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[Il Palazzo]

g is hard.

This lands us a healthy 1.8593106x10^18.

Looks like a far cry from the healthy 10m/s you would normally expect, eh?

[wierd]

Air resistance on mt everest is 75% thinner (less viscuous) than at sea level.

I'd say a bound possibility of that scale is significant.

As I said though, I am too busy to do it right now.

[misko27]

g is hard.

This lands us a healthy 1.8593106x10^18.

Looks like a far cry from the healthy 10m/s you would normally expect, eh?

Oh dear. Forgot to square the radius.
EDIT: Actually, I appear to have also multiplied the radius into the other numbers. This is what happens when I can't write things down on paper.

[misko27]

Actually, I messed up a lot of things. I guess that is why they tell me to show unit conversions.

g is actually around 9.81804500245 m/s. That means a rounded 184041.21718 J.

[10ebbor10]

Air resistance on mt everest is 75% thinner (less viscuous) than at sea level.

I'd say a bound possibility of that scale is significant.

As I said though, I am too busy to do it right now.

It's an exponential thingy though. With low altitude differences, it barely matters.

Besides, it's not worth modeling the small change in air pressure, or the small change in gravity, if you're just going to assume the Earth is a homogenous glob of matter (Gravity changes by location due to the density of the stone beneath it, after all*) and if you're abstracting the air resistance anyway (Doubt you're going to calculate the aerodynamic properties of a slam dunking basket ball player.).

*And yes, the differences of that are larger than going 100 meters up in the sky.

[wierd]

Ok-- I have a really silly one to ask.

If we presume that neutron stars can and do collide, and that this can result in smaller clumps of electron degenerate matter forming--

What is the smallest possible volume for a sample of electron degenerate matter to retain the gravitational attractiveness needed to stay degenerate, and-- if these objects are flung into deep space from such collisions, what impact would a captured fragment have on planetary formation in an infant planetary system?

[MonkeyHead]

Well, there is a minimum limit to the mass that an object must have to remain electron degenerate due to gravitiational compression as defined by the Chandrasekhar limit - 1.44 solar masses. From there it is trivial to determine the rough radius of an object based on the density of the colliding neutron stars in question. Now, a 1.44 solar mass object tearing through a newly formed or forming solar system I assume would be fairly catastrophic...

[wierd]

I was meaning more like, it becomes gravitationally bound with a protostar, and starts orbiting the planetary accretion disc.

[MonkeyHead]

Hmm. I suspect it would disrupt accretion, though am unsure as to how signifignatly. Though it is worth noting that there are noteable binary systems of orthadox stars that have planets, and planets are known to orbit pulsars, so its not unreasonable to assume it is possible to have a stable configuration.

[Kicior]

These questions aren't really "crazy" but I can't figure out the answers
There is a limit to how heavy a nuclide can be (and not decay more or less instantly) because protons repel each other or maybe it's just magic. Is it possible to "glue" neutrons together (guess it is because nuclear force works on protons and neutrons alike) and make... dunno, a bunch of neutrons being held together by nuclear force?

How does mass defect work? You take two light nuclides, smash 'em together and poof - some of their mass changes into energy (because what you get is lighter than the "ingredients"). Mkay, I'm cool with that but then you can take some heavy nuclide, smash it with a neutron or something and poof you get some energy again - it means that the products are lighter than the heavy nuclide, right? I suppose it has something (or everything) to do with potential energy...

Power transmission lines use high voltage to decrease energy losses due to heat because "the amount of heat produced is I^2*R (Joule heating)" - but I^2*R is the same as U*I or U^2/R so what is the difference? Do some AC shenanigans come into play?

How does one calculate escape velocity for a binary star system when a spaceship is in the barycenter?

I've never written about physics in English before so I probably sound like a caveman

[MonkeyHead]

Mass defect relies on good old E=mc^2. Lose some m, get lots of E. Or, use lots of E to get some m. There is a limit to nucleus size for the resons you mentioned (electromagentic repulsion drops off with range slower than nuclear forces), though the "island of stability" may interest you. In some ways, neutron stars are kind of what you describe, but there it is the gravitational forces that dominate, though the dentisy matches that of an atomic nucleus.

