Gravity in a hollow sphere

[DavidHooper]

I've just finished a piece of college work on gravity in a hollow sphere and come to the conclusion... that there isn't any.

By way of explanation: the sphere is hollow (filled with a vacuum), has a thin (read: thickness=0) shell of mass M with radius and all the usual properties.

Anyway, I've come to the conclusion that a body inside the shell is subjected to zero external force. Which means if you hollow out, say, a planet and jump inside then, wherever you are - even next to the shell - you are not attracted to part of any side.

I posted because it's interesting that this goes against common intuition. I originally speculated that the body would accelerate towards a side of the shell.

I remember that Guv is quite into problems of this nature? If he's around could he comment how he'd go about this problem? I split the sphere into rings of mass m' and considered the force towards the centre of each ring and then went on from there.

[Guv]

It can be proven that there is no net field inside a hollow sphere of uniform density.

Almost any calculus text that deals with gravity can provide the proof, which I have long since forgotten. I remember that it involves a simple integration.

It is something like the integral of some function is the potential function, and the derivative of the potential is the gravitational force. For inside a sphere, the integeral is a constant, so the derivative is zero.

It would be interesting to fool around inside a hollow sphere filled with an atmosphere. With some simple wings, you could fly. You could move around slowly with just ping pong paddles.

Another cute one would be a small planet with a hole drilled through the center. If you jumped in, you would speed up until you got to the center, then slow down, and be hardly moving when you got to the other side.

[Staifour]

i'm not a scientist, and i'm really young, but i have something to say (it may be wrong)
as i know when dealing with gravity, you normally deal with the center of the object (not the center in volume but the center in mass, don't know what it is called in english (english is not my native language)), that's why if you calculate the gravity of an object with a building and calculate the gravity of that object with the erath without the biulding, then find the force that does the same as those two force, you'll have the same force if you calculate the earth with the building on it (i know it doesn't really look sensible, but if you put a really massive object instead of the biulding, you'll understand what i mean)

(here i wanted to proove the Guv had something wrong, but then i found i was wrong)

what i'm wondering is what is the force that will a planet will effeact an object that has it center in the center of the planet, if we use Maths, i think it'll be infinitive

BTW: i'm 15 years old

[DavidHooper]

Almost any calculus text that deals with gravity can provide the proof, which I have long since forgotten. I remember that it involves a simple integration.

Chuckle, I was quite chuffed with myself! (Note to self: Must try harder, must try harder, must try harder, repeat to fade...)

what i'm wondering is what is the force that will a planet will effeact an object that has it center in the center of the planet

Sounds like zero to me, by Newton's Law of Gravitation:
F=Gmm'/r^2

[kedaman]

Gauss law states something similar about electromagnetic flux, so I assume you can apply the same on gravitational flux, that is the gravity field that flows trough an area, the gravity field is reverse proportional to the square of the distance, while the the area to surround would on the other hand be proportional to the square of the distance, and thus the gravity field would be independent of distance. In other words, not only spheres, but whatever malformed shapes you can think of (that has the same density across its surface) will not induct any gravitational field within.

[sql_lall]

I noticed that you said "hollow out a sphere". Does this imply that you now have a vaccum inside, or that there is stuff in it??

[Guv]

With or without an atmosphere, the gravity due to a hollow spherical shell (of uniform density & thickness) would be zero though out the interior. From outside, it would act like a point mass at the center.

An interior atmosphere would result in a gravitational field centered on the center of the sphere. This field would be small but noticeable for a planet sized hollow sphere. The gravity due to the atmosphere would be zero in the center and increase as you moved away from the center.

Assume an earth sized thin hollow spherical shell filled with air at the same density as at the surface of the earth. I think the maximum gravitational force would be about 1/1000 of earth gravity at the surface. I think an accurate value could be calculated by dividing the density of the atmosphere by the density of the earth. My estimate is based on some guesses about these densities.

[Judd]

Does that mean that a Dyson Sphere is practically impossible? - Lack of gravity making it very difficult to live on the inside....

