[alancalverd]

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?

Apparently. But you will observe a red shift if you are at a higher gravitational potential than the source. And if M is so large that the escape speed exceeds c, the red shift will be infinite and no electromagnetic energy will escape.   

Note that v_esc = sqrt(2GM/R), i.e. is independent of the mass of the escaping entity, therefore applies equally to photons of no mass.

Re: cannonballs on trampolines

Some energy is lost to air movement, more to heat in the springs of the trampoline. A perfect reflector in vacuo will return both cannonballs to their starting height. But if all the energy is lost in a single impact, both must come to rest at the same time.

Galileo presented a very simple argument. Take a big stone and a little stone. Suppose the little stone falls slower than the big one. Tie them together. Does the little one slow the big one, or does the big one speed up the little one? Since you can't have it both ways, they must fall at the same rate.  Galileo was forced to recant for stating the obvious in the face of Papal authority. 

[timey (OP)]

Yes I have read that Galileo was given a right hard time for being of progressive mind.  Poor dude!

Anyways's, just to double check - you say that with a perfect reflector in a vacuum g-field caused by M, there would be no energy loss...and the cannonball's would bounce back to original height?

[timey (OP)]

In addition to post above...

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.

This equation has been a part of my thought path last few posts here:

https://www.thenakedscientists.com/forum/index.php?topic=69592.0

...and if you are saying that it is only air resistance and the trampoline's springs energy absorption that causes a cannonball 'not' to continue bouncing forever...? 

[alancalverd]

Yes. Not just air resistance, but energy loss as sound waves too. And some heat dissipated in the cannonball itself.

Alas, sordid reality rarely approximates to pure physics, which is why engineers exist.

[timey (OP)]

So - without these energy robbers then both cannonball would bounce on their trampolines to same height indefinitely, and therefore the 'suck' of gravity itself cannot be considered as an energy robber.

A light ray pointed away from M will carry on going indefinitely, unless M is so great that escape velocity exceeds the speed of light...

... m in relation to M will fall back towards M unless it can achieve the associated escape velocity for that value M...
This being the same value of escape velocity for all values of m.

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? 

[alancalverd]

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.

"moment of inertia" means something quite different! But there will be a point where v=0 so p = 0

If p = mv, and v = 0, then does p = m? 

NO!!!!  bank balance = hours x salary. If salary = 0 it doesn't mean that balance = hours....we'd all be zillionaires! Anyway p is a vector and m is a scalar, so you can't have p = m under any circumstances. 

[timey (OP)]

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?

After looking at simple pendulum maths last night, I've been dreaming about the idea of torque around the pivot to the scenario, with the pivot being M.
So in light of your post, I daresay the 'whatever it is called' cannot be mathematically described by Inertia = m*d^2?  Which is what had somehow scrambled together in my head by the time I woke up.

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.

*

Please excuse me, but the notion of bank balance is entirely alien to me these days, but having now investigated the concept of scaler versus vector, it occurs perhaps lack of momentum might well have been that my p=problem, in that it's hard to get going when stuff holds you back.

I did suspect that where v=0 that p=0, but have been a little confused by you're saying that p is not caused by a, and now see clearly that p cannot be caused by m either if p=0 when v=0.
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.

P.S.  I read a link provided by Bored Chemist a while back about a Terry Pratchet character's notion that the use of exclamation marks is inverse to the sanity of the person using them.  More explanation marks, less sanity.
It's been of some psychological interest to me that I've noticed myself using less of them since reading that, (chuckle), despite knowing full well it's a joke.

[timey (OP)]

In addition to post above:

So G is the gravitational constant, and 9.807m/s^2 is the 'acceleration' of g at sea level earth (an approximation).
And at distance of 2 radii from centre Earth g will be 2.45m/s^2...
...Where in both cases the measurement is being held relative to the SI unit of the standard second.

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?
i.e: That it takes a longer amount of time for that 'constant' velocity to cover a meter at that radius from Earth?

[timey (OP)]

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.
Even if we made the lighter cannonball the same size in spatial proportions as the 100kg cannonball to equalise air drag, the 10kg cannonball will still be making less of a dent in the trampoline's fabric than the 100kg cannonball, where consequently there will be less energy lost to the springs, and to the sound waves resulted from impact.

