Cookie Posted October 31, 2012 Share Posted October 31, 2012 (edited) Correct me if I'm wrong, but in each layer you re-treat the gravitational pull from the inner layers as a point source right? As you go deeper down into the earth the layers above you are irrelevant when it comes to the gravitational pull. perhaps this is a clearer image for everyone else Edited October 31, 2012 by Cookie Link to comment Share on other sites More sharing options...
Nipple Posted October 31, 2012 Share Posted October 31, 2012 yes laz it's still treated as a point source whose mass is equal to that of the total mass of the sphere from your radius out from the center. Link to comment Share on other sites More sharing options...
LazaHorse Posted October 31, 2012 Share Posted October 31, 2012 (edited) MFW I was incorrectly using Laplace's equation when I should have been using Poisson's. Cookie and Nipple are right about gravitational potential, I was reading too far into it looking for the wrong answer. XD Gravitational potential (and the resulting gravitational force) decreases as you get closer to the Earth's surface as less and less mass is enclosed within the Gaussian surface. Potential is non-zero at the center (the conclusion I was trying to get at but messed up along the way), but force IS zero at the center where it then changes direction and slowly increases in magnitude as the object crosses the center. Sorry, I'm clearly horrible at physics. Hence why I'm a Government double major! XD TO SUM UP THIS THREAD! 1. The object would oscillate indefinitely between the two ends of the hypothetical "tunnel" unless you're taking account air-resistance, which would slowly dissipate the ball's energy until it came to rest at the center of the Earth. 2. You can apply Poisson's equation and Gauss' law which relate the potential or force at a location within a spherical mass density to the mass within the appropriately defined volume. 3. I failed to do number 2 properly; Cookie and Nipple corrected me. 4. I suck at physics; Perry is worse. 5. This is why I'm going to law school. ... Hopefully. Real. Edited November 1, 2012 by LazaHorse Link to comment Share on other sites More sharing options...
BranHorse Posted October 31, 2012 Share Posted October 31, 2012 Question answered GG wp report perry for feeding. Link to comment Share on other sites More sharing options...
Nipple Posted October 31, 2012 Share Posted October 31, 2012 MFW I was incorrectly using Laplace's equation when I should have been using Poisson's. Cookie and Nipple are right about gravitational potential, I was reading too far into it looking for the wrong answer. XD Gravitational potential (and the resulting gravitational force) decreases as you get closer to the Earth's surface as less and less mass is enclosed within the Gaussian surface. Potential is non-zero at the center (the conclusion I was trying to get at but messed up along the way), but force IS zero at the center where it then changes direction and slowly increases in magnitude as the object crosses the center. Sorry, I'm clearly horrible at physics. Hence why I'm a Government double major! XD TO SUM UP THIS THREAD! 1. The object would oscillate indefinitely between the two ends of the hypothetical "tunnel" unless you're taking account air-resistance, which would slowly dissipate the ball's energy until it came to rest at the center of the Earth. 2. You can apply Poisson's equation and Gauss' law which relate the potential or force at a location within a spherical mass density to the mass within the appropriately defined volume. 3. I failed to do number 2 properly; Cookie and Nipple corrected me. 4. I suck at physics; Perry is worse. 5. This is why I'm going to law school. ... Hopefully. the problem was not hard to begin with, i just think ron paul thought he could be cool and look smart after he learned something in class. the reason why (i believe) the acceleration changes (in accordance to what laz was confused about and cookie was explaining).. if you look at the force on an object due to gravity. F=Gm1M2/(r^2) G = gravitational constant, m1 being mass of the earth, and M2 being the ball, and r being the radius of the ball away from the earth. as you go towards the earth there are two variables in this equation, m1 (the mass of the earth) and r (the distance you (or the ball) are from the center. you can almost always treat a spherical mass as a point source. so at the surface of the earth gravity is around 9.8 m/s/s/ As you make your way into the earth you are decreasing r, so you are increasing F by 1/r^2. also as you go into the earth you are decreasing the mass (m1) of the earth because you ONLY count the mass of the sphere at the radius you are at, the other mass will cancel itself out. So it's a matter of what effects the force, F more. 1/r^2 or the change in m1? the beginning part of the earth (as shown by cookie's pictures) is not very dense at all, so you lose very little mass and gain more in the 1/r^2 term, so the Force (and therefore the acceleration) MUST increase. but as we get even deeper in the earth the large loss of mass due to it being more dense will overpower the 1/r^2 term, and the force then MUST decrease. Link to comment Share on other sites More sharing options...
enigma# Posted October 31, 2012 Share Posted October 31, 2012 All these answers assume a uniform definition of a ball. Consider this, a single graviton "ball" (assuming this elementary particle even exists) is "dropped" through the same hole. During that time, it is interacting with the Earth (hollow but it would have weak interactions with the Earth itself). Link to comment Share on other sites More sharing options...
