Gravitational potential of a point within a hollow sphere

[eep]

Hi,
I know that if you have a hollow sphere of some finite mass, when you place an object inside it, it experiences no gravitation force. Does this mean that the point also has no gravitational potential energy?

thread title should be "Gravitational" not "Gravitional"

 

[rcgldr]

A "point" is an abstract concept, it's infinitely small and has no mass. A real object would have mass and create a gravitational field on it's own, regardless of whether or not it was inside a hollow sphere.

 

[eep]

Well say you've got some tiny object inside of a hollow sphere. What's the gravitation potential energy of the object due to the surrounding mass of the hollow sphere?

 

[rcgldr]

Edit: correcting my posts in this thread.

The potential energy within the hollow sphere is constant, regardless of the objects position within the sphere.

It's the net gravitation forces on the objects that sum up to zero, regardless of the objects position within the shpere.

 

[eep]

But how can it's gravitation potential energy due to the spherical shell be 0 if the potential is defined as

$$\frac{-GMm}{|r|}$$

How do you add a bunch of negative numbers together and get 0?

 

[rcgldr]

How do you add a bunch of negative numbers together and get 0?

Edit: update to reply:

If you add up all the forces on the object, the forces sum up to be zero at any location within the sphere. If there is no net force on the object, then the potential energy should be zero, or at least a constant independent of position.

I don't understand how potential energy from multiple sources could be summed without taking direction (location of those sources) into account. It seems to me that you'd need to sum up vectors, not scalars.

 

[neutrino]

Hi,
I know that if you have a hollow sphere of some finite mass, when you place an object inside it, it experiences no gravitation force. Does this mean that the point also has no gravitational potential energy?

The potential has to be a constant (since $-\nabla V = \vec{F} = 0$)but, need not be zero. You have to calculate the potential as two parts...from the reference point (which, in this case, is usually taken at infinity) to the surface of the sphere, and the other from the surface to the point under consideration. In each part (an integral) you have to use the form of the field (which is essentially the force) which exists in that particular region.The second integral will vanish since the field is zero inside, and the first part will come out in such a way that it is a constant.

 

[vanesch]

But how can it's gravitation potential energy due to the spherical shell be 0 if the potential is defined as

$$\frac{-GMm}{|r|}$$

How do you add a bunch of negative numbers together and get 0?

As the previous poster said, you do not have to get 0. You only have to get a constant, independent of position.

 

[Rach3]

If the object is spherical, I would assume that the gravitation potential energy from the hollow sphere is zero, since you can assume that all the mass is at the center point of the object.

This assumes one is outside the sphere, not inside.

 

[eep]

Because you're adding up vectors, not numbers, and the vectors have directions. The net result is the difference in direction for the force vectors cause them to sum up to zero, no matter where the spherical object is inside the hollow sphere.

Potential energy is a scalar, not a vector.

 

[pervect]

Hi,
I know that if you have a hollow sphere of some finite mass, when you place an object inside it, it experiences no gravitation force. Does this mean that the point also has no gravitational potential energy?

thread title should be "Gravitational" not "Gravitional"

As others have mentioned, the condition for zero force is only that the gravitational potential be constant. It does not matter to the force what the magnitude of the potential is, the force is proportional to the rate of change of potential (dV/dx).

In classical mechanics, the "zero point" of potential is arbitrary in any event - one can add or subtract any constant number from the potential function without changing the physics.

A good way of looking at this: the escape velocity inside a hollow sphere will be the same as the escape velocity on the surface of a hollow sphere. An object will be at escape velocity when it has just enough kinetic energy to reach infinity.

 

[rcgldr]

Potential energy is a scalar, not a vector.

If you add up all the forces on the object, the forces sum up to be zero at any location within the sphere. If there is no net force on the object, then the potential energy should be zero, or at least a constant independent of position (within the sphere).