When a 100kg cannonball and a 10kg cannonball are dropped into free fall onto their trampolines at exactly the same moment in time, the 100kg cannonball should come to rest sooner than the 10kg cannonball...
Shouldn't it?

Would the 100kg cannonball come to rest sooner than the 10kg cannonball?
If so - what is the proportionality that describes the difference of energy loss over time difference?
And if the cannonball's do actually come to rest at same time - then why?

[alancalverd]

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?

maximum altitude

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?

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.

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.

You can ignore relativistic effects for escape for all known planets

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

integral, not 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.

dpdx=h/2pi. That's all there is to it.

[alancalverd]

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?

m/s² is an acceleration, not a velocity.

[alancalverd]

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.

Obviously. but it started with more.

[jeffreyH]

GO Timey!

[timey (OP)]

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?

maximum altitude

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?

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.

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.

You can ignore relativistic effects for escape for all known planets

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

integral, not 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.

dpdx=h/2pi. That's all there is to it.

Maximum altitude.  Ok thanks!

*

I'll have to look into u and s to get a pictorial understanding of what those maths terms mean.  Visualising the physical actions of the maths being the only way that I can understand them...
Yes - That equation forms part of the parabolic trajectory considerations last few posts of "My model of a cyclic universe continued again" thread.

*

Not where light is concerned.

*

Sorry - preemptive text error

*

h being Plancks h constant*2 pi is h bar, where p = hbar*k, and k is angular velocity.
So d being distance, but x?
distance*p*distance*x= h/2pi
Back to my You Tube 'visual' tutorials for me me thinks!

[timey (OP)]

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?

m/s² is an acceleration, not a velocity.

I meant that 'something' travelling at a constant velocity in the g-field, such as light perhaps...

[timey (OP)]

GO Timey!

Awww Jeff, stop please, I'm in danger of becoming emotional.

[timey (OP)]

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.

Obviously. but it started with more.

Hmmm... Clearly we are not referring to losing any rest mass energy.  The balls are travelling at same speed, so no additional kinetic energy there, but 100kg cannonball has more momentum.
Are you referring to more kinetic energy associated with greater momentum?
Or are we talking potential energy?

[alancalverd]

Sorry about the dpdx confusion, but we stll don't have Greek characters available so I couldn't do a proper "delta". dpdx is Heisenberg's indeterminacy product.

Cannonballs falling onto trampolines has nothing to do with rest mass energy. We are talking simple classical mechanics where mgh =  mv²/2 ;  potential energy being converted into kinetic energy as a body falls. If you converted the rest mass energy of a 10 kg cannonball into kinetic energy you would vaporise the trampoline and most of the planet.

The amount of matter converted to energy in the atomic bomb dropped on Hiroshima was about 700 milligrams

[timey (OP)]

Which is why I said

Clearly we are not referring to losing any rest mass energy.

So the 100 kg cannonball has more m for the calculation than the 10kg cannonball, and has more potential energy to be converted into kinetic energy.
Converting potential energy into kinetic energy is caused by the acceleration 9.807m/s^2 (or whatever the acceleration is at greater height from M), and this conversion hasn't got anything to do with momentum?
But... p being useful in the maths as an integral that is a calculation of slices of the value m*v (or slices of the differing means of defining p) at any moment in time as coordinates change?

So - do the cannonballs come to rest at the same time?

[alancalverd]

1. No. p = mv at every instant. But if m and v are variable, then their instantaneous values are the integrals of mass change and acceleration up to that instant.

2. Probably not. The cannonballs will only come to rest if the trampolines are not perfect reflectors. If we don't know the nonlinearities of real trampolines we can't predict which will come to rest first. It's unlikely to  the point of practically impossible that any two trampolines will behave identically under very different loads.

Since kinetic energy is mv²/2 and momentum is mv they are obviously related but ke is a scalar and p is a vector, so although you can derive ke from p if you know m, you can't derive the p vector direction from ke unless you happen to know the direction of travel as well. You can always tell a mathematician by the way he ends every sentence with x1 (note the bold type) when giving directions to a stranger.