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tommac
2008-Nov-18, 05:04 PM
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.

grant hutchison
2008-Nov-18, 05:17 PM
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

tommac
2008-Nov-18, 05:31 PM
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?

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

tommac
2008-Nov-18, 05:39 PM
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.

grant hutchison
2008-Nov-18, 05:49 PM
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.

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.

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

tommac
2008-Nov-18, 06:56 PM
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?

Tim Thompson
2008-Nov-18, 07:23 PM
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.

Per prior discussions on black holes ...
For the moment, forget about black holes. You can't mix Newtonian gravity and general relativity like that.

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.

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'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.

tommac
2008-Nov-18, 07:28 PM
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?

Durakken
2008-Nov-18, 07:33 PM
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.

grant hutchison
2008-Nov-18, 07:44 PM
So how does that differ from a black hole?The black hole has no surface.

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.

and how about beyond the surface?More redshift.

Grant Hutchison

grant hutchison
2008-Nov-18, 07:52 PM
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.

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

John Mendenhall
2008-Nov-18, 07:53 PM
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.)

StupendousMan
2008-Nov-18, 07:58 PM
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 :-)

PraedSt
2008-Nov-18, 08:02 PM
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? :confused:

Durakken
2008-Nov-18, 08:06 PM
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

tommac
2008-Nov-18, 08:08 PM
7 minutes right ... between any two points on the surface of the earth.

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.)

Tim Thompson
2008-Nov-18, 08:12 PM
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'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?
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 (http://adsabs.harvard.edu/abs/1980SoPh...66..245L). 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).

tommac
2008-Nov-18, 08:12 PM
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? :confused:

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.

tommac
2008-Nov-18, 08:16 PM
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 (http://adsabs.harvard.edu/abs/1980SoPh...66..245L). 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?

phunk
2008-Nov-18, 08:40 PM
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.

tommac
2008-Nov-18, 08:55 PM
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 ?

Tim Thompson
2008-Nov-18, 09:11 PM
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

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.

(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 :)

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.
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.

grant hutchison
2008-Nov-18, 09:13 PM
OK ... and this would decrease ( become more negative ) by 1/r^3 or 1/r^2 ?What would decrease? Where?

Grant Hutchison

phunk
2008-Nov-18, 09:15 PM
I think we're using different definitions of gravitational potential here.

grant hutchison
2008-Nov-18, 09:45 PM
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

mugaliens
2008-Nov-18, 09:45 PM
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

grant hutchison
2008-Nov-18, 09:47 PM
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 moonYou're talking about force, I take it, rather than potential?

Grant Hutchison

cosmocrazy
2008-Nov-18, 10:13 PM
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.

cjameshuff
2008-Nov-18, 10:32 PM
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)

grant hutchison
2008-Nov-18, 10:50 PM
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/r² 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 r² 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

cjameshuff
2008-Nov-18, 10:59 PM
This makes me wonder just how homogenous Earth's inner core actually is. There's a lot of pressure, but little gravitational gradient. The buoyancy effects that sort materials of different densities should be quite small...

undidly
2008-Nov-18, 11:16 PM
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

I understand the problem but cannot work out an answer.
The G does not change sign but is at a maximum at the Earth surface.

The question is does the red shift depend on the magnitude of the G or on the depth of the G well.
If it is the magnitude then maximum shift is at the surface.
If it is the depth then maximum shift is at the center.

Einstein said G is equivalent to acceleration.If acceleration is the cause of shift (as well as, but different to Doppler) then there will be none at the center.

grant hutchison
2008-Nov-18, 11:26 PM
The question is does the red shift depend on the magnitude of the G or on the depth of the G well.
If it is the magnitude then maximum shift is at the surface.
If it is the depth then maximum shift is at the center.Gravitational redshift is due to the change in gravitational potential, so for a light source at the centre of the Earth static observers at all radial distances, inside and outside the Earth, will (with sensitive enough instruments) observe a red-shift, because they are all at higher potentials than the source.
(I'm assuming there are no other masses around. Of course in practice we'd need to worry about the gravitational potential of the sun, etc, once we got far enough away.)

Grant Hutchison

cjameshuff
2008-Nov-18, 11:31 PM
The question is does the red shift depend on the magnitude of the G or on the depth of the G well.

