the flood, mammoths, elphants, and food.

by Crazyguy 280 Replies latest watchtower bible

• 

  Viviane

  To follow up, Bohm, you are extremely intelligent, everyone should be able to recognize that. What happened, in my opinion, and of course I could be wrong, is that you did not read the entire thread where I DID make a claim and show myself wrong with my own math. I'll go ahead and cut-n-paste it for you...

  In the meantime, the current gravitational acceletation for the surface of the earth is 9.8m/s^2 (including the water). The average density of the earth is 5.52g/cm^3. The average density of water is about 1 g/cm^3. The current volume of the earth is about 10.8e+12 m^3. Doing some quick work in Wolfram Alpha show that that to add enough volume to the earth to cover Mt. Everest would be a difference of about 4557e+12 m^3, almost as much mass as the Earth has, so therefore... hmmm, interesting. I'm wrong. By orders of magnitude.

  I am wrong about the gravity. The orders of magnitude are completely off on the density of mass without a corresponding or greater increase in the radius.

  ... and...

  Plugging the mass of the earth, additional mass of the water MINUS the approximate volume dry land on Earth PLUS the additional 8848 meters into Wolfram Alpha, shows a gravitational acceleration at Everest with all of the water as... 9.94 m/s^2, slightly higher gravity.

  I was still wrong, but at least I know know why. All of this, BTW, is rough math. Point out any errors I may have made...

  So, clearly I have done some math. I've shown myself wrong. What you missed is prologos making all sort of claims and saying he has done the math but, when asked to show it, claiming it's stupid to even be talking about his claims. So, when you pop in and say:

  Well actually it is true (It is known as the "shell theorem" which I was not aware of when I wrote my post) and you can find a proof in any elementary book on classical mechanics, c.f. wikipedia. I will be happy to help you with the required integration if neccesary:

  ... you didn't read the thread, didn't know the context, weren't aware of the theorem you are so passionately discussing and what it's for, gave a snarky offer to "help" with math and then wondered what was going on when you got called out on showing math and extreme precision in a discussion where the very core of it is someone making claims, constantly switching terms and what they are claiming and making false claims about having done the math.

  It's like sticking your hand in between a hungry lion and a steak and wondering why your hand is missing.

  Now that I've said all of that, a conversation I had with my son the other day seems salient. He's extremely intelligent and has recently been caught saying he knows things he doesn't. I cautioned him on that, telling him that because he is so very smart, he doesn't have to work hard to keep up and can often jump in, catch up and pass everyone else in short order. I know I've been guilty of that in the past. The issue, as least as I framed it to him, is that day he gets caught because he does that to someone that is at least as smart as him and gets called out on it.

  If I am wrong, I am wrong. But you gotta show me.
• 

  Twitch

  It's like sticking your hand in between a hungry lion and a steak and wondering why your hand is missing.

  lmao
• 

  prologos

  Viv, your error was to have been told that, and ignore the post and keep trying to keep people worried whether they please you or not.

  The equations were about RATIOS, no higher math required. as you said, all you had to plug in the data to the published equations, and the answer pop out. Why have pages and pages of wrangling to arrive at that? It is FUN remember? it is FUNNY. smile

  ALL.
• 

  Viviane

  Spreading ignorance under the guise of knowledge is never funny. Please stop doing it.

  And show the math out claimed to have done. Not one thing you've mentioned can be done without calculus. Please, proceed to educate us, prologos.
• 

  bohm

  Howdy viviane!

  • Bohm: i only expect you to tell me what you think i have done wrong. You hinted this had something to do with a radius and i asked the radius of what?
  • Viv: You brought up shell theorem without understanding it. YOU brought the theorem up AND even included it in some rudimentary equations you wrote down. I can't read the theorem for you!

  You are entitled to think I do not understand the shell theorem. Not only does it seem quite pointless for me to correct that misconception, you seem so invested in it I cannot see how I could do it in principle. Would a derivation serve? At any rate this is not my main point. I was referring to the following exchange:

  • Bohm: so returning to the mineshaft, if we assume the earth is a body of uniform density the gravitational pull at a radius r will (per 1 of the shell theorem) scale as the mass (proportional to r^3) divided by square of distance (newtons low of gravitation) and so scale as r.
  • Viv: So, even using classical mechanics, it's not, as you said, "no acceleration in a hollow sphere".

  or

  • Viv: BTW, for those about to say "B-b-b-but Shell Theorem!", it's got a LOT to do with the radius of the objet under questions. It's not as simple as reading Wikipedia.
  • Bohm: the radius of which object? what part of wikipedia is wrong?
  • Viv: Yes. "No gravity" isn't what the shell theorem says.