I don't understand how potential energy from multiple sources (in this case, the integral sum of all the points of the hollow shpere) could be summed without taking direction (location of those sources) into account. It seems to me that you'd need to sum up vectors, not scalars.

it experiences no gravitation force

Edit: The object experiences no gravitation force. Original and incorrect reply: The object inside a hollow shpere will experience a pull in all directions, expanding the object slightly (assuming that the object isn't decompressable).

What I don't know for sure is if the object inside a hollow shpere isn't spherical, if the potential energy is still independent of where in the sphere that the object is located. If you divide the object into smaller and smaller components, then as the size of the components approach being points, then each point's potential energy is independent of it's location, so it would seem that the integral sum of all these points would be independent as well.

 

[Doc Al]

The object inside a hollow shpere will experience a pull in all directions, expanding the object slightly (assuming that the object isn't decompressable).

Since the gravitation field within the hollow sphere is zero, the object will experience no gravitational force at all. It will not be pulled in all directions; it will experience no stress or strain, and will not be stretched or expanded.

 

[rcgldr]

the object will experience no gravitational force at all

Agreed, I just realized that when I considered the integral sum of all the points that make up the object inside the sphere. Each point experiences zero force, so the entire object experiences zero force. I need to take more time in composing replies instead of editting them afterwards, in the words of Ronald Reagon, I'll try to avoid "shooting from the hip".

 

[rcgldr]

This was brought up before but never clarified. What gravitational effect would a third object, outside the sphere, have on an object inside the sphere? My intution is that it would be the same as if the hollow sphere wasn't there at all.

 

[Hootenanny]

This was brought up before but never clarified. What gravitational effect would a third object, outside the sphere, have on an object inside the sphere? My intution is that it would be the same as if the hollow sphere wasn't there at all.

Indeed, this would the case, as long as the object inside the hollow sphere did not collide with the hollow sphere, note that the gravitational field of the third object will also infuence the hollow sphere.

 

[Doc Al]

I don't understand how potential energy from multiple sources (in this case, the integral sum of all the points of the hollow shpere) could be summed without taking direction (location of those sources) into account. It seems to me that you'd need to sum up vectors, not scalars.

If you are adding up the gravitational field contributed by each mass element, then of course you'd need to add them as vectors. But if, instead, you added up the potential contributed by each mass element, only distance, not direction, would matter since the potential is a scalar. (That's one of the advantages of using potential--it's a scalar.) And if you set up the calculation and do the integral you'll find that the potential at any point is independent of position within the hollow sphere.

Since field and potential are intimately related (given the potential function, one can derive the field), either method must lead to the same conclusion.

This was brought up before but never clarified. What gravitational effect would a third object, outside the sphere, have on an object inside the sphere? My intution is that it would be the same as if the hollow sphere wasn't there at all.

Exactly. The field contributions from all masses just add up. Since the sphere contributes nothing, all you have is the field from that third object.

 

[reilly]

A pedantic point or two: first, you must assume that the mass distribution of the sphere is spherically uniform, it can be radially nonuniform. Second, you must assume that the inside mass has no influence on the mass distribution of the sphere -- unlike, say, the same problem for a charged spherical shell, isolated or grounded. And yes, the charge outside will give a -GmM/s portential for the inside charge, where s is the distance between the two masses. But, is this the correct total potential?

If the shell has mass M', what is the potential for the outside charge?

Regards,
Reilly Atkinson

 

[rcgldr]

My point about potential energy from multiple sources was meant to be more general than the hollow sphere.

Take an object that is between the Earth and the Moon right at the point where it does not fall towards either. In this case, the potential energy is probably indeterminate. If the object moves and falls towards the Moon it's potential energy is far less than if the object falls and moves towards the Earth. How is this situation handled with scalar math?

 

[Doc Al]

Take an object that is between the Earth and the Moon right at the point where it does not fall towards either.

The gravitational field at that point is zero; the potential has a local maximum; an object placed there will be in unstable equilibrium.