[Guv]

A Dyson sphere is a far out concept about enclosing an entire solar system (galaxy?) or part of it. Since there would be a star (or stars) inside the Dyson sphere, there would be gravity inside the Dyson sphere.

The purpose of a Dyson sphere is to allow a civilization to utilize almost all of the energy produced by stellar radiation.

Note that the earth receives about .000 000 00185 of the radiation from the sun. 540 million times as much radiation goes past us, most of it going into outer space.

We could eliminate the need for fossil fuel or nuclear energy if we could harness all of the sun’s energy for useful purposes.

BTW: Building a Dyson sphere is likely to be one of the SciFi projects that will never be done. It seems to be an incredibly formidable task, if not impossible. If it is possible, a civilization with the technology and resources required to construct a Dyson sphere is likely to decide it is not an economically feasible project. They are very likely to find something better to do with all those resources and technology. Like make their entire planet a generation space ship.

[Staifour]

what i'm wondering is what is the force that will a planet will effeact an object that has it center in the center of the planet

Sounds like zero to me, by Newton's Law of Gravitation:
F=Gmm'/r^2

well
as i see this equation, and i remember it as F = GMm/d^2 (d=distance), if the object is at the center of the planet, the distance will be 0, 0^2 = 0, soo F=Something/0, there fore F=INFINITIVE , now my question is how a force can be INFINITIVE, and if what i'm saying is right, then what Guv said about the drilled planet is wrong, cause when the object is exactly at the center of the planet, it couldn't move out of there again

am i right ???

[kedaman]

well
as i see this equation, and i remember it as F = GMm/d^2 (d=distance), if the object is at the center of the planet, the distance will be 0, 0^2 = 0, soo F=Something/0, there fore F=INFINITIVE

The formula only applies to masses centered at points, and when you enter the sphere, you have forces that undermine each other, so you should perform partial integration on each axis, on the sphere.
btw, x/0 is undefined, not infinite.

[DavidHooper]

And it's definitely not infinitive

[Guv]

Ignoring the higher mathematics, there is an intuitive explanation for the gravitational force at the center of a sphere.

Image being inside a small spherical space (say 2 yards or meters in diameter) at the very center of the earth. Inside that small space, all of the mass would be surrounding you. Any gravitational force would be directed toward the surface, not toward the center. Furthermore, all of the forces would cancel each other if the surrounding mass was symmetric with respect to the center.

The planet with the hole drilled through it (say from the North to the South Pole), would be a fun amusement park ride. If you were at the center, you would stay there unless the mass was not uniformly distributed.

If you jumped into the hole at the North Pole, the initial gravitational force would be a maximum. Your speed would increase as you fell toward the center, but the force would decrease to zero at the center. At the center your speed would be at a maximum and the force would be zero. As you started toward the South Pole, a small but increasing force would start pulling you toward the center. At or near the South Pole, the force pulling you toward the center would be a maximum, while your speed would be zero. You would act like a yo-yo, oscillating between the North and South Poles.

If there were no atmospheric friction, other frictional forces, or other energy losses, you would yo-yo forever. With losses due to atmospheric friction (or other effects), your yo-yo motion would be damped, and you would eventually end up motionless at the center.

An interesting problem would be to determine atmospheric equilibrium if a planet with an atmosphere were to be hollowed out and the mass concentrated in a thin spherical shell with a diameter equal to the original diameter. Suppose the shell were porous, allowing the atmosphere to easily pass from inside to outside & vice versa. The pressure at the shell would be approximately the same inside & outside. In the case of the earth, I think that most (perhaps all) of the atmosphere would end up inside.

For some parameters, there might be no atmosphere outside the shell and none in the outermost part of the inside. For other parameters, there would be atmosphere outside the shell. There are no initial conditions which would allow some atmosphere outside, with none inside.

I am pretty sure that the equilibrium conditions would vary with the mean temperature of the system. If the temperature fluctuated, the system would breathe. As the temperature rose, more atmosphere would be outside and vice versa with decreasing temperature.