Relative depths of the emitter and receiver. Observers deeper within the well see it as being more blue shifted, as if it were gaining EM energy by falling into the well and losing potential energy, and observers further out of it see it as being red shifted, as if it were losing EM energy as it gained potential energy on the way out, analogous to the gain and loss of kinetic energy of an object falling through the field. Though as mentioned in another recent conversation, the photon is actually unchanged, its measured wavelength and energy just depend on the frame you're measuring it in.

grav
2008-Nov-19, 05:45 AM
Something in this thread seems backwards to me. I always thought that the gravitational potential energy was just the same as the amount of kinetic energy that would be aquired if an object were dropped from a particular distance, but just assigned a negative value since the kinetic part has not actually happened yet. So if an object were dropped at the center of the Earth, it wouldn't go anywhere, and gain no kinetic energy, so the potential energy there would be zero. The greater the distance from Earth we drop the object, however, the greater the kinetic energy that would be gained when the object is let go, so the greater the absolute value of the potential energy, or the more negative when we assign the negative sign. Approaching infinity, it would approach a limit, and that would be the most negative it can get. Is that not right?

grav
2008-Nov-19, 06:01 AM
Hmm. This Wiki page (http://en.wikipedia.org/wiki/Potential_energy) shows the formula for gravitational potential energy. About a third of the way down, it shows one formula that is U = m g h, where h is the height above the surface of the Earth, only for short distances, though, no doubt. With g negative, it would give a result that increases the absolute value with distance. This is the way I tend to think of it. But immediately after that, it gives another definition that is U = - G m1 m2 / R. The absolute value of that one would decrease with greater distance, obviously figuring for the kinetic energy that would be gained if an object were dropped from infinity to R instead of dropped from R, and of course, that particular part of the formula would only apply to R is greater than the surface of the mass. But those are two completely different formulas. Depends upon the convention, then, I guess.

grant hutchison
2008-Nov-19, 08:34 AM
You need a formulation in which the potential energy decreases as the kinetic energy of a freely falling body increases. So potential energy has to increase with increasing radial distance from a mass. Your mgh example does this if we use positive acceleration.
But it's a common convention to set potential energy to zero at infinity, which gives you negative values everywhere else. Hence -GMm/r2, which tends to zero at infinity.

Grant Hutchison

Jeff Root
2008-Nov-19, 12:19 PM
My understanding of potential energy:

The maximum value is at infinity. The closer you get to the center of mass,
the lower the value. In most discussions of potential energy, especially for
engineering analysis, a value of zero is assigned to a reference level, such
as the ground surface. This is often chosen to be the minimum value likely
to be encountered. In cosmology, it is more common to assign a value of
zero to the potential at infinity, making zero the maximum value.

-- Jeff, in Minneapolis

dmassey
2008-Nov-19, 02:11 PM
This is an A Level physics answer:
The zero of gravitational potential energy is conventionally set at infinity. This makes sense because when an object is 'at infinity' it will no longer be affected by gravitational fields (we assume it is 'at infinity' from everything), so we have a definite value to compare other gravititational potentials to. If we pick another value for zero, for instance the surface of the earth - which might seem attractive at first glance - we get problems with calculating the gravitational potential on, for instance, Mars, since the postition of Mars changes with respect to the surface of the Earth constantly. On the other hand it is always the same distance away from infinity.

Gravitational potential at a point is then defined as 'the energy per unit mass required to move an object from that point to infinity (zero)'. This means that gravitational potential will get greater but be negative, the closer you get to the surface of a massive object.

The gravitational potential INSIDE a uniform sphere decreases as you get towards the centre. This is because it is only the mass in the sphere below you which matters. Think of the solid sphere being made of an infinite number of thin layers, onion like, in towards the centre. Gauss' Law means that there is NO CONTRIBUTION to gravitational potential for all the layers which are further out from the centre than you are.

Ignoring non-sphericity of the Earth and any non-uniformity throughout the Earth, there will be zero gravitational potential due to the Earth, at the Centre of the Earth.

grant hutchison
2008-Nov-19, 03:02 PM
Ignoring non-sphericity of the Earth and any non-uniformity throughout the Earth, there will be zero gravitational potential due to the Earth, at the Centre of the Earth.I think you're mixing force and potential, here.
Potential continues to decrease as you descend into the substance of the Earth, and reaches a minimum at the centre of the Earth. If this were not so, objects would not gain kinetic energy when dropped down a hole. if the potential started to rise towards zero inside the Earth, the Earth would eject objects from holes in its substance; they would "fall" outwards towards the surface, just as objects outside the Earth fall inwards towards the surface.

Coming at it another way:
Force is the negative gradient of potential. Since force doesn't change sign as we pass through the surface of the Earth, this implies that the potential gradient continues in the same direction beneath the surface of the Earth. So potential continues to get lower as we get deeper.