In this case, the potential energy is probably indeterminate.

Not at all. See below.

If the object moves and falls towards the Moon it's potential energy is far less than if the object falls and moves towards the Earth.

This is true.

How is this situation handled with scalar math?

Very simply. The gravitational potential at a point outside a spherical object (assume the Earth and moon are spherically symmetric--close enough) is given by:
$$-G\frac{M}{R}$$
where, as is typical, we take the potential as zero at infinity, and R is the distance to the center of the gravitating object.

So the potential at some point due to the gravity of Earth and Moon is just the sum of each contribution:
$$-G (\frac{M_{earth}}{R_{earth}} + \frac{M_{moon}}{R_{moon}})$$

 

[rcgldr]

So the potential at some point due to the gravity of Earth and Moon is just the sum of each contribution

I didn't do the math, but the change in kinetic energy will correspond to the change in the sums of these two potential energies as the object moves towards one or the other? I can see that it's velocity would be more if there was only one object instead of two, but does it work out as I just mentioned that kinetic energy will equal change in the sum of potential energy?

 

[Doc Al]

Sure, mechanical energy (KE + PE) is conserved. And, yes, you must include the PE due to both objects.

 

[rbj]

This was brought up before but never clarified. What gravitational effect would a third object, outside the sphere, have on an object inside the sphere? My intution is that it would be the same as if the hollow sphere wasn't there at all.

actually, i had mistakenly brought it up in that other thread:

https://www.physicsforums.com/showthread.php?t=121120"

and your intuition is correct. since there is no "negative" or opposite gravitational charges (from positive mass charge) as there is for E&M, there is no gravitational sheilding.

 

[eep]

How about the potential that the spherical shell has due to itself? I've been trying to calculate this but end up with problems as I inevitably run into a point where the distance between the two masses is 0, giving me an infinity. Also, I understand that setting the 0 of energy is arbitrary. However, for a system, it must be the same zero for the entire system. If we have one spherical shell, and set the potential inside the shell to be zero, the potential inside a slightly larger sphere couldn't possibly be 0, since the potential energy at a point inside a spherical shell depends upon the radius of the shell, correct?

 

[rbj]

How about the potential that the spherical shell has due to itself? I've been trying to calculate this but end up with problems as I inevitably run into a point where the distance between the two masses is 0, giving me an infinity. Also, I understand that setting the 0 of energy is arbitrary. However, for a system, it must be the same zero for the entire system. If we have one spherical shell, and set the potential inside the shell to be zero, the potential inside a slightly larger sphere couldn't possibly be 0, since the potential energy at a point inside a spherical shell depends upon the radius of the shell, correct?

the place where you set potential energy to zero is sort of arbitrary since it is only the difference of PE that matters. the math is cleanest if PE is zero at the point out at infinity (which is why the PE is negative for any point closer than that).

the way to determine what the net gravitational field is inside a spherical shell, due to the shell itself (with no other objects around inside or outside the shell), is to use Gauss's Law (which works for any inverse-square field) and assume spherical symmetry. for any spherical surfaces fully inside the shell, there is no mass contained so the total outward gravitational flux crossing the surface must add to zero. once you're outside the shell, then there is no difference between the field due to the shell or if all of the mass of the shell were concentrated at the center point. it will be a net inverse-square field.

 

[pervect]

How about the potential that the spherical shell has due to itself? I've been trying to calculate this but end up with problems as I inevitably run into a point where the distance between the two masses is 0, giving me an infinity. Also, I understand that setting the 0 of energy is arbitrary. However, for a system, it must be the same zero for the entire system. If we have one spherical shell, and set the potential inside the shell to be zero, the potential inside a slightly larger sphere couldn't possibly be 0, since the potential energy at a point inside a spherical shell depends upon the radius of the shell, correct?

The potential of a spherical shell is equal to -GM/r outside the shell, and zero inside the shell.