As for power lines... its not a raw number of volts you would put into those equations. Its the voltage drop that you would put in - the potential difference over the length of the cable. R can be determined from the dimensions and resistivity of the material, and I can be determined from the properties of the transformer. Just as the "voltage" in a power line is 275000 V does not mean that all 275000 are lost in transmission.

It is much easier to consider complex objects as one simplified body that exists at the centre of mass of the system - after all, any objects orbiting both the srats in the binary system would do so around the centre of mass... or you could always do it in terms of gravitational forces (or potential) and do the vector sum, but that way might lead to a 3 body problem depending on how you define your reference frames.

[Kicior]

Mass defect relies on good old E=mc^2. Lose some m, get lots of E. Or, use lots of E to get some m.

It's rather intuitive when we are talking about fission: there's a heavy nucleus (M) that divides into some smaller ones (let's say m1 and m2) and the energy is (M-m1-m2)*c^2
And during fusion it's 2 smaller ones (m1,m2) forming a bigger one (M) which for some reason is lighter than the two combined  (m1+m2-M)*c^2
The question: what kind of sorcery is this? I guess it has something to do with binding energy per nucleon?

As for power lines... its not a raw number of volts you would put into those equations. Its the voltage drop that you would put in - the potential difference over the length of the cable. R can be determined from the dimensions and resistivity of the material, and I can be determined from the properties of the transformer. Just as the "voltage" in a power line is 275000 V does not mean that all 275000 are lost in transmission.

Well, I've never really figured out how transformers work (or electricity at all to be honest )
I guess that the power on both sides has to be equal, right (assuming there are no losses)? So if it's 1000V 100A  (100000W) on the input then there has to be 100000W on the output? I know that the voltage depends on the number of coils but I have no idea about amperage. Can the voltage be 100000V and the amperage 1A or 1V and 100000A (theoretically speaking of course)? Doesn't Ohm's law apply here?

It is much easier to consider complex objects as one simplified body that exists at the centre of mass of the system - after all, any objects orbiting both the stars in the binary system would do so around the centre of mass

Yeah but wouldn't the x in the V=-GM/x formula be equal to 0?

[Helgoland]

Mass defect relies on good old E=mc^2. Lose some m, get lots of E. Or, use lots of E to get some m.

It's rather intuitive when we are talking about fission: there's a heavy nucleus (M) that divides into some smaller ones (let's say m1 and m2) and the energy is (M-m1-m2)*c^2
And during fusion it's 2 smaller ones (m1,m2) forming a bigger one (M) which for some reason is lighter than the two combined  (m1+m2-M)*c^2
The question: what kind of sorcery is this? I guess it has something to do with binding energy per nucleon?

The sum is greater than its parts - or lighter. It depends on the sum, really. The reason for the mass defect is the binding energy of the nucleus. You can imagine it like the binding energy in a molecule: It takes energy to rip a molecule apart, and energy is released when a new one is formed.

[Kicior]

Well guys, I apologise for brainfarting - it turns out that instead of reading physics textbooks (which are quite shitty to be honest, you can only understand them when you already know what you are supposed to learn from them...), browsing wikipedia in 2 languages and reading about nuclear physics I should have just googled "how is that both fusion and fission produce power" .

[Il Palazzo]

Say we're still angry with Moon for its tides, and we are unhappy with the poor performance of our Giant Doom Laser of Doom. So we try a different approach.
We've got ourselves a very strong, very light tether or rope, running between a pair of blocks. One block is on the North pole, the other on the Moon. Together they form a sort of conveyor belt, able to transport stuff one way or the other.

We've got a henchman on the Moon with a spade, who's going to load the Moon bit by bit onto the conveyor belt.

Now, we're doing it sneakily to surprise everybody, so we need to improvise with the power issues. After disassembing an electric floor fan to get its 100W motor, we've attached it to the conveyor belt, and plugged it in in Santa Claus' shed when he wasn't looking.

With this sure-fire setup, how long will it take to move the Moon in its entirety to Earth, and what will be the electricity bill Santa will need to pay on top of his regular power consumption(your local fees apply)?

No friction, circular orbit and all that jazz are a given.