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Ask any question you want about Physics

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posted on Sep, 24 2015 @ 06:49 AM
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a reply to: Arbitrageur

Thanks for the reply. My question is why is the apple slide even there? The apple can never be red shifted because it's always a mass at rest without some force that provides momentum - just like a rocket ship. Is it just assumed that there's some force moving the apple? Kinetic and potential energies are different for the apple and the photon - isn't that right?
I guess I'm just confused why the apple slide is there or how it's supposed to demonstrate red shift.




posted on Sep, 24 2015 @ 07:24 AM
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a reply to: Phantom423
OK now I understand. I think the topic is gravitational redshift, so he used that header on the slides.

The example of the apple is something he started out with to show how an apple can lose energy to give students a concept they can easily visualize.

It's a lead-in to the topic of light similarly losing energy, it's not an example of redshifting of the apple. The idea of gravitational redshift might be unfamiliar to some students, so he's trying to make the concept seem more familiar to them by associating it with something they are more familiar with, and since both the apple and the light lose energy they are related in that way.

Of course I'm just speculating on his intent but it's an educated guess.




posted on Sep, 24 2015 @ 07:42 AM
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a reply to: Arbitrageur

Ok, got it. The inclusion of the slide just didn't make sense to me. Thanks.



posted on Sep, 24 2015 @ 11:43 AM
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a reply to: Arbitrageur

Ok, I had a weird epiphany this morning and I'm wondering your thoughts on it.

Postulation: A planet (lets call it Earth) has a hollow center, has uniform density and we can reach the large center void.

It seems to me this void would operate a bit like space, except instead of having no gravity, you would have gravity pulling you with similar force in all directions. Near equal force pulling in all directions would create almost no movement, but you would move. You would be drawn to whichever wall had the strongest pull based on your proximity. Unless you were smack-dab in the middle and the pull was completely cancelled out, you eventually would reach a wall.

Do you think this is a safe and accurate assumption?

If it is, you would eventually land on a wall and could stand up. But, you would be standing upside down, at least relative to people standing on the Earth's crust 3000 or so miles directly under your feet.

So what happens when you start digging? It seems to me you would have to reach a point when down became up. You could dig and reach a point (other than the direct center of the void) where your gravitational pull was the same in all directions You would float as if you were in the center, but you would be somewhere in the "mantle".

If these are all correct assumptions (as to me they logically seem to be) a hollow planet would have more than one gravitational center. A theoretical hollow Earth would have a gravitational center right at the center, but there would also be a spherical plane outside the center void where gravity was equal to zero.

I guess I don't really have a question other than what are you thoughts on this theory? I don't imagine it's validity would have any effect on other scientific theories, it was just something I had never considered and thought others might find interesting to think about. But I guess it's all relative

edit on 9/24/2015 by scojak because: (no reason given)



posted on Sep, 24 2015 @ 12:56 PM
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a reply to: scojak
I've actually thought about that myself.

There's a nice graphic of Earth's gravity in the Wiki related to that question; best guess is probably the blue line shows the gravity at various depths all the way to the center where it becomes zero:

en.wikipedia.org...


The orange area to the left is all inside the Earth. If you could dig a shaft to the Earth's center, the gravity in the shaft would be along these lines.

Note there's no reversal of "up" as you descend down the shaft (the graph would have to cross the X-axis for that to happen and it doesn't).

Your scenario is a little bit different where you hollow out a cavity of unspecified size at the earth's center. Of course this probably can't be done since the temperatures and pressures would exceed the capabilities of known materials to maintain the cavity, but let's assume you can do the impossible and make a cavity.

That won't change the location of the center of mass of the Earth, all it will do is reduce the total mass of the Earth by whatever amount you removed. So, I think what would happen is the gravity versus depth curve would still be quite similar to what's shown above, but the height would be lowered according to how much mass was removed, and the slope of the curve would change at the edge of the hollowed out area.