You may be interested in not the potential, but the gravitational self-energy aka gravitational binding energy of the shell. That would be the amount of energy needed to disassemble the shell.

(If you're not, just skip this part below).

If you imagine pulling the shell apart, you can see that you would need to apply a force (GMdm/r^2) to every mass element dm, from r to infinity.

As the shell expands, the force equation won't change - to calculatse the required force, you can still use the simplification that the force is the same as if all the mass were at the center of the shell.

This intergal gives a binding energy of GM^2/r for a uniform shell.

Contrast this to a sphere, where the binding energy is (3/5)GM^2/r

http://en.wikipedia.org/wiki/Gravitational_binding_energy

The binding energy of a sphere is not just the sum of the binding energies of its shells, because the shells interact with each other.

 

[rcgldr]

So the potential at some point due to the gravity of Earth and Moon is just the sum of each contribution

Sure, mechanical energy (KE + PE) is conserved. And, yes, you must include the PE due to both objects.

Ok, I'm having a problem with this. Say the initial condition is an object about 100,000 km from the earth, which is allowed to free fall towards the earth. The object is directly between the moon and the earth. The velocity when the object reaches the Earth's atmosphere would be higher if there were no moon behind the objects path than if there was a moon behind the object (because the moons gravity would slow the object somewhat). Yet getting back to the initial condition at 100,000km from the Earth the PE of just the Earth would be less than the sum of the PE from the Earth and the moon. Are the PE values signed so that the PE component from the moon has the opposite sign of that of the PE from the earth?

 

[rbj]

The potential of a spherical shell is equal to -GM/r outside the shell, and zero inside the shell.

small bone: the force inside the shell is zero, but the P.E. is constant and equal to whatever the P.E. is at the inside surface. if the shell is infinitely thin with a radius of R, the P.E. inside the shell is -GM/R everywhere inside. it would be quite a force indeed if the P.E. changed suddenly from -GM/R to 0 right at the radius of the inside surface, no?

also, i wonder if the binding energy for a solid sphere of uniform mass density is the maximum binding energy for the same mass and density of any other shape? i'll bet that is true and that is fundamentally why planets that are large enough form in the shape of a sphere. nowhere farther for the material (that makes up the planet) to fall.

 

[pervect]

small bone: the force inside the shell is zero, but the P.E. is constant and equal to whatever the P.E. is at the inside surface. if the shell is infinitely thin with a radius of R, the P.E. inside the shell is -GM/R everywhere inside. it would be quite a force indeed if the P.E. changed suddenly from -GM/R to 0 right at the radius of the inside surface, no?

Ooops, definitely an error on my part.

The potential energy at a point outside the shell is is -GM/r where r is the the distance of the point from the center of the shell, and is -GM/r0 inside the shell, where r0 is the radius of the shell.

also, i wonder if the binding energy for a solid sphere of uniform mass density is the maximum binding energy for the same mass and density of any other shape? i'll bet that is true and that is fundamentally why planets that are large enough form in the shape of a sphere. nowhere farther for the material (that makes up the planet) to fall.

For a uniform density (and no rotation) I think your'e right.

 

[Doc Al]

Ok, I'm having a problem with this. Say the initial condition is an object about 100,000 km from the earth, which is allowed to free fall towards the earth. The object is directly between the moon and the earth. The velocity when the object reaches the Earth's atmosphere would be higher if there were no moon behind the objects path than if there was a moon behind the object (because the moons gravity would slow the object somewhat). Yet getting back to the initial condition at 100,000km from the Earth the PE of just the Earth would be less than the sum of the PE from the Earth and the moon. Are the PE values signed so that the PE component from the moon has the opposite sign of that of the PE from the earth?

What matters is the change in PE as the object falls. In your scenario, as the object falls the change in PE due to the Earth decreases while the change in PE due to the moon increases. So the changes in PE due to Earth and moon have different signs.

