Page 1 of 2 12 LastLast
Results 1 to 30 of 56

Thread: What is the potential gravity ( gravitational potential ) at the center of the earth?

  1. #1

    What is the potential gravity ( gravitational potential ) at the center of the earth?

    What is the potential gravity ( gravitational potential ) at the center of the earth?

    I realize that the net gravitational force at the center of the earth is 0 ... but I since there is a net pull at all points towards the center the center must have the max potential gravity right?

    ATM Warning: depending on how this is answered it could end up in ATM very quickly.

  2. #2
    Join Date
    Jul 2005
    Posts
    9,765
    Potential energy, as conventionally used, levels out to a minimum at the centre of the Earth. An object in flat space, infinitely far from all mass, would have zero potential energy; objects in the vicinity of mass concentrations have negative potential energies.
    To the conventional picture of a funnel-shaped gravity well surrounding a massive object, you can add a rounded bottom to the funnel, like an inverted dome, which plots the potential inside that object. The precise shape of the inverted dome will vary with the radial mass distribution.

    I have no idea of the exact value at the centre of the Earth in joules per kilogram. Is that important?


    Grant Hutchison

  3. #3
    Grant ...

    Per prior discussions on black holes we used the letter W to represent colliding black holes.

    The point in the center of the two black holes is the center of the W and the bottom of the V parts is the black hole itself ( really there would be not tip of the V but this is close enough ).


    So can we use a U to represent the gravitational potential of the Earth?

    If so wouldnt the center of the earth have the max gravitational potential? Or am I using my words wrong here. Wouldnt the center of the earth be in the deepest part of the gravitational well?














    Quote Originally Posted by grant hutchison View Post
    Potential energy, as conventionally used, levels out to a minimum at the centre of the Earth. An object in flat space, infinitely far from all mass, would have zero potential energy; objects in the vicinity of mass concentrations have negative potential energies.
    To the conventional picture of a funnel-shaped gravity well surrounding a massive object, you can add a rounded bottom to the funnel, like an inverted dome, which plots the potential inside that object. The precise shape of the inverted dome will vary with the radial mass distribution.

    Grant Hutchison

  4. #4
    If you drilled a perfect hole to the center of the earth. Then shined a light to the surface of the earth. Would there be a) very slight blue shift b) very slight red shift or C) absolutely no red shift or blue shift due to gravity.

  5. #5
    Join Date
    Jul 2005
    Posts
    9,765
    Quote Originally Posted by tommac View Post
    So can we use a U to represent the gravitational potential of the Earth?
    Very roughly, if it had sloping sides, and if it represented the potential inside the Earth.

    Quote Originally Posted by tommac View Post
    If so wouldnt the center of the earth have the max gravitational potential? Or am I using my words wrong here. Wouldnt the center of the earth be in the deepest part of the gravitational well?
    Deepest part, yes. So most negative value of the potential. So minimum.

    Quote Originally Posted by tommac View Post
    If you drilled a perfect hole to the center of the earth. Then shined a light to the surface of the earth. Would there be a) very slight blue shift b) very slight red shift or C) absolutely no red shift or blue shift due to gravity.
    I'm guessing the light source is at the centre of the Earth. If so, there would be a progressively larger redshift at the light detector as it moved farther from the centre of the Earth and closer to the Earth's surface.

    Grant Hutchison

  6. #6
    Quote Originally Posted by grant hutchison View Post

    Deepest part, yes. So most negative value of the potential. So minimum.
    So how does that differ from a black hole?

    Basically we are saying that both a black hole would be like a V and the earth would be probably more like a U ( the difference really being that the black hole would be in an infinitely deep gravitational well at the bottom of V )

    But when you say minimum you dont mean 0 do you? You mean that the center is deeper in a gravitational well than at the surface, right?

  7. #7

    Lightbulb Matters of Gravity

    Quote Originally Posted by tommac View Post
    I realize that the net gravitational force at the center of the earth is 0 ...
    This would be true only for a mass distribution that has exact spherical symmetry. The real Earth is not exactly spherically symmetrical, and I am quote sure nothing else in the universe is exactly & truly spherically symmetrical either. So the net gravitational acceleration at the center of the Earth is only almost zero, but not exactly zero. And so it is for all the planets & stars that really exist in the real (non ideal) universe.