You would be drawn to whichever wall had the strongest pull based on your proximity.
I haven't tried to model this but my guess is, it depends. If the shell had a very asymmetrical density, you might be attracted more to a very high density spot on the shell from either the inside or the outside of the shell. However if the density of the shell was radially symmetrical, I think you'd always be attracted toward the center of mass, which is at the center, so "You would be drawn to whichever wall had the strongest pull based on your proximity" wouldn't happen without some significant asymmetry in the density of the shell, which would relocate the center of mass of the shell from the center of the shell to the inside of one of the walls of the shell.

edit on 2015924 by Arbitrageur because: clarification



posted on Sep, 24 2015 @ 06:57 PM
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a reply to: scojak

For a sphere, as you descend into it, the mass above you has no effect at all anymore.

So for hollow spheres, if you are in the interior, no, you are not accelerated toward the inner surface. This is one of those proofs you have to do in calculus based physics about semester 2.

This is also why you can't have a "central sun", it's an unstable configuration. There's no force to keep it centered. When you understand why the Ringworld is unstable in the plane, you'll see why the central sun is unstable in the sphere.



posted on Sep, 24 2015 @ 10:05 PM
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a reply to: Arbitrageur

a reply to: Bedlam

Thank you both. I didn't understand that the pull of mass above was effectively cancelled out. I did some research though and it all makes sense now.



posted on Sep, 25 2015 @ 12:30 AM
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a reply to: scojak

It's not intuitive that it should perfectly balance out but it does.



posted on Sep, 25 2015 @ 01:12 AM
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originally posted by: Bedlam
a reply to: scojak

For a sphere, as you descend into it, the mass above you has no effect at all anymore.

So for hollow spheres, if you are in the interior, no, you are not accelerated toward the inner surface. This is one of those proofs you have to do in calculus based physics about semester 2.

This is also why you can't have a "central sun", it's an unstable configuration. There's no force to keep it centered. When you understand why the Ringworld is unstable in the plane, you'll see why the central sun is unstable in the sphere.




I'm going to say your both wrong you two forgot to include the moon in your duscussions the reason you see the sharp decline is earth's center mass is about 2000 km under us. There is no point in the earth with zero gravity. There will always be pull towards the moon in any location. Let's find out ok

Mass of the Earth is 6.00*10^24 kg.
Mass of the is 7.35*10^22 kg
And they are 3.80*105 km apart roughly.
So let's figure for earth's center mass using our starting point as earth's center.

ecm = ((0*6.00*10^24) + (3.80*108 * 7.35*10^22)) m / (6.00*10^24 + 7.35*10^22)= 4598666m = 4600 km.

Ok so from the center of earth the center mass is 4600 km the radius of the earth is 6,371 kilometers again roughly. Subtract 4600 km from 6371km we get earth's center mass as being 1771 km below us and no where near the center of the earth. Also means the chart he posted would be wrong. Because everything would pull towards that point meaning there would be. Multiple different answers depending on location of the moon.

So back to the original question if there was a hollow area and you were there there greater mass would be pulling you up and the mass below you pulling you down decreases. So in the center I would suspect if we ran a computer simulation we would find you make circles lagging behind the moon. How ever as was pointed out do to heat and pressure this scenario is impossible.


edit on 9/25/15 by dragonridr because: (no reason given)



posted on Sep, 25 2015 @ 04:43 AM
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a reply to: stormbringer1701

While you are absolutely correct that infrastructure is necessary to make use of and distribute the energy, I'm not sure the same financial system is going to exist to require payments for this. To even survive long enough as a species to develop the tech, serious social changes would have to occur.



posted on Sep, 25 2015 @ 05:08 AM
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originally posted by: pfishy
a reply to: stormbringer1701

While you are absolutely correct that infrastructure is necessary to make use of and distribute the energy, I'm not sure the same financial system is going to exist to require payments for this. To even survive long enough as a species to develop the tech, serious social changes would have to occur.
nobody works for nothing. we found that out again in the early american colonies. there was no personal property and no stake in production and they damned near starved. Humans are not wired that way. no one wanted to work hard when they saw their more lazy fellows not only shirking but getting an even share from thier own hard labor.

but really this is a lesson that repeats through out our history. and man will not evolve out of it. nor is really in our interest to do so.



posted on Sep, 25 2015 @ 10:54 AM
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a reply to: dragonridr

Ok, hang on a sec. I think I might see what KrzYma was getting at with the energy of light and wavelengths. At least partially. And KrzYma, please feel free to jump in and respond as well.
Ok, if I'm grasping this, we normally measure the energy of light, be it radio, ELF, or gamma, by what it can impart upon a target in a fixed amount of time. (Not getting into the time debate here, just need an standard exposure window). So, for example, ELF can impart x×10 gev/second. Then microwaves can impart y×10 gev/second and Gamma can impart z×10 gev/second. I know this is rather crude, but please bear with me.
If you envision the light at each wavelength striking the target, ELF is very long waves, as close to flat as you'll be able to use for practical purposes. Microwaves have many times more waves in a given meter, and again the increase with gamma. Like visualizing one continuous line in the classic school drawing of light waves, each increase in frequency getting wavier.
But what I think your getting at KrzYma is quantizing the light not by waves per meter, but a meter of flat baseline energy (still light) then compressed (wave folded?) into whatever length the flat baseline becomes at the different frequencies.
Like using a piece of string one meter long (yes, I realize you could not represent elf with only one meter of string) and meandering it into the given frequencies shapes. There's still only one meter of string, so you only have one fixed quantity of energy regardless of what frequency you broadcast it at.

Sorry this is such a stumbling attempt. I have slept about 9 hours in the past 4 days. My employers, I think, are trying to see if it's possible to make me physically implode.



posted on Sep, 25 2015 @ 11:06 AM
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a reply to: stormbringer1701

Yeah, you make a good point. My mind was drawn to the Star Trek currency-less ideal. Except the Ferengi, of course.



posted on Sep, 25 2015 @ 03:25 PM
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originally posted by: Bedlam
a reply to: scojak

It's not intuitive that it should perfectly balance out but it does.


Indeed, this calculation or a related one (showing mass of a uniform spherical object could be treated externally as a point for gravitation) is a critical part of Newton's Principia. And there, Newton did it with triangles & in Latin.

[He kept back calculus for himself for many years as his secret weapon, producing extraordinary results and publicly proving with then accepted geometric and algebraic methods after he figured out the answer from calculus]



posted on Sep, 25 2015 @ 05:57 PM
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originally posted by: dragonridr
Ok so from the center of earth the center mass is 4600 km the radius of the earth is 6,371 kilometers again roughly. Subtract 4600 km from 6371km we get earth's center mass as being 1771 km below us and no where near the center of the earth. Also means the chart he posted would be wrong. Because everything would pull towards that point
I have about the same numbers for the barycenter location and the Earth's radius so we agree on those. If your assertion that everything pulled toward the Earth-moon barycenter was correct, would that also include a plumb bob?

If I apply your model to the plumb bob, I get a right triangle formed by the radius of the Earth on the adjacent side and the distance from the center of the Earth to the barycenter on the opposite side, with angle of 36 degrees. See sketch:


So, either I don't understand your model, or I don't agree with it, because I don't think you'll see a plumb bob at a 36 degree angle (no more than 1 degree I suspect). Therefore I also don't think the graph I posted of the Earth's gravity is wrong, well not THAT wrong, maybe slightly wrong since it's only an estimate.


originally posted by: pfishy
There's still only one meter of string, so you only have one fixed quantity of energy regardless of what frequency you broadcast it at.
How would this model explain the photoelectric effect?



posted on Sep, 25 2015 @ 07:00 PM
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a reply to: Arbitrageur

No earth's gravity will always over power anything here of course. However at no point can we have zero gravity at the center. As earth's gravity is not the only body we have to be concerned with. So say theoritaically at earth's center gravity from earth drops to zero. If the moon wasn't there. Because the moon there can be no location on earth that isn't going to feel gravity. So in this sphere we create you would be dragged along the wall following the moon. There is no way to have zero gravity inside the earth.

