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Natural Science Forum / Physics / Relativity / December 2006Graviational potential inside the Earth
Wolfram used to have the general formula for the gravitational potential inside the Erath at an arbitrary depth d from the surface. (NOT the potential of the shell but of the filled sphere). Anybody that has it at his fingertips? > Wolfram used to have the general formula for the gravitational > potential inside the Erath at an arbitrary depth d from the surface. > (NOT the potential of the shell but of the filled sphere). > Anybody that has it at his fingertips? Are you thinking of the Earh as being a solid sphere of uniform mass density? If so then the relationship is linear and can easily be decuded by Gauss' law. Mass density = rho = M/V => M = rho*V Integral g*dA = G*M = G*rho*V = G*rho*(4*pi*r^3/3) g(4*pi*r^2) = G*rho*(4*pi*r^3/3) g = (G*rho*/3)r You'd better double check this in case I screwed up the units. Best wishes Pete >> Wolfram used to have the general formula for the gravitational >> potential inside the Erath at an arbitrary depth d from the surface. [quoted text clipped - 18 lines] > > Pete Oops! That was for acceleration, not potential. The force is given by F = mg = m(G*rho*/3)r = -grad U U = -m(G*rho*/3) Hmmm. A constant potential? That doesn't seem quite right. What did I do wrong folks? Pete >>> Wolfram used to have the general formula for the gravitational >>> potential inside the Erath at an arbitrary depth d from the surface. [quoted text clipped - 27 lines] > Hmmm. A constant potential? That doesn't seem quite right. What did I do > wrong folks? Okay. I got it U = -(1/2)m(G*rho*/3)r^2 > >>> Wolfram used to have the general formula for the gravitational > >>> potential inside the Erath at an arbitrary depth d from the surface. [quoted text clipped - 31 lines] > > U = -(1/2)m(G*rho*/3)r^2 Thank you, Pete, the units don't seem right (the presence of r^2) >> >>> Wolfram used to have the general formula for the gravitational >> >>> potential inside the Erath at an arbitrary depth d from the surface. [quoted text clipped - 35 lines] > > Thank you, Pete, the units don't seem right (the presence of r^2) You're right. I think that the constant of proportionality in Gausses law that I used has to be wrong. To find that constant consider a point charge at the origin and chose the Gaussian surface as a sphere centered at the origin. Pete > >> >>> Wolfram used to have the general formula for the gravitational > >> >>> potential inside the Erath at an arbitrary depth d from the surface. [quoted text clipped - 42 lines] > > Pete I found something close, done for shell: http://hep.ph.liv.ac.uk/~green/mechanics/lectures/l15notes.pdf The same formalism should apply to non-shell but the integral needs to change from a surface (double) integral to a volume (triple) integral. I did it years ago , I was just hoping that someone has the results at his fingertips. Dang, Wolfram took his reference down, U don't know why. > >>> Wolfram used to have the general formula for the gravitational > >>> potential inside the Erath at an arbitrary depth d from the surface. [quoted text clipped - 31 lines] > > U = -(1/2)m(G*rho*/3)r^2 I think I found the correct solution: http://www.madsci.org/posts/archives/2002-01/1011285522.Ph.r.html >> >>> Wolfram used to have the general formula for the gravitational >> >>> potential inside the Erath at an arbitrary depth d from the surface. [quoted text clipped - 34 lines] >I think I found the correct solution: >http://www.madsci.org/posts/archives/2002-01/1011285522.Ph.r.html For a uniform density the solution is very simple inside the earth: g(r) = g0*r/R where r is distance from center of the earth of radius R and g0 is surface gravity. It linearly increases from the center. The shell theorem shows that it.s only the mass beneath you that counts. So g is proportional to r^3/r^2 = r Mass = k* R^3, and the usual 1/r^2. John Polasek > For a uniform density the solution is very simple inside the earth: > g(r) = g0*r/R > where r is distance from center of the earth of radius R and g0 is > surface gravity. It linearly increases from the center. This is quite clearly wrong. The potential at the center must be much less than the potential at infinity, where it is zero, and less than at the surface. But this is correct for the Newtonian gravitational force, so you were presumably answering a different question. Koobee Wublee wrote: > ** U c^2 = 2 G M / R - G M r^2 / R^3 In matter > ** U c^2 = G M / r Free space This has the correct limiting behavior, at both r=0 and r=R, but is wrong by numerical factors. The sign is also wrong -- gravitational potential is NEGATIVE near a massive body, and zero at infinity. The c^2 is also omitted in conventional units (c never appears in Newtonian mechanics). The correct answer is: phi(r) = - (G M / R) (3/2 - (1/2) r^2 / R^2) for r<=R = - G M / r for r>=R > Force = m c^2 del U In conventional units there is no c^2, and there must be a minus sign (objects get pushed from regions of higher potential energy to regions with lower potential energy). > At the center of earth, the gravitational potential is the highest. > Notice I choose to define gravitational potential as positive. You think that objects fall up??? Or perhaps energy is not conserved??? > Credit should go to Dr. Roberts. I don't remember it.... And I doubt I would have made those elementary errors. I certainly would not have made the sign errors, as the signs are of fundamental importance. Tom Roberts > > For a uniform density the solution is very simple inside the earth: > > g(r) = g0*r/R [quoted text clipped - 40 lines] > > Tom Roberts Thank you, Tom Appreciate it. > The correct answer is: > > phi(r) = - (G M / R) (3/2 - (1/2) r^2 / R^2) for r<=R > = - G M / r for r>=R Thanks for the correction. > > Force = m c^2 del U > > In conventional units there is no c^2, There is no conventional agreement. I have clearly implied the following. ** U = G M / r / c^2, for r >= R > and there must be a minus sign > (objects get pushed from regions of higher potential energy to regions > with lower potential energy). Since there are no such separate quantities as the kinetic energy and the potential energy, I can choose to define them as positive or negative. As long as I am consistent and make it very clear to others readers. You need to read more carefully what I wrote. > > At the center of earth, the gravitational potential is the highest. > > Notice I choose to define gravitational potential as positive. > > You think that objects fall up??? No. What gives you that idea? > Or perhaps energy is not conserved??? Conservation of energy is a universal, fundamental phenomenon. It even applies to spacetime in general in which GR is just a special case of. <shrug> > Are you thinking of the Earh as being a solid sphere of uniform mass > density? If so then the relationship is linear and can easily be decuded by [quoted text clipped - 9 lines] > > You'd better double check this in case I screwed up the units. So, Gauss' law fails you, eh? Welcome to the real world. Try ** U c^2 = 2 G M / R - G M r^2 / R^3 In matter ** U c^2 = G M / r Free space Force = m c^2 delU At the center of earth, the gravitational potential is the highest. Notice I choose to define gravitational potential as positive. Credit should go to Dr. Roberts. >> Are you thinking of the Earh as being a solid sphere of uniform mass >> density? If so then the relationship is linear and can easily be decuded by [quoted text clipped - 21 lines] > >Credit should go to Dr. Roberts. The problem is complicated by the fact that you must specify potential DIFFERENCE. If m falls from infinity, then U is negative as the mass absorbs energy on its way to the surface under inverse square force. If computed from the center of the earth, U is positive: energy is put into the mass to raise it, now with a force proportional to radius. The negative U achieved at the surface is further increased on its passage down a hole to the center. del U is already force. U is already potential. What is the mc^2 for? John Polasek |
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