[kedaman]

the gravitational field in the entire shell is uniform, none in other words, to explain this, holding up a coin, with the same thickness as the shell, towards the shell, wherever you are and in whichever direction, the area covered by the coin in your point of view will have the same gravitational effect on you as the coin itself. Thus, extrapolating this in every angle around you, you find that, no matter what kind of shape the container you are enclosed within, under the circumstance that the shell has uniform thickness. ie, this example with X at your position:

____________________________________
|                                   |
|                           X       |
|                                   |
|___________________________________|

There is an error marginal caused by asymetry and thickness of the shell though, which is specific to any shape depending on where you are located, since the thickness should be calculated normal from your position, and not to the surface. ie by rotating the above to the right:

      | |/
      | /
      |/|
      /b|
     /| |
    X | |
   /  | |
__/___| |
_/a_____|

The gravity field caused by the mass at b is stronger than the mass at a. Although if you redefined the thickness to be constant normal to your position, it would cancel the field.

[Guv]

Kedaman: Your argument is unconvincing. I am almost certain that only a hollow spherical shell of uniform density has the zero internal gravity property. To prove your contention, you need something more formal than what you have presented.

To me, it seems intuitively obvious that inside a rectangular parallelepiped, there would be a net force toward the nearest surface. Only at the center would there be zero gravitational force. The force is proportional to the inverse square of the distance. For a rectangular parallelepiped, every part of the nearer face is closer to an off-center point than the corresponding part of the opposite face. An analysis for other shapes is more difficult.

As far as I know, a spherical shell of uniform density is the only shape with zero gravity every where inside. It and a sphere (with radially symmetric density) are the only shapes which act exactly like a point mass on external objects.

[kedaman]

Gravity field is produced by mass per square of distance. My argument is that area projected at any distance to a surface with uniformly distributed thickness density is constant.

Consider this extreme case:

||\
|| \
||  \
||   \
||    \
||     \
||      \                              
||       \    /||
||        \  / ||
||         \/  ||
||         /\  ||
||        /  \ ||
||       /    \||
||      /
||     /
||    /
||   /
||  /
|| /
||/

You are positioned at the x in the middle, with 2 quadratic plates of metal at distance of 1 km and 3 km, with the size 1 km x 1 km = 1 km^2 and 3 km x 3 km = 9 km^2. Lets say the density thickness is uniform and 100kg/km^2 for both surfaces

According to Gauss, as the radial electric field is reverse proportional to the square of distance, and the areal is proportional to the square of distance (refer to earlier presentation), and thus the electric flux trough the surface of a hollow shape encapsulating the electric charge, is independent of distance and thus form of the shape, as long as its surface covers all sight from inside. This is formally known as Gauss Law.

Now assuming you had gravity concentrated in a surface with uniformly thickness and density, or thickenss density, you could extrapolate it for gravitational flux.

The gravity field in the above example will be G * m/r^2 = G * ( 100kg/km^2 * 9 km^2 / 9 km^2 + 100kg/km^2 * 1 km^2 / 1 km^2 ), which should resolve to 0 kg/s^2 (gravity field, not gravitational force, by multiplying your mass you will get gravitational force)

This case doesn't take into account that the radial thickness (the thickness normal to x) is not uniform across the surface, but the result should be 0 in any case since the net field produced by the opposite surfaces match each other at any angle.

If you have understood me correctly, then you should understand the coin projected on surface point I made.

However about the hollow asymetric shapes with density, I stand corrected, and Guv is correct that spheres have this property although I doubted it until I did an analysis of it, due to symmetry of thickness in any direction. Although assuming that at any specific position in any hollow shell of any asymetric shape with the criteria of uniform density and symetric thickness normal to the position in all directions, should produce no gravitational field.

[Judd]

However about the hollow asymetric shapes with density, I stand corrected, and Guv is correct that spheres have this property although I doubted it until I did an analysis of it, due to symmetry of thickness in any direction. Although assuming that at any specific position in any hollow shell of any asymetric shape with the criteria of uniform density and symetric thickness normal to the position in all directions, should produce no gravitational field.

