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Originally Posted by OngBonga
Not really. Newtonian mechanics is just an excellent approximation, but it's a flawed understanding of what's really happening.
Fine, but I kinda suck at GR.
Originally Posted by OngBonga
I'm talking about anything. Something in freefall is not subject to any forces other than gravity, which is an inertial force, that is one that emerges as a result of frame of reference. An orbit is freefall.
It's still a force and it still changes the velocity w.r.t. whatever it is you're orbiting about.
Just because you can't measure the difference inside the closed box, doesn't mean you can't measure the difference outside the closed box.
Originally Posted by OngBonga
I don't agree. If there are two objects in the universe, they orbit each other. That orbit might be ridiculously huge, but it's an orbit. Something moving away from Earth will give up kinetic energy and gain potential energy. This is simply inevitable.
A thousand times no, dude.
If the total energy in the system is negative, then the system is bound. If it is non-negative, it is unbound.
Escape velocity, man. I thought you got that much.
Let's dig it, then:
Empty universe. Just mass M in it. Just M and you, m, all the way out at infinity, with no speed at all relative to M.
That's the initial condition.
What's your speed as you fall toward M?
Go.
The force accelerating you toward M is
F = GMm/r^2
Your initial speed at r_inf is
v_0 = 0 m/s
The Work-Energy Theorem says that the work done on an isolated system is equal to the change in energy. In this case, the work done is 0 J, as there is no outside agency aside from the internal energy in the system, whatever it is. This is a purely mechanical system, i.e. there's no pressure changes, no entropy, no chemical changes, no heat, no sound, blah, blah... just Grav. PE and the KE of the relative motion.
delta_W = delta_KE + delta_PE = 0 J
Good, we've proved that KE + PE is constant, and that means the total Energy is constant.
We can further state
delta_KE = -delta_PE
We can apply the same Work-Energy theorem on the isolated system of just the PE and state
delta_W = delta_PE
but delta_W is the change in work, or the work acquired by moving an object parallel to a force.
delta_W = F*dr
(technically that's a dot product, but since in our case the motion is always exactly parallel to the force, we can straight multiply)
Integrate both sides
W = PE = integral(GMm/r^2 dr)
Good to go.
KE = -integral(GMm/r^2 dr) r = inf to r_0
where r_0 is the distance between M and m.
KE = -GMm * integral(1/r^2 dr) r = inf to r_0
KE = -GMm * integral(r^-2 dr) r = inf to r_0
KE = -GMm * [-1/r] | r = inf to r_0
KE = -GMm * (-1/r_0 - -1/inf) = -GMm * (-1/r_0 + 0)
KE = GMm/r_0
A familiar gem, and not too bad to get there from first principles.
KE = 1/2 m |v|^2 = GMm/r_0
|v|^2 = 2GM/r_0
|v| = sqrt(2GM/r_0)
We've answered the question. That may not be as familiar to you, but that is the equation for escape velocity.
OK. Now. If starting at infinity distance and approaching to any finite distance r_0 gets you up to v_escape, then moving in exactly the opposite direction at v_escape is enough to barely make it back to infinity. If you go any faster at all, then you are only escaping to infinity even faster. The only part of v_escape that's even relevant is that it is finite.
Not all orbits are closed. QED
If we open this problem to allow all possible initial positions and velocities, then all the possible solutions are conic sections. Circles and ellipses, yes, but also parabolas and hyperbolas. The parabolas are the ones that are just exactly at v_escape for all r_0. The hyperbolas are the ones going above v_escape at all r_0. The circles and ellipses are the ones going below v_escape for all r_0.
Originally Posted by OngBonga
I need to think more about this.
We could flip the prior problem around and talk about how you're sitting still while M starts at inf distance away. It's the same problem. Then of course, we don't see you accelerating, but M accelerating, and in order to understand acceleration, we have to know what it's in relation to.
Since position, x, isn't absolute, then dx/dt and d^2x/dt^2 aren't either.
It's probably incorrect language on my part, but look at it from the F = ma ; F/m = a perspective.
Force and mass need to be understood relativistically, but once they are, all observers agree on the force and rest mass experiencing the acceleration... so they must ultimately agree on the acceleration. All Einstein really talked about was F/m if you read the way he did his own math.
Originally Posted by OngBonga
Not true. Mass changes depending on motion, and motion is relative. Therefore, we will disagree on the mass of an object from different frames of reference.
I should have said rest mass. Once relativity is accounted for, all observers agree on rest mass.
Originally Posted by OngBonga
I'm not so sure. If KE is changing into PE, with no net loss of energy to the system, then no acceleration happened. Why? Because KE and PE are the same thing.
F = -grad(PE)
This is a hugely broad-sweeping statement that holds in all cases. For every potential energy function, an object interacting with it experiences a force of minus the gradient of the potential. Nothing to do with KE, there.
That PE may turn into KE, but there are all kinds of other forms of energy when we move beyond a purely mechanical system.
Within a purely mechanical system with no losses, KE = -PE and the only way KE can change is if m changes or v changes. When we assert that m is constant, then v is the only variable that can change. a = dv/dt. If v is changing, a is happening.
Originally Posted by OngBonga
If I throw a ball and then run alongside it, does the ball have kinetic energy from my frame of reference? If I catch it, it doesn't exert the same force on my hand compared to someone who catches it from a standing position. From two different frames of reference, we disagree on the amount of kinetic energy an object has, but we'll never disagree on its KE + PE.
Yes. It has KE.
No. The force is different. You're in a moving frame w.r.t. the other person, and applying the Galilean frame transformations is plenty enough for you to agree.
You don't agree on its KE + PE, because you agree on its PE, but differ on its KE. You agree that KE + PE = const., but you observe different E_tot as a constant.
They observe only velocity in the up-down direction. You also observe additional motion in the not-up-down direction, so you observe greater velocity.
All of this is assuming you move a constant velocity w.r.t. the line traced out by the ball's trajectory.
Originally Posted by OngBonga
This is rather like electricity and magnetism. If you move with the same velocity as an electron flowing along a wire, you won't observe an electric flow, you'll observe a magnetic field.
I know what you mean, but no. The equations which describe how energy is stored in a magnetic field and electric field are different in form from the equations that describe how energy is stored as potential and kinetic energy. You can't just draw this parallel with any 2 forms of energy.
FYI
By invoking a wire, there is no reference frame in which there are no moving charges.
Whether you see electrons flowing against a stationary background of cations or you see electrons sitting still while a stream of cations flow past in the opposite direction, the conventional current is the same. There is always a magnetic field, and its orientation (direction it rotates around the wire) is agreed upon by all observers.
I.e. you will either see electrons going one way or protons going the other way, which is equivalent in E&M equations.
(obv. you could see inf possibilities of both moving, but the relative speed is the same, once relativity is taken into account.)
Originally Posted by OngBonga
No, but it does imply no net change in force.
dF/dx = 0 at apoapse and periapse... so even a broken clock..?
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