    Quote Originally Posted by tommac View Post
    Per prior discussions on black holes ...
    For the moment, forget about black holes. You can't mix Newtonian gravity and general relativity like that.

    Quote Originally Posted by tommac View Post
    If so wouldnt the center of the earth have the max gravitational potential?
    No. For any spherically symmetric distribution of mass, the maximum gravitational potential will be at the surface. As you move in from infinity to the surface, and use a point like test mass so we can ignore tides, then the acceleration felt by the test mass will increase like 1/r2, where r is the distance between the point mass and the center of the spherically symmetric mass distribution (i.e., planet, star & etc.). But once you reach the surface and keep going down you are feeling the attraction of only the mass inside the spherical shell your test mass is sitting on. That acceleration would continue to increase like 1/r2 if the mass inside the sphere were constant, but it is in fact decreasing like 1/r3 (i.e., like the volume of the shrinking sphere). So the acceleration is actually decreasing like r until you reach the center, where r=0 and so the acceleration is zero.

    You cannot compare this to what happens with black holes. We know exactly what's going on gravitationally inside the Earth, or any other real mass distribution, all the way down. But we do not know exactly, or really even approximately, what is going on inside the event horizon of a black hole. We don't even know how to assess the fundamental reality of space & time inside the event horizon of a black hole. So the two gravitational potentials (black hole vs real mass distribution) have nothing in common.

    Quote Originally Posted by tommac View Post
    If you drilled a perfect hole to the center of the earth. Then shined a light to the surface of the earth. Would there be a) very slight blue shift b) very slight red shift or C) absolutely no red shift or blue shift due to gravity.
    Quote Originally Posted by grant hutchison View Post
    I'm guessing the light source is at the centre of the Earth. If so, there would be a progressively larger redshift at the light detector as it moved farther from the centre of the Earth and closer to the Earth's surface.
    Grant's answer looks right to me. It's just the reverse of the falling down scenario I already described. As the light moves upwards towards the surface, there is an ever increasing sphere of mass behind it, so it will feel an ever increasing gravitational potential, and therefore present an ever increasing redshift, all the way to the surface.

  8. #8
    Quote Originally Posted by Tim Thompson View Post

    Grant's answer looks right to me. It's just the reverse of the falling down scenario I already described. As the light moves upwards towards the surface, there is an ever increasing sphere of mass behind it, so it will feel an ever increasing gravitational potential, and therefore present an ever increasing redshift, all the way to the surface.
    and how about beyond the surface?

  9. #9
    Join Date
    May 2008
    Posts
    798
    Actually a black hole would be more like a ")(" or a "\ /" with a non connected bottom.

    Also... it would only be slightly heavier... nothing unbearable... the pressure might give an altered result though.

  10. #10
    Join Date
    Jul 2005
    Posts
    9,765
    Quote Originally Posted by tommac View Post
    So how does that differ from a black hole?
    The black hole has no surface.

    Quote Originally Posted by tommac View Post
    But when you say minimum you dont mean 0 do you? You mean that the center is deeper in a gravitational well than at the surface, right?
    I mean minimum: the lowest value of potential energy. As I said, the potential energy is negative everywhere inside the gravity well, it gets more negative as you move towards the centre, and the centre is the most negative. The minimum.

    Quote Originally Posted by tommac View Post
    and how about beyond the surface?
    More redshift.

    Grant Hutchison

  11. #11
    Join Date
    Jul 2005
    Posts
    9,765
    Quote Originally Posted by Durakken View Post
    Actually a black hole would be more like a ")(" or a "\ /" with a non connected bottom.
    I think you're perhaps thinking about wormhole diagrams in popsci. We're plotting potential energy, here.

    Quote Originally Posted by Durakken View Post
    Also... it would only be slightly heavier... nothing unbearable... the pressure might give an altered result though.
    I'm not sure what any of this means.

    Grant Hutchison

  12. #12
    Join Date
    Jun 2006
    Posts
    2,698
    Quote Originally Posted by Tim Thompson View Post

    For the moment, forget about black holes. You can't mix Newtonian gravity and general relativity like that.
    Quite correct.