  This went back and forth a bit. You have not tried to clarify the point about the radius:

  Viv: The shell theorem shows that gravitational acceleration between two bodies can be calculated using the center of those bodies because, effectively, the TOTAL gravitational acceleration inside nets out to zero as can be shown using two perfectly spherical but hollow bodies and how, relative to the internal coordinates, gravitational acceleration is calculated relative to ANOTHER body either inside or external to the spherical hollow mass (that's the radius you were missing, the second body). That doesn't mean there IS NO GRAVITY or that it is not stronger inside the body in one corrdinate than in another place. It just means that it NETS to zero.

  There are some minor issues in this description I wont touch upon here. I do think i now understand the confusion regarding the second body. Firstly, notice that as best as i can tell what you wrote is in no way in contradiction to what I wrote. I claimed the acceleration of an object inside a hollow sphere was zero and you wrote: it's not, as you said, "no acceleration in a hollow sphere". I am still puzzled where you think the contradiction lie.

  Now as regard to your point with the second radius. The shell theorem describes the vector field of force (the gravitational field) in any point inside or outside the hollow sphere due to the gravitational attraction of the hollow sphere. Can we agree upon this basic point?

  Thus, the radius of this sphere matters insofar as it determines when an object is inside or outside the hollow sphere, however asides this trivial point the only thing that matters is the mass of the hollow sphere (see the wikipedia description or my description). There is furthermore no "other" radius as you seem to suggest ("that's the radius you were missing, the second body") since the geometry of the other object does not matter insofar as the shell theorem is concerned, as long as it is either inside or outside the hollow sphere (if it is on the boundary the shell theorem still applies to the parts on either side of the boundary seperately). This is simply because the force exerted at the other object is, independent of it's geometry, the integral of the vector field computed by the shell theorem over the density function of the object. Can we agree upon this point too?

  Thus your characterization of the shell theorem as having to do with two *spherical* objects where the radius matters is an unecesary restriction of the version found at the wikipedia page (or as described by me) and that was what confused me.

  • Viv: It's like sticking your hand in between a hungry lion and a steak and wondering why your hand is missing.

  So you are comparing yourself to a hungry lion?

  • Now that I've said all of that, a conversation I had with my son the other day seems salient. He's extremely intelligent and has recently been caught saying he knows things he doesn't.

  I hope he was not "caught" in the same sence I was, that would be very frustrating for him, and that you did not act like a hungry lion to him.
• 

  Caedes

  So... how is there no gravity at the center if gravity is decelerating the object? See my previous post.

  Once past the centre your velocity would decrease as gravity started to decelerate you. Eventually the gravity would reduce your velocity to zero and then would start accelerating you back towards the centre. Hence your velocity would still be at it's maximum when you go through the centre.

  Because at the instant you were at the centre (bearing in mind it's an instant due to your velocity) the gravity would be acting equally on each point of your body pulling towards a point you already occupy. The net effect would be that your acceleration due to gravity would be zero, not that there would be no gravity. Once your velocity is taking you away from the centre the acceleration due to gravity would be acting in the opposite direction (decelerating you) and reducing your velocity. Eventually that deceleration would reduce your velocity to zero and then start pulling you back towards the centre.

  Obviously this is all assuming you can engineer a tube that could be run all the way through the earth of course! Which is quite impossible with any technology we currently possess.
• 

  prologos

  To add to that, during such a fall, inside a preferable evacuated shaft, the downward acceleration would be greatest near the surface, and at the center the traveller would be coasting. Because of any friction, the almost weightless floater would start falling from lower and lower heights from the center, would be moving slower and slower, come to a standstill or hangstill at the center, ceased to be accelerated, for there is no gravitational pull in either direction at the center of such an ideal sphere. Now, in Earth, you would be able to sense where the baricenter of the Earth/ Moon sytem was at the moment, but

  not whether the water canopy had fallen or not. It would not matter,

  it never has.
• 

  bohm

  Caedes:

  • Viv: So... how is there no gravity at the center if gravity is decelerating the object? See my previous post.
  • Caedes: The net effect would be that your acceleration due to gravity would be zero, not that there would be no gravity.