You can calculate the change in PE using the formulas I gave earlier. Just input the values of the distance at the points you want to compare, compute the value of the PE at those points, then subtract to find the change.

 

[eep]

You may be interested in not the potential, but the gravitational self-energy aka gravitational binding energy of the shell. That would be the amount of energy needed to disassemble the shell.

(If you're not, just skip this part below).

If you imagine pulling the shell apart, you can see that you would need to apply a force (GMdm/r^2) to every mass element dm, from r to infinity.

As the shell expands, the force equation won't change - to calculatse the required force, you can still use the simplification that the force is the same as if all the mass were at the center of the shell.

This intergal gives a binding energy of GM^2/r for a uniform shell.

Contrast this to a sphere, where the binding energy is (3/5)GM^2/r

http://en.wikipedia.org/wiki/Gravitational_binding_energy

The binding energy of a sphere is not just the sum of the binding energies of its shells, because the shells interact with each other.

This is what I've been trying to calculate, because the derivation I saw didn't make any sense to me. Here's what I saw:

http://scienceworld.wolfram.com/physics/SphereGravitationalPotentialEnergy.html

What they're doing is taking a spherical shell of radius r, thickness r+dr, and calculating the potential between that spherical shell and all the mass containing within the shell. What *I* don't get is why can they ignore all the mass outside the shell, and why can they ignore the potential energy due to the interactions of the mass that lies *on* the shell.

I've been trying to do it without somebody just giving me the answer, which is why I didn't ask what the self-binding energy was outright :)

EDIT: Also, in pulling the shell apart, how can it be GMdm/r^2, as all the little bits that make up M are not equi-distance from dm. I also don't see how you can simplify things by saying you can concentrate all the mass at the center of the shell, since we're trying to calculate the force *on* the shell, due to the other parts of the shell, not inside or outside of it.

 

[pervect]

This is what I've been trying to calculate, because the derivation I saw didn't make any sense to me. Here's what I saw:

http://scienceworld.wolfram.com/physics/SphereGravitationalPotentialEnergy.html

What they're doing is taking a spherical shell of radius r, thickness r+dr, and calculating the potential between that spherical shell and all the mass containing within the shell. What *I* don't get is why can they ignore all the mass outside the shell, and why can they ignore the potential energy due to the interactions of the mass that lies *on* the shell.

I've been trying to do it without somebody just giving me the answer, which is why I didn't ask what the self-binding energy was outright :)

EDIT: Also, in pulling the shell apart, how can it be GMdm/r^2, as all the little bits that make up M are not equi-distance from dm. I also don't see how you can simplify things by saying you can concentrate all the mass at the center of the shell, since we're trying to calculate the force *on* the shell, due to the other parts of the shell, not inside or outside of it.

Take a spherical mass. The amount of energy required to remove any small piece of the mass from its surface to infinity will be

E = G*M*dm/r

where r is the distance of the piece of the mass element dm from the center.

This comes from the gravitational potential of a sphere.

Now, to remove a thin spherical shell, the energy required is just

E = G*Mencl*(rho*4*Pi*r^2*dr)/r

from the above formula. This works because the shell is so thin that it doesn't have any significant self energy. (Can you see why the self energy of just the shell is on the order dm^2 when the shell is small, while the energy due to the shell coupling to the rest of the mass is of the order M*dm?)

Mencl is the mass of the enclosed sphere, which is just

Mencl = rho*(4/3)*Pi*r^3

We make r vary downard from R to 0, R being the radius of the mass. As we peel away each layer of the sphere, Mencl drops.Put this all together, mash it around, do the intergal, and you should get the desired result, after back-substituing for the total mass M in terms of density, rho, and radius, R.

 

[eep]

Ah, now I understand. A much better explanation. I thought that perhaps you could say the that due to the thinness of the shell it doesn't have any significant self-energy, but it makes much more sense because the self energy of the shell will be on the order of dm^2. Thank you.