Surely, a shell with symmetric thickness normal to an internal point in ALL directions is a hollow sphere?

[kedaman]

I guess. Its intuitive, but intuition can be a fault, and I didn't have time to find a proof

[kedaman]

No matter what I think gauss law of electric flux can be applied on gravity flux though, since it can be applied on magnetic flux

[Staifour]

sorry to disturbe all
but
if the equation is (F = GMm/d^2) how come an object at the center of a planet has a gravitional force of 0 (x/0 doesn't equal 0)??? what is D in this equations, i mean i know it is ditance but distance from where to where ???
and about the hollow sphere, an object not in the middle of the hollow sphere will be closed to one part of the sphere then another, if D is the distance between the center of object and center of shpere, how come it has no gravitional force ????
BTW, X/0 is really undefined, but english is not my native language, so i didn't know what it is called in english, so i thought infinite will be the closest i can think of
thanks in advance

[Guv]

Staifour: F = GMm/d^2 is only applicable when you are outside a spherical object or a spherical shell.

If you dug a hole from the North to the South Pole, gravity would get weaker as you traveled toward the center. The mass above counteracts some of the mass below. The exact center is surrounded by a symmetric distribution of mass, causing all the forces to cancel. Of course, this makes certain assumptions about density and mass distribution being symmetric with respect to the center.

Note also that F = GMm/d^2 is not applicable to irregular shapes. It becomes a better and better approximation as the irregular objects move farther and farther apart, but is exact only for spherical objects.

Kedaman & Others: My calculus is too rusty to set up the integrands and do the integrations required to determine the potential functions. The following is an intuitive analysis of this problem.

Imagine a hollow sphere of radius R and a hollow parallelepiped both centered on the origin of an XYZ coordinate system. For simplicity, let the parallelepiped have the its top and bottom rectangles tangent to the North & South Poles of the sphere (Poles assumed to be on Z-Axis). The sides are assumed equidistant from the XZ-Plane, with the front & back equidistant from the YZ-Plane.

For purposes of the following discussion, the actual dimension do not matter, only the symmetries.

Imagine moving up and down the Z-Axis, staying inside the sphere and the box.

• Both the sphere and the box are symmetric with respect to the Z-Axis. Therefore there is no net gravitational force in any direction parallel to the XY-Plane.
  
• As the point moves toward the North Pole of the spherical shell, more of the mass of the shell is below the point and less is above. The point is closer to the mass above it and farther from the mass below it. Each bit of mass above being closer exerts more force. Each bit below exerts less due to be farther away. Intuition cannot determine the net effect of the greater mass farther away versus the nearer mass which is closer.
  
• As the point moves toward the top of the box, it gets closer to all of the mass in the top and farther from all of the mass in the bottom. If the sides, front, and back of the box are distant from the origin compared to the top and bottom, they will contribute little force parallel to the Z-Axis.

To me, the above is a convincing argument indicating that there is a net gravitational force inside the box. The above is inconclusive as regards the sphere, but allows for the possibility of a zero field.

Those whose calculus is in better shape than mine, might want to consider the following.

Potential(x, y, z) = G*Integral(Density*dv / r)
GravitationalForce(x, y, z) = DirectionalDerivative[ Potential(x, y, z) ]

Where G is the gravitational constant, dv is an infinitesimal of volume, and r is the distance from (x, y, z) to the little piece of volume.

If the density is constant, it can be brought outside the integral as a multiplicative factor.

As mentioned above, my calculus is too rusty to deal with the above. I remember doing the work for a spherical shell as an exam question about 50+ years ago. I remember the potential being constant everywhere inside the shell. The derivative of a constant is zero, resulting in zero gravitational force inside the shell. The integral for outside the shell indicated that the shell acted as a point mass on external objects.

I do not remember any discussion of other hollow shapes, but my intuition strongly suggests that none have the zero gravity property of the hollow sphere. I am also pretty certain that only a spherical shell or a sphere act as point masses on external points. At large distances, any shape is usually treated as a point mass, but this is an approximation. The approximation is excellent at large distances, but questionable when close to irregular shapes.