    Slightly off topic, bore the hole clear through. Evacuate the air. Neglect rotation. Drop a rock through. What is the travel time to the other side?

    Extra credit. What is the travel time through such a borehole, to any point on the surface, neglecting all friction?

    (This has been considered for an L.A. to Las Vegas tunnel.)
    Last edited by John Mendenhall; 2008-Nov-18 at 07:54 PM. Reason: clarity

  13. #13
    Join Date
    Sep 2006
    Posts
    1,391
    Quote Originally Posted by John Mendenhall View Post
    Quite correct.

    Slightly off topic, bore the hole clear through. Evacuate the air. Neglect rotation. Drop a rock through. What is the travel time to the other side?

    Extra credit. What is the travel time through such a borehole, to any point on the surface, neglecting all friction?

    (This has been considered for an L.A. to Las Vegas tunnel.)
    About 42 minutes. It's the same for any tunnel connecting any two points on the Earth's surface, not just two points on opposite sides of the sphere.

    I ask this question sometimes on our sophomore physics oral exam :-)

  14. #14
    Join Date
    Sep 2008
    Posts
    5,892
    Quote Originally Posted by StupendousMan View Post
    About 42 minutes. It's the same for any tunnel connecting any two points on the Earth's surface, not just two points on opposite sides of the sphere.

    I ask this question sometimes on our sophomore physics oral exam :-)
    Hang on. Does this tunnel have to go through the centre of the Earth? Or do you mean a chord (or the 3D equivalent)? How would all chords give you the same time?

  15. #15
    Join Date
    May 2008
    Posts
    798
    Grant...somewhat true... Black holes are supposed to have infinite gravity or whatever so a diagram would have no connection...

    And I'm taking Tommac as asking what the gravity at the center of the earth would be...and that would be slightly greater than on the surface... and I'm not sure how pressure effect red shift so not sure how that might alter the results of testing via a beam of light

  16. #16
    7 minutes right ... between any two points on the surface of the earth.

    Quote Originally Posted by John Mendenhall View Post
    Quite correct.

    Slightly off topic, bore the hole clear through. Evacuate the air. Neglect rotation. Drop a rock through. What is the travel time to the other side?

    Extra credit. What is the travel time through such a borehole, to any point on the surface, neglecting all friction?

    (This has been considered for an L.A. to Las Vegas tunnel.)

  17. #17

    Lightbulb Small changes.

    Quote Originally Posted by tommac View Post
    If you drilled a perfect hole to the center of the earth. Then shined a light to the surface of the earth. Would there be a) very slight blue shift b) very slight red shift or c) absolutely no red shift or blue shift due to gravity.
    Quote Originally Posted by grant hutchison View Post
    I'm guessing the light source is at the centre of the Earth. If so, there would be a progressively larger redshift at the light detector as it moved farther from the centre of the Earth and closer to the Earth's surface.
    Quote Originally Posted by Tim Thompson View Post
    Grant's answer looks right to me. It's just the reverse of the falling down scenario I already described. As the light moves upwards towards the surface, there is an ever increasing sphere of mass behind it, so it will feel an ever increasing gravitational potential, and therefore present an ever increasing redshift, all the way to the surface.
    Quote Originally Posted by tommac View Post
    and how about beyond the surface?
    In principle, the redshift will continue to increase until the gravitational potential drops to zero, which means an infinite distance, since 1/r2 will never go to zero for any finite value of r. In practice, and in our case of light climbing out of the Earth, once the light passes through the surface, and the mass behind it now remains essentially constant, the increase in gravitational redshift with distance will be too small to measure. So we can treat the gravitational redshift as unchanging beyond the surface and make no significant error as a result, as long as we are far from the surface. This is the case, for instance, for observing the gravitational redshift of sunlight due to the Sun's gravity (i.e., Lopresto, Chapman & Sturgis, 1980. Lopresto tells me this is still the most precise measure of solar gravitational redshift, but it is only good enough to show that the correct theory of gravity must be metric, and not good enough to show that general relativity in particular is that metric theory).