And I'm not sure your point really with the plumb bob I know you know greater mass will always be down. The position of the plumb bob we can use the same equation to find its center mass with the earth moon. It will always be down.
edit on 9/25/15 by dragonridr because: (no reason given)



posted on Sep, 26 2015 @ 06:35 AM
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a reply to: Arbitrageur

I never said it explained anything. I was trying to see if that was part of what KrzYma was trying to describe concerning the 1m and 1cm wavelengths ultimately containing the same energy.



posted on Sep, 26 2015 @ 07:07 AM
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originally posted by: dragonridr
a reply to: Arbitrageur

No earth's gravity will always over power anything here of course. However at no point can we have zero gravity at the center.


What you WON'T have is attraction to the inside surface of a hollow earth, which is the point of the original question.



posted on Sep, 26 2015 @ 12:07 PM
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originally posted by: dragonridr
And I'm not sure your point really with the plumb bob I know you know greater mass will always be down.
The point was primarily that I wanted you to see that graph I posted isn't really wrong (you said it is).


There is no way to have zero gravity inside the earth.
You've got to be careful with statements like that. Is the ISS zero gravity or 89% of Earth's surface gravity?

Trick question because it depends on your inertial frame. From an inertial frame inside the ISS, microgravity is all they experience. But an observer on the moon would see the ISS accelerating toward the Earth so it doesn't look like microgravity from that inertial frame. We could also ask why when you chose to consider bodies external to the Earth, you chose the moon instead of the sun, even though the gravitational force from the sun is 175 times greater. Again, the Earth orbiting the sun is somewhat like the ISS orbiting the Earth where centripetal force is balanced by the fictitious centrifugal force from inertia, so depending on your inertial frame gravitational force can seem very large or very small.

This is also why the plumb bob doesn't get attracted to the Earth/moon barycenter, because you can't do a static analysis on a dynamic system and get the correct answer. The inertial frame of the Earth isn't just the orbital motion around the sun, it's also the "wobble" because of the moon's orbit.

You raised a good point about the moon but it's a little trickier to analyze than just saying everything will be attracted to the Earth-moon barycenter, because of the dynamics.



posted on Sep, 26 2015 @ 12:12 PM
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originally posted by: Arbitrageur

originally posted by: dragonridr
And I'm not sure your point really with the plumb bob I know you know greater mass will always be down.
The point was primarily that I wanted you to see that graph I posted isn't really wrong (you said it is).


There is no way to have zero gravity inside the earth.
You've got to be careful with statements like that. Is the ISS zero gravity or 89% of Earth's surface gravity?

Trick question because it depends on your inertial frame. From an inertial frame inside the ISS, microgravity is all they experience. But an observer on the moon would see the ISS accelerating toward the Earth so it doesn't look like microgravity from that inertial frame. We could also ask why when you chose to consider bodies external to the Earth, you chose the moon instead of the sun, even though the gravitational force from the sun is 175 times greater. Again, the Earth orbiting the sun is somewhat like the ISS orbiting the Earth where centripetal force is balanced by the fictitious centrifugal force from inertia, so depending on your inertial frame gravitational force can seem very large or very small.

This is also why the plumb bob doesn't get attracted to the Earth/moon barycenter, because you can't do a static analysis on a dynamic system and get the correct answer. The inertial frame of the Earth isn't just the orbital motion around the sun, it's also the "wobble" because of the moon's orbit.

You raised a good point about the moon but it's a little trickier to analyze than just saying everything will be attracted to the Earth-moon barycenter, because of the dynamics.


And also one of the reasons they figure the core has not cooled is the moon. Though id argue hasn't been sufficient time either.



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