[kedaman]

Guv and Others
I don't like messing with calculus either, and usually find another way around it.

The reason why it only works for spherical (I don't know if this actually is correct, but probably proovable but with much more effort than I want to spend) shells is because 1. mass is distributed over volume, not area, therefore tilting a surface away from the normal to the point of which we want to evaluate the gravitational field, to increase the density volume per area (or density thickness). This happens with all shells including the sphere, but the sphere supports another symmetry,

Cut a spherical water melon in any direction across any point, you will see the radial symmetry of the thickness of the shell, and assuming the center of the section is the point you want to calculate the gravity field for, you will notice that the shell is equally thick in opposite sides, and in whichever direction you cut, you will find the same, and trough whichever section you cut. This symmetry is not possible on other shells.

[Staifour]

i have an idea that will solve the whole problem
i think that F=GMm/D^2 applies to ALL kinds of objects, if you are dealing with each molecule apart, so if you take each molecule in the first object with all the molecule in the second and find the result, you should have a right result.
now if you work with what i said, you'll find that the hollow sphere really doesn't have any gravitation inside (now i think i uderstand why), cause if you are at any point inside the sphere, you can be nearer to a part of the shell then another part, but that part will have less mass, cause it will be smaller, so it'll have less molecules inside (if the mass of the sphere is uniformly distributed)
so at the end all the forces will cancel each other
(i know that you have all said what i'm saying, but i'm just saying HOW i understood it)
and using my explanation of the formula (F=GMm/D^2) you can also deal with irregular objects, but it won't be practical.
hope i got it right

[sql_lall]

Is this sphere perfectly still, or is it spinning??

If it is spinning, wouldn't you have to assume that the gravity would pull you away from the centre. (For after all, isn't this how they intend to create gravity in space-shuttles???)

[kedaman]

Staifour
The formula applies to pointformed objects masses, and applies to spheres with uniform densities as well. For other shapes you'd have to divide it into fundamental geometrical objects and then use partial integrals to evaluate the field to to each, and then sum them up in superposition.
Sql_lall
A spinnig it won't contribute to gravity field, the force you are thinking of is centrifugal and does not exist, but is the illusion of another, the centripetal force that pulls you towards center. Centripetal force can be used to simluate gravity, but they have nothing to do with each other.

[sql_lall]

So what you are saying is that if someone makes a sphere, and gets inside, that there will be no gravity imparted on them??

Or, what i think you are saying, is it that the sphere itself imparts no gravity, it's just that the earth the sphere is resting on does??

But isn't this also implying that the only time that the inside of a sphere truly has no gravitational fields is when it is far enough away from large objects that impart their own grvitational field onto the sphere??

[Guv]

Sql_lall: You are correct, we are talking about gravitational forces due to the hollow sphere. If the hollow sphere were near or resting on a large object, there would be forces felt inside the hollow sphere due to the other object. This is unlike an external electric charge or magnetic field, which would not be felt inside a metal box of any shape.

You would think there was gravity inside a spinning sphere. It gets complicated if the sphere is spinning on more than one axis. It is easier to think about a cylinder spinning about its axis of symmetry.

The inertia associated with an object inside a spinning cylinder, makes an object inside tend to be motionless or move in a straight line. If in contact with the cylinder, the object is forced to move in a circular path. This forced motion, which fights inertia, acts like gravity directed perpendicular to the surface of the cylinder.

BTW: General Relativity (not Special Relativity) was developed by Einstein because he thought deeply about the equivalence between gravitational forces and the inertia forces due to acceleration.

[sql_lall]

Who is sql_loll???

I mean, you have 10 posts. Is this just an obsession in trying to seem smart (in which case, since we don't know who you are, you can't claim credit )

I actually asked the question because, for simialar reasons that many people ask questions, i didn't know something, and wanted clarification.

If this truly is your name, and it is just a strange coincidence, then i apologise, and find it unfortunate that we have similar names. (But still don't understand your post

BTW: Guv, thanks for answering.

[DavidHooper]

Where's the moderators around here?!