out, or change its direction?So perhaps I can look upon the scenario as v causing p and a causing v?Although I get the concept of your abstract that p is an integer
of force over time, I'm just trying to understand the mathematical 'structure' here in order to understand the action of p when Planck's h constant is in play in the maths.
Is this saying that a 'constant' velocity will travel 9.807m/s^2 or thereabouts near earth, and then having achieved a distance, at 'constant' velocity, of 2 radii from centre of earth, the 'constant' velocity will be travelling 2.45m/s^2?
In addition to last 2 posts:Ok - back to the trampolines...Surely the 100kg cannonball will be being robbed of more energy every touch down than the 10kg ball.
Quote from: timey on 08/03/2017 12:00:44So - if not moment of inertia, then what is the moment called when m not travelling at v(escape) stops going upwards when v=0? maximum altitudeQuoteDoes v reduce by the inverse square law with distance from m, I guess is what I'm interested in......and a quick Google shows me that Jeff asked this question here at Naked Scientist's in 2013...So based on what I just read there, a velocity does not reduce by the inverse square law to distance because of time contraction and spatial dilation, but if these are both accounted for, by what proportion to a straight upward distance (from Earth) does a velocity reduce by?Is it -9.807m/s^2? for a projectile, v2 = u2 + 2as in the school textbooks. It's a bit more complicated going upwards because acceleration a (i.e. g) is a function of altitude s. If you are interested in escape speed, it's sqrt (2GM/R) where R is the distance from the centre of the earth. QuoteAnd where the distance is spatially and temporally extended, by what proportion is it extended by?Is it 1.5?i.e: In that escape velocity from radius is 1.5 times orbit velocity at that radius...or have I got that wrong?And surely if the distance is spatially and temporally extended on the outbound, it will be the exact equivalent on the inbound?...Hence my thought path of inbound=+9.807m/s^2, and outbound=-9.807m/s^2. You can ignore relativistic effects for escape for all known planets QuoteI know that you have told me that p is not really caused by anything, but if a property requires a force to cancel it out, or change its direction?So perhaps I can look upon the scenario as v causing p and a causing v?Although I get the concept of your abstract that p is an integer integral, not integer Quoteof force over time, I'm just trying to understand the mathematical 'structure' here in order to understand the action of p when Planck's h constant is in play in the maths. dpdx=h/2pi. That's all there is to it.
Quote from: timey on 08/03/2017 14:02:47Is this saying that a 'constant' velocity will travel 9.807m/s^2 or thereabouts near earth, and then having achieved a distance, at 'constant' velocity, of 2 radii from centre of earth, the 'constant' velocity will be travelling 2.45m/s^2?m/s2 is an acceleration, not a velocity.
GO Timey!
Quote from: timey on 08/03/2017 14:55:26In addition to last 2 posts:Ok - back to the trampolines...Surely the 100kg cannonball will be being robbed of more energy every touch down than the 10kg ball. Obviously. but it started with more.
The amount of matter converted to energy in the atomic bomb dropped on Hiroshima was about 700 milligrams
Clearly we are not referring to losing any rest mass energy.