  18. #18
    Quote Originally Posted by PraedSt View Post
    Hang on. Does this tunnel have to go through the centre of the Earth? Or do you mean a chord (or the 3D equivalent)? How would all chords give you the same time?
    The amount of gravity in the "downward direction" is reduced when the hole is not parallel with gravity.

    If the hole is parallel with gravity you get the most acceleration. If the hole is at a 45% angle then some of that momentum is counteracted by the walls of the hole. So things would move through slower.

    I know my wording is wrong. But I think I got the concept.

    Now can we go back to the OQ?

    if we shine a light from the center of the earth out into space. The light would redshift until it hits the receptor.

  19. #19
    Quote Originally Posted by Tim Thompson View Post
    In principle, the redshift will continue to increase until the gravitational potential drops to zero, which means an infinite distance, since 1/r2 will never go to zero for any finite value of r. In practice, and in our case of light climbing out of the Earth, once the light passes through the surface, and the mass behind it now remains essentially constant, the increase in gravitational redshift with distance will be too small to measure. So we can treat the gravitational redshift as unchanging beyond the surface and make no significant error as a result, as long as we are far from the surface. This is the case, for instance, for observing the gravitational redshift of sunlight due to the Sun's gravity (i.e., Lopresto, Chapman & Sturgis, 1980. Lopresto tells me this is still the most precise measure of solar gravitational redshift, but it is only good enough to show that the correct theory of gravity must be metric, and not good enough to show that general relativity in particular is that metric theory).

    Sorry to take very babysteps here but I want to make sure that I understand exactly what is being said.

    if A is the center of the earth and B is the surface of the earth and C is a point in space.

    We would see some redshift from B-C but we would see much more redshift from A-B right?

    This is from the gain in potential energy on its journey right?

  20. #20
    Join Date
    Dec 2005
    Posts
    982
    Quote Originally Posted by Tim Thompson View Post
    No. For any spherically symmetric distribution of mass, the maximum gravitational potential will be at the surface.
    I have to disagree. The maximum force will be at the surface (assuming uniform density), but the maximum potential is an infinite distance away.

  21. #21
    Quote Originally Posted by grant hutchison View Post
    I mean minimum: the lowest value of potential energy. As I said, the potential energy is negative everywhere inside the gravity well, it gets more negative as you move towards the centre, and the centre is the most negative. The minimum.

    More redshift.

    Grant Hutchison
    OK ... and this would decrease ( become more negative ) by 1/r^3 or 1/r^2 ?

  22. #22

    Lightbulb Stuff

    Quote Originally Posted by tommac View Post
    if A is the center of the earth and B is the surface of the earth and C is a point in space. We would see some redshift from B-C but we would see much more redshift from A-B right?
    Correct
    Quote Originally Posted by tommac View Post
    This is from the gain in potential energy on its journey right?
    The easiest way to think of it is that the photons lose energy doing work against the gravitational field.

    Quote Originally Posted by John Mendenhall View Post
    (This has been considered for an L.A. to Las Vegas tunnel.)
    A tunnel that passes through the San Andreas & San Jacinto fault zones, and maybe the Garlock fault zone as well? Not to mention the myriad local faults perpendicular to the tunnel route that characterize the block tectonics of southern California? The Baker fault? the Death Valley fault? Sure, I'll pay to ride that puppy

    Quote Originally Posted by Tim Thompson View Post
    No. For any spherically symmetric distribution of mass, the maximum gravitational potential will be at the surface.
    Quote Originally Posted by phunk View Post
    I have to disagree. The maximum force will be at the surface (assuming uniform density), but the maximum potential is an infinite distance away.
    Yes, sloppy language on my part. The gravitational potential (-GM/r) is everywhere negative, and it's maximum possible value at infinite distance is zero. The maximum absolute value for the potential is at the surface, where the acceleration will be a maximum. It is this maximum of the absolute value that I had in mind.

  23. #23
    Join Date
    Jul 2005
    Posts
    9,765
    Quote Originally Posted by tommac View Post
    OK ... and this would decrease ( become more negative ) by 1/r^3 or 1/r^2 ?
    What would decrease? Where?

    Grant Hutchison

  24. #24
    Join Date
    Dec 2005
    Posts
    982
    I think we're using different definitions of gravitational potential here.