I've graphed the relevant equations in an earlier post (http://www.bautforum.com/1368328-post30.html).

Grant Hutchison

John Mendenhall
2008-Nov-19, 03:15 PM
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 :)

Hmm, must be why they didn't do it.

Yes, chords count.

mugaliens
2008-Nov-19, 08:57 PM
You're talking about force, I take it, rather than potential?

Grant Hutchison

Let's assume for a moment that Earth is a perfect sphere with a uniform density throughout. In empty space (absent the sun, moon, and other planets), the center of the Earth would indeed be a location of dynamically stable zero potential. Add the sun and assume the earth is in orbit around the sun, and the center of the Earth is no longer the zero point, as the orbital sphere is curved, not flat. Thus, half the Earth's mass is inside it, and half outside. At it's point of closest approach to Earth's center, the orbital sphere is a tad further away from the sun. Thus, a object at Earth's center, inside the orbital sphere, but travelling around the sun at the same velocity as Earth, would tend towards the Sun.

Thus, the point of zero potential is a little closer to the sun than Earth's geometric center.

tommac
2008-Nov-19, 08:59 PM
For the purpose of the OP ... lets assume a perfect sphere.

Let me re-word OP slightly:

For a mass in the shape of a perfect sphere ( homogenous)

What is the gravitational potential in the following locations:

1) at infinity and beyond
2) on the surface
3) at the center of the mass

Note I am not talking about force. I am talking about the gravitational potential ( I am not sure if I am using the correct terminology ... please correct me if I am wrong )

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.

grant hutchison
2008-Nov-19, 09:08 PM
Let's assume for a moment that Earth is a perfect sphere with a uniform density throughout. In empty space (absent the sun, moon, and other planets), the center of the Earth would indeed be a location of dynamically stable zero potential.We seem to be using different definitions of the word "potential".
I'm talking about gravitational potential as treated in, for instance, this derivation (http://cnx.org/content/m15108/latest/) of the same potential curve I provided earlier (http://www.bautforum.com/1368328-post30.html). That gravitational potential (the integral of the force) reaches a non-zero minimum at the centre of the Earth.

Grant Hutchison

tommac
2008-Nov-19, 09:08 PM
My understanding of potential energy:

The maximum value is at infinity. The closer you get to the center of mass,
the lower the value. In most discussions of potential energy, especially for
engineering analysis, a value of zero is assigned to a reference level, such
as the ground surface. This is often chosen to be the minimum value likely
to be encountered. In cosmology, it is more common to assign a value of
zero to the potential at infinity, making zero the maximum value.

-- Jeff, in Minneapolis

Per your descritption above wouldnt then there would still be a potential energy at the surface and 0 at the center?

grant hutchison
2008-Nov-19, 09:11 PM
For the purpose of the OP ... lets assume a perfect sphere.

Let me re-word OP slightly:

For a mass in the shape of a perfect sphere ( homogenous)

What is the gravitational potential in the following locations:

1) at infinity and beyond
2) on the surface
3) at the center of the mass

1) Zero
2) -GMm/R
3) -3GMm/2R

Where G is the universal gravitational constant, M is the mass of the sphere, R is the radius of the sphere, and m is the mass of your test object.

See here (http://cnx.org/content/m15108/latest/) for the standard textbook derivation.

Grant Hutchison

tommac
2008-Nov-19, 09:11 PM
We seem to be using different definitions of the word "potential".
I'm talking about gravitational potential as treated in, for instance, this derivation (http://cnx.org/content/m15108/latest/) of the same potential curve I provided earlier. That gravitational potential (the integral of the force) reaches a non-zero minimum at the centre of the Earth.

Grant Hutchison

I think I agree with you here Grant.

Now the question I have is what is the relationship between the potentials at the surface and the center. the center is the min. the surface is greate than the center. Is this relationship 1/r^2 ( distance based) or 1/r^3 ( volume based ) or niether?

grant hutchison
2008-Nov-19, 09:18 PM
Now the question I have is what is the relationship between the potentials at the surface and the center. the center is the min. the surface is greate than the center. Is this relationship 1/r^2 ( distance based) or 1/r^3 ( volume based ) or niether?Neither. As I've said already, the gravitational potential varies with r2 inside a homogeneous sphere. That r2 curve is offset in the negative direction by the fact it's already at the bottom of a gravity well, so you need to subtract a constant from it, driving the whole thing negative.
The maths is given in my previous link (http://cnx.org/content/m15108/latest/), and I gave the values for surface and centre potential of a homogeneous sphere in my previous post (http://www.bautforum.com/1369030-post46.html). In the case of the homogeneous sphere, the potential at the sphere centre is 1.5 times the potential at the sphere surface, as was shown in my second graph (http://www.bautforum.com/1368328-post30.html).