  You are correct in your analysis of the well.

  However with regards to there being no gravity however I would tend to disagree. I also think this is a point that has been confused on this thread and may also be the cause of some of the misunderstanding between me and Viviane.

  Actually I would say there was no gravity inside the shell (or at the center of the earth), not just that it "cancelled" out, however clearly the answer depends on what we mean by there "being gravity"

  In classical (Newtonian) physics, which is all I have discussed so far and seem relevant to the OP, there being gravity would to my mind mean the gravitational field was not zero -- however the gravitational field is zero inside a hollow shell (or in the center of the earth) per the shell theorem.

  More interesting is perhaps to ask the question in the context of general relativity. The general formulation of the shell theorem in a relativistic setting is Birkhoffs theorem and according to it the space-time metric inside a hollow sphere is flat Minkowski, and a Minkowski metric is also that of empty space; this is not to say the space is globally flat as there would still be time-dilation effects comparing the inside of the sphere to the outside, also I am not touching upon the black-hole limit. These statements are consistent because the Minkowski metric on the inside and outside may differ, i.e. there is no continious *globally* flat Minkowski metric. It is a very strange and interesting situation actually.

  when we intuitively insist on there being gravity (this is also what my intuition insists on BTW) this is merely because we "know" there must be gravity near a large object. However from a perspective of physics, we should first define what gravity (precense or absence) mean and then determine the answer. If we define gravity as being a vanishing gravitational field or Minkowski space-time metric then there is "no gravity"; mind in the context of GR this definition is far more questionable and an observer on the outside would disagree as there would be e.g. time dilation. Perhaps some GR geeks could weight in with their oppinion.

  (updated).
• 

  Viviane

  You are entitled to think I do not understand the shell theorem. Not only does it seem quite pointless for me to correct that misconception, you seem so invested in it I cannot see how I could do it in principle. Would a derivation serve? At any rate this is not my main point. I was referring to the following exchange:

  You offered to "help" me with the math on something you had just discovered existed. Please, DO show me the math. It's a shame something that wasn't your main point has now become a focus, but, you brought it up, not me.

  Firstly, notice that as best as i can tell what you wrote is in no way in contradiction to what I wrote. I claimed the acceleration of an object inside a hollow sphere was zero and you wrote: it's not, as you said, "no acceleration in a hollow sphere". I am still puzzled where you think the contradiction lie.

  Because it's not "no acceleration on an object inside the sphere", it's "total acceleration is equal to zero". Very different things.

  The shell theorem describes the vector field of force (the gravitational field) in any point inside or outside the hollow sphere due to the gravitational attraction of the hollow sphere. Can we agree upon this basic point?

  YES. Now you are getting it.

  Thus, the radius of this sphere matters insofar as it determines when an object is inside or outside the hollow sphere, however asides this trivial point the only thing that matters is the mass of the hollow sphere (see the wikipedia description or my description).

  Please explain to me how the radius of a second body in gravitational computations that explicitly REQUIRE the radius of the objects in in question canbe trivial.

  There is furthermore no "other" radius as you seem to suggest ("that's the radius you were missing, the second body") since the geometry of the other object does not matter insofar as the shell theorem is concerned, as long as it is either inside or outside the hollow sphere (if it is on the boundary the shell theorem still applies to the parts on either side of the boundary seperately).

  Well, you were getting it. Read a paper on it rather than wikipedia.

  So you are comparing yourself to a hungry lion?

  No, I'm comparing you to someone that doesn't understand what's going on but should.

  Actually I would say there was no gravity inside the shell (or at the center of the earth), not just that it "cancelled" out, however clearly the answer depends on what we mean by there "being gravity"

  Again, there is not "no gravity". The total gravitational acceleration is zero.
• 

  Caedes

  Because of any friction, the almost weightless floater would start falling from lower and lower heights from the center, would be moving slower and slower, come to a standstill or hangstill at the center, ceased to be accelerated, for there is no gravitational pull in either direction at the center of such an ideal sphere.

  Prologos,

  Not at all, as I have said previously your acceleration would be zero, however your velocity would be at it's maximum at the centre.