  25. #25
    Join Date
    Jul 2005
    Posts
    9,765
    Quote Originally Posted by Tim Thompson View Post
    Yes, sloppy language on my part. The gravitational potential (-GM/r) is everywhere negative, and it's maximum possible value at infinite distance is zero. The maximum absolute value for the potential is at the surface, where the acceleration will be a maximum. It is this maximum of the absolute value that I had in mind.
    I'm not getting this.
    While the acceleration reaches a maximum at the surface of the Earth, it doesn't change sign. So the potential continues to grow more negative all the way to the centre of the Earth, albeit at a decreasing rate. The absolute value of the potential should therefore be maximum at the centre of the Earth.

    Grant Hutchison

  26. #26
    Join Date
    Dec 2005
    Posts
    14,315
    It's very nearly zero, but it's not quite zero, for a couple of reasons:

    1. Earth's mass is not uniformly distributed.

    2. Even if it were, at the geographic center of the Earth, there's a slight gravitational attraction towards the sun, and to a lesser extent, the moon

  27. #27
    Join Date
    Jul 2005
    Posts
    9,765
    Quote Originally Posted by mugaliens View Post
    It's very nearly zero, but it's not quite zero, for a couple of reasons:

    1. Earth's mass is not uniformly distributed.

    2. Even if it were, at the geographic center of the Earth, there's a slight gravitational attraction towards the sun, and to a lesser extent, the moon
    You're talking about force, I take it, rather than potential?

    Grant Hutchison

  28. #28
    Join Date
    Mar 2008
    Posts
    5,195
    Quote Originally Posted by grant hutchison View Post
    You're talking about force, I take it, rather than potential?

    Grant Hutchison
    yes i do believe he is.

    3. The inside of the earth is a very dynamic place, mass is constantly being shifted around i would suspect that the centre of gravity is changing position slightly all the time.

    sorry for going slightly off topic.

  29. #29
    Unless I misunderstand him, he's talking about actual potential. There's likely several local minima, not all of them points, due to the slight oblateness of the planet and the various mass concentrations. They would be small in extent and very, very close to the center of a best-fit sphere around Earth.

    As for the moon and sun...the Earth is within their gravity wells, yes, but if you want to include their influence, the center of Earth is just another local minimum of the solar system's gravity well. The same goes for the Milky Way as a whole, for the Local Group of galaxies, etc...you'll never actually reach zero potential. (barring a discovery of some weird knot in spacetime or something)

  30. #30
    Join Date
    Jul 2005
    Posts
    9,765
    I knocked off a couple of quick graphs to show the variation of gravitational force and potential with radial distance.

    The first shows how force (F) varies with radius (r) inside and outside a homogeneous sphere. Horizontal (r) scale is in units of sphere radius, the vertical is scale arbitrary. Force varies with 1/r≤ outside the sphere, and directly with r inside.

    The second is the corresponding graph of gravitational potential (U). Same horizontal distance units, vertical scale again arbitrary. There's a -1/r dependency outside the sphere, and an r≤ dependency inside, with a smooth transition at the surface.

    I may come back and tart these up a little more for clarity, but hopefully they get the gist across.

    Grant Hutchison
    Last edited by grant hutchison; 2008-Nov-19 at 12:11 AM. Reason: Attached improved graphs to post

Similar Threads

  1. Replies: 5
    Last Post: 2011-Oct-11, 11:18 AM
  2. gravitational force vs gravitational potential ( of a sphere )
    By tommac in forum Space/Astronomy Questions and Answers
    Replies: 47
    Last Post: 2011-Apr-18, 09:15 PM
  3. Potential energy and gravity
    By Anders Starmark in forum Space/Astronomy Questions and Answers
    Replies: 35
    Last Post: 2008-Nov-25, 03:23 PM
  4. Gravitational potential energy of white dwarfs
    By ExpErdMann in forum Space/Astronomy Questions and Answers
    Replies: 13
    Last Post: 2006-Dec-16, 02:22 AM
  5. Gravitational Potential Energy and Redshift
    By ExpErdMann in forum Against the Mainstream
    Replies: 17
    Last Post: 2004-Aug-08, 03:15 PM

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
here
The forum is sponsored in-part by: