Why does gravity cancel out for all points inside a sphere?

[Chaos' lil bro Order]

Hi,

I've heard that gravity cancels out for all points that are inside a sphere. How is this proven mathematically?

ty

 

[ImagineLab]

yes, all you need is an intergral

 

[rbj]

whomever you heard that from is not to be trusted for accurate knowledge of physics.

 

[Lyuokdea]

whomever you heard that from is not to be trusted for accurate knowledge of physics.

Perhaps your source was correct and you misheard:

Gravity does cancel out for all points inside a hollow spherical shell. That is, assume there is no air inside a soccer ball, the gravity from the soccer ball on an ant floating anywhere inside the ball will sum to zero. This can be shown by adapting the integral form of Gauss's Law replacing the magnetic field with the gravitational field. If you are not yet familiar with vector calculus in order to use Gauss's Law directly, then you can simply think about it this way. In the exact middle of the hollow shell, there will be a gravitational pull from every point on the shell, but each point will have a different direction. In fact, for every point with a gravitational pull in one direction, there is a point on the exact opposite side of the ball with the same amount of pull but in the opposite direction.

As you move from the middle to some other point inside the ball, you move closer to one edge of the ball than the other, and thus the forces from one side become stronger than the other. However, there is a competing change, as you move towards one side of the ball the amount of material behind you gets large, as you get closer to the material in front of you, these two competing changes will tend to cancel each other. What Gauss's Law beautifully proves, and what you may not guess intuitively, is that these two competing changes actually cancel each other exactly. Thus there is no net gravitational force anywhere inside the hollow spherical shell

But for a solid sphere, like, for instance, the earth. Gravity certainly does not cancel out. What would this mean for tunnels? And given that the Earth is not a perfect sphere what does this mean for valleys and other places of lower radius?

~Lyuokdea

 

[pervect]

Hi,

I've heard that gravity cancels out for all points that are inside a sphere. How is this proven mathematically?

ty

see for example

http://www.merlyn.demon.co.uk/gravity1.htm#FiSSh

 

[Doc Al]

But for a solid sphere, like, for instance, the earth. Gravity certainly does not cancel out.

Just to expand on this a bit. If you pretend the Earth has a spherically symmetric distribution of mass, then at any point within the Earth (say a distance r from the center) the gravitation field depends only upon the mass contained within the sphere of radius r. The mass at points > r can be considered a collection of spherical shells and, as you explained, they contribute nothing to the field at points inside them.

 

[arildno]

And, if you want to prove it mathematically using Newton's general law of gravitation and calculus techniques, here is most of it:
1. In spherical polar coordinates, let the position of a mass particle inside the ball be given by (\hat{r},\theta,\phi) (measured from the C.M)
where \phi is the angle between the vertical and the particle's position vector.
Let the density be constant for all sphere particles and the radius of the sphere R.


2. Consider a test particle P having mass m and position vector r\vec{k}, i.e, a distance r along the "vertical"

3. We need to sum up all forces acting on P from sphere particles, i.e, compute the integral:
\vec{F}= -G \rho{m} \int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi} \frac{\hat{r}^{2} \sin \phi((r-\hat{r}\cos\phi) \vec{k} - \hat{r} (\sin \phi ( \cos \theta\vec{i} + \sin \theta \vec{j} ))} {( \hat{r}^{2} + r^{2} - 2r \hat{r} \cos \phi )^{ \frac{3}{2} } } d \theta {d \phi} d \hat{r}
where G is the universal gravitation constant and \rho is the density of sphere particles.

4. It is easy to see that the horizontal plane components vanishes; the \phi-integration is then best handled by integration by parts.
In the \hat{r} integration, take care of whether you have r<\hat{r} or r>\hat{r}

 

[jasc15]

I love this stuff.. So suppose the Earth was a hollow sphere, there would still be gravity on the outside surface, but what if you dug a hole and jumped inside? you would have some velocity from gravitational acceleration while you were above the surface, so would you continue at a constant velocity inside the sphere? also, could air be pulled inside the sphere?

 

[DaveC426913]

I love this stuff.. So suppose the Earth was a hollow sphere, there would still be gravity on the outside surface,

It would be VERY small though.

To make a shell ~100miles thick, you'd need to remove 90% of the Earth's mass. g would drop proportionally - it would be ~ 0.98m/s^2.

but what if you dug a hole and jumped inside? you would have some velocity from gravitational acceleration while you were above the surface, so would you continue at a constant velocity inside the sphere? also, could air be pulled inside the sphere?

Now that we see gravity would be very small, we can see that our fall through the hole would be quite slow, as would the air.

I'd be interesteed in edge effects though. If your acceleration were 0.98m/s^2, how fast would you be going after a 100 mile fall? And once you exited, yes, I guess you'd continue at that speed.

 

[quantumdude]

G would drop proportionally - G would be ~ 0.98m/s^2.

Just to avoid confusion: It's not "BIG G" (Newton's universal gravitation constant) but rather "little g" (the acceleration due to the Earth's gravity) that changes.

 

[rbj]

Gravity does cancel out for all points inside a hollow spherical shell.

you're absolutely correct. i was thinking of gravitational sheilding like there is for E&M. since there is no negative gravitational charges, there is no gravitational sheilding.

but, being an inverse-square field, Gauss's Law applies and inside any hollow sphere, there is no net graviational field (unless there is some other big object nearby, the hollow shell does not sheild the field from that.

 

[Loren Booda]

How does the calculation of force within a charged sphere differ from that within a massive sphere? Does not the gravitational case only cancel for geometrical symmetries?

 

[Doc Al]

How does the calculation of force within a charged sphere differ from that within a massive sphere?

Since both are governed by inverse square forces, the calculations are essentially identical. The electric field within a spherically symmetric shell of charge and the gravitational field within a spherically symmetric shell of mass are both zero.

Does not the gravitational case only cancel for geometrical symmetries?

Yes, the shell theorem applies to spherically symmetric distributions of mass (or charge).

 

[Loren Booda]

How does the calculation differ between the two cases for a point outside the sphere?

 

[Doc Al]

Again, the calculations are essentially identical. The gravitational field outside a spherically symmetric mass distribution is given by GM/r^2 (where M is the total mass of the sphere); similarly, the electric field outside a spherically symmetric charge distribution is given by kQ/r^2 (where Q is the total charge of the sphere).

 

[Chaos' lil bro Order]

What if gravity obeyed an inverse-cubed law, would the force inside the sphere still net to zero?

 

[scott1]

What if gravity obeyed an inverse-cubed law, would the force inside the sphere still net to zero?

I thik you mean inverse-square law. I don't think there's no such thing as an inverse cubed law.

 

[dav2008]

He was asking a hypothetical.

The point is that the gravitational force is always radially outward from the center of the sphere. No matter how rapidly it dies off, it still has the same value at any given point at the surface of a spherical shell of some arbitrary radius r.

 

[Greg825]

What if gravity obeyed an inverse-cubed law, would the force inside the sphere still net to zero?

I doubt it. I don't know enough 3-d math to prove it but just doing a mental expiriment... If gravity obeyed, say, an inverse r^10 law, it seems to me that being very near the edge of the sphere would create a (relatively) huge gravitational force toward that part of the sphere, while the points on the sphere surface that are really far away would have that much less of an effect.

 

[Chaos' lil bro Order]

Why were 2 of my comments deleted?

I made 3 posts around 10:45.
ONLY one is showing.
None were offensive or redundant.

Since when do moderators sensor legitimate comments on PF?

What the hell is going on here?

 

[arildno]

Why were 2 of my comments deleted?

I made 3 posts around 10:45.
ONLY one is showing.
None were offensive or redundant.

Since when do moderators sensor legitimate comments on PF?

What the hell is going on here?

I'm sure they didn't. There must have been a bug.

 

[ZapperZ]

Why were 2 of my comments deleted?

I made 3 posts around 10:45.
ONLY one is showing.
None were offensive or redundant.

Since when do moderators sensor legitimate comments on PF?

What the hell is going on here?

You should check with the Moderators first before making such accusation. There has been NO post by anyone deleted on this thread. I've double-checked!

Zz.

 

[arildno]

Well, but I can understand Chaos Order's frustration that some of his posts aren't to be found here. Could it be the unfortunate result of some server weakness?

 

[ZapperZ]

For that, someone will have to ask Greg. That's nothing any of the Mentors are able to check.

Zz.

 

[arildno]

Chaos lil bro Order:
As to your other question:

Well, it is a matter of simple calculation. Doing the \theta-integration in my previous post, and adding a "p" for deviation from an inverse square law, we basically should calculate:

\vec{F}=-2 \pi {G} \rho {m} \vec{k} \int_{0}^{R} \int_{0}^{\pi} \frac{ \hat{r}^{2} \sin \phi ((r - \hat{r} \cos \phi) \vec{k} }{( \hat{r}^{2} + r^{2} - 2r \hat{r} \cos \phi )^{ \frac{3+p}{2} }} {d \phi} d \hat{r}

This is definitely "doable"; I'm sure you'll find out that it won't cancel out for other values of p than zero.
Take care to delimit the range of acceptable p-values for which the integral does not diverge.

 

[pervect]

For the calculations above, I think it would be much easier to integrate the potential inside the sphere than the forces.

I should add that if one does a GR analysis rather than a Newtonian analysis of the spherical shell problem, Birkhoff's theorem applies.

http://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity)

Another interesting consequence of Birkhoff's theorem is that for a spherically symmetric thin shell, the interior solution must be given by the Minkowski metric; in other words, the gravitational field must vanish inside a spherically symmetric shell. This agrees with what happens in Newtonian gravitation.

The above remarks woudl apply to an expanding spherical shell in GR as well.

 

[rbj]

What if gravity obeyed an inverse-cubed law, would the force inside the sphere still net to zero?

if it weren't for the inverse-square behavior of gravity, then you would not be able to apply Gauss's Law (with nothing inside) to determine that the net force is zero no matter where you are inside the sphere.

if gravity was inverse-cubed, the mass of the shell of the sphere that is closer to you will have greater effect relative to the mass on the other side. what this means is, if you were in exactly the dead center of the sphere, because of symmetry all 1/r³ gravitational forces will pull you equally in all directiions. but as soon as you move away from the center position, there will be a net force pulling you toward the inside surface of the shell that is closest to you.

 

[Farsight]

Is there a simple conceptual explanation involving a circle and elastic? I'v got to go, so I can't look for it on Google.

 

[RandallB]

I love this stuff.. So suppose the Earth was a hollow sphere, there would still be gravity on the outside surface, but