Grant Hutchison

tommac
2008-Nov-19, 10:11 PM
Thanks! It is making sense now.

Sorry for a newbie question ... but what are the units of that U ( gravitational potential ) in your graph?

Neither. As I've said already, the gravitational potential varies with r2 inside a homogeneous sphere. That r2 curve is offset in the negative direction by the fact it's already at the bottom of a gravity well, so you need to subtract a constant from it, driving the whole thing negative.
The maths is given in my previous link (http://cnx.org/content/m15108/latest/), and I gave the values for surface and centre potential of a homogeneous sphere in my previous post (http://www.bautforum.com/1369030-post46.html). In the case of the homogeneous sphere, the potential at the sphere centre is 1.5 times the potential at the sphere surface, as was shown in my second graph (http://www.bautforum.com/1368328-post30.html).

Grant Hutchison

Hornblower
2008-Nov-19, 10:17 PM
Thanks! It is making sense now.

Sorry for a newbie question ... but what are the units of that U ( gravitational potential ) in your graph?If we use kilograms, meters and seconds, U will be joules.

grant hutchison
2008-Nov-19, 10:20 PM
Sorry for a newbie question ... but what are the units of that U ( gravitational potential ) in your graph?As I said in the accompanying post, they're arbitrary. I just set G, M and R equal to one. As it turns out, that had the handy effect of making F=1 and U=-1 when r=1. So the potential is plotted in units equal to Earth's surface potential.

The conventional units for gravitational potential are J.kg-1, which is what is being used in the link I gave. But for a given test mass you can simply plot its gravitational potential energy in joules, as Hornblower says.

Grant Hutchison

hhEb09'1
2008-Nov-19, 11:19 PM
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.Potential, in the way we're using it, wouldn't be greater than zero. And the force of gravity is the derivative of the potential, so not only would the force be zero at those points but at a point between, and at points between those points. That can actually happen--the entire interior of a symmetric hollow sphere has that character.

grant hutchison
2008-Nov-19, 11:44 PM
Potential, in the way we're using it, wouldn't be greater than zero. And the force of gravity is the derivative of the potential, so not only would the force be zero at those points but at a point between, and at points between those points. That can actually happen--the entire interior of a symmetric hollow sphere has that character.My link to the maths (http://cnx.org/content/m15108/latest/) also derives and graphs this result. (The hollow spherical shell, that is. I don't know where you're going with cjameshuff's local minima.)

Grant Hutchison

cjameshuff
2008-Nov-20, 12:47 AM
Potential, in the way we're using it, wouldn't be greater than zero. And the force of gravity is the derivative of the potential, so not only would the force be zero at those points but at a point between, and at points between those points. That can actually happen--the entire interior of a symmetric hollow sphere has that character.

I'm also wondering where you're going with this. Non-spherical distributions of matter can certainly have multiple minima that are not connected. The Earth-Moon system, for one...potential energy decreases from zero (I at least did not intend to imply it was ever higher than zero, and am not sure why you mentioned it) at infinite distance to one minimum near the moon's center, and another at Earth's center. Sufficient "lumpiness" in density at the cores of either could form multiple minima at those locations, though close together and with very shallow potential barriers between them. And even with perfect homogenity, the oblateness of Earth might make the minimum a disc or ring of finite size rather than a point. I attempted to stress that for almost all purposes, these could be considered points at the centers of the bodies involved. Even the Earth-Moon system, at a sufficient distance, can be considered as having a single minimum at the barycenter of the two bodies.

rodin
2008-Nov-22, 01:21 AM
A thought

we think of descending into a gravity well as a loss of potential energy. And that work must be put into the system to lift that body back out of the well

Right?

But the energy required to this is stored as pressure

Pressure can be released one way or another...

Hornblower
2008-Nov-22, 02:26 AM
A thought

we think of descending into a gravity well as a loss of potential energy. And that work must be put into the system to lift that body back out of the well

Right?

Yes.

But the energy required to this is stored as pressureOr heat, chemical energy, etc.

Pressure can be released one way or another...Yes, it can be released in a manner that does no work on the desired body. What does that have to do with the question at hand?