PerryThePlatypus Posted October 31, 2012 Share Posted October 31, 2012 I was right according to branflakes video ,u kids blow no wonder I'm 2k elo Link to comment Share on other sites More sharing options...
LazaHorse Posted October 31, 2012 Share Posted October 31, 2012 (edited) the problem was not hard to begin with, i just think ron paul thought he could be cool and look smart after he learned something in class. the reason why (i believe) the acceleration changes (in accordance to what laz was confused about and cookie was explaining).. if you look at the force on an object due to gravity. F=Gm1M2/(r^2) G = gravitational constant, m1 being mass of the earth, and M2 being the ball, and r being the radius of the ball away from the earth. as you go towards the earth there are two variables in this equation, m1 (the mass of the earth) and r (the distance you (or the ball) are from the center. you can almost always treat a spherical mass as a point source. so at the surface of the earth gravity is around 9.8 m/s/s/ As you make your way into the earth you are decreasing r, so you are increasing F by 1/r^2. also as you go into the earth you are decreasing the mass (m1) of the earth because you ONLY count the mass of the sphere at the radius you are at, the other mass will cancel itself out. So it's a matter of what effects the force, F more. 1/r^2 or the change in m1? the beginning part of the earth (as shown by cookie's pictures) is not very dense at all, so you lose very little mass and gain more in the 1/r^2 term, so the Force (and therefore the acceleration) MUST increase. but as we get even deeper in the earth the large loss of mass due to it being more dense will overpower the 1/r^2 term, and the force then MUST decrease. You bring up an interesting point. When looking at an Earth with non-uniform mass (layers of decreasing density as you go out from the center), you can treat the total force as the gravity sum of two point masses at the Earth's center. One has constant mass but decreasing r, and so increases since F = G*(m1m2)/r^2. The other has a mass that increases with radius by r^3 (m1 = desntiy*r^3/R^2), and so the total force decreases as r decreases (F = G*(density/R^3)*m2*r). If the force from the first term increases faster than the force from the second term decreases (i.e. the density of the inner cores is much greater than the density of the outer shell), then you can get an increase in gravitational force through the outer layers as you move toward the center. All these answers assume a uniform definition of a ball. Consider this, a single graviton "ball" (assuming this elementary particle even exists) is "dropped" through the same hole. During that time, it is interacting with the Earth (hollow but it would have weak interactions with the Earth itself). Wow you would go quantum on us. SMH. I was right according to branflakes video ,u kids blow no wonder I'm 2k elo The world despises your existence. Branflake I had to do problems involving that Corliolis force for a month in my mechanics course last semester. Most confusing crap I've ever done. Imaginary forces, reference frame forces, and centripetal acceleration EVERYWHERE!!! Btw, we should have more scientific/philosophical discussions like this. This is fun. :3 재미있다 ^^ Real. Edited November 1, 2012 by LazaHorse Link to comment Share on other sites More sharing options...
enigma# Posted October 31, 2012 Share Posted October 31, 2012 Macrophysics gives us a great picture of everyday basic interactions. Sadly if you're looking at more theoretical and more technologically advanced approaches to everyday situations, our infant "quantum physics" is the way to go. e.g. femtophotography... holy crap xD Link to comment Share on other sites More sharing options...
LazaHorse Posted October 31, 2012 Share Posted October 31, 2012 (edited) Macrophysics gives us a great picture of everyday basic interactions. Sadly if you're looking at more theoretical and more technologically advanced approaches to everyday situations, our infant "quantum physics" is the way to go. e.g. femtophotography... holy crap xD I got a C in Quantum 1 and a B in Quantum 2. My TA explained it to me that I probably overcame the "mindfuck" of quantum mechanics in the first course enough to do well in the second course. His words, not mine. Lol... Real. Edited November 1, 2012 by LazaHorse Link to comment Share on other sites More sharing options...
PerryThePlatypus Posted October 31, 2012 Share Posted October 31, 2012 I do not expierience this hate therefore it doesn't exist. I would make and assumption that your not happy because I'm physics #1 Link to comment Share on other sites More sharing options...
Cookie Posted October 31, 2012 Share Posted October 31, 2012 (edited) Btw, we should have more scientific/philosophical discussions like this. This is fun. :3 재미있다 ^^ I think it would be more fun talking about discrepant events rather than reposts like ops. Edited October 31, 2012 by Cookie Link to comment Share on other sites More sharing options...
enigma# Posted October 31, 2012 Share Posted October 31, 2012 Thanks for your opinion... moving on.... Link to comment Share on other sites More sharing options...
Slowpoke Posted November 1, 2012 Share Posted November 1, 2012 what would happen if i put my dick in it Link to comment Share on other sites More sharing options...
BranHorse Posted November 1, 2012 Share Posted November 1, 2012 what would happen if i put my dick in it It wouldn't be able to feel it either Link to comment Share on other sites More sharing options...
LazaHorse Posted November 1, 2012 Share Posted November 1, 2012 what would happen if i put my dick in it It wouldn't be able to feel it either Wrecked Real. Link to comment Share on other sites More sharing options...
Oreo Posted November 1, 2012 Share Posted November 1, 2012 what if the ball had abs? Link to comment Share on other sites More sharing options...
Slowpoke Posted November 1, 2012 Share Posted November 1, 2012 It wouldn't be able to feel it either You a heartbreaker. Link to comment Share on other sites More sharing options...
Ron Paul Posted November 4, 2012 Author Share Posted November 4, 2012 I just came back to this thread. New things added in OP. Link to comment Share on other sites More sharing options...
Cookie Posted November 4, 2012 Share Posted November 4, 2012 I think I'll give you best I answer I can give you in layman's terms; if you drop the ball through the hole it's going to have a constant speed for the first ~8 minutes, then gradually get a little faster, then slower tell it it reaches the other side of the planet at ~21 minutes, to where it will be pulled back down to the earth and will do the same ~21 minutes (it would be like someone dropping it the same as you just at the other side of the hole), then after it made its way back, it will go back and forth from Chicago to Beijing taking ~42 minutes to complete each journey where it will continue this ~42 minutes indefinitely unless something stops it. Link to comment Share on other sites More sharing options...
LazaHorse Posted November 4, 2012 Share Posted November 4, 2012 unless something stops it. Enter slowpoke's penis. Real. Link to comment Share on other sites More sharing options...
PerryThePlatypus Posted November 4, 2012 Share Posted November 4, 2012 No his penis is fake , I've checked. Fake. Link to comment Share on other sites More sharing options...
SuperChargerFan Posted November 4, 2012 Share Posted November 4, 2012 The equation should be simple to understand. Consider: Lets assume there is a retracting platform over the hole. You stand 5 feet tall and drop the ball. The ball at the instantanious second you drop the ball is at with no momentum. Hence, we find when the ball is at the max point, momentum is 0. As the ball desends, the momentum increases. When the ball reaches the center, the momentum carries the ball through, but the momentum decreases because the ball encounters resistance from gravity. The force created by gravity up to this point is equal to the gravity against the ball now, so it will rise 5 feet above the surface on the other side, and decend again, returning to the same point, and going back down in continuous and infinite, given there ARE NO OTHER VIABLES THAT ALTER THE BALLS PATH, as in wind, pressure, or a prick that steals your ball in China. Need proof: Ah, I have you focus now! Heres how it works on a smaller model. The ball starts at its max, and distributes kinetic energy into the second ball, all the way down to the last ball, launching the last ball with equal force, causing it to rise and desend the same distance and repeat infinitly. Ahhh, makes sence, doesn't it? Link to comment Share on other sites More sharing options...
LazaHorse Posted November 4, 2012 Share Posted November 4, 2012 This was over a while ago guys. Lol... ctrl+f Cookie for simple and correct answer. Real. Link to comment Share on other sites More sharing options...
Slowpoke Posted November 5, 2012 Share Posted November 5, 2012 my dick inside da earth Link to comment Share on other sites More sharing options...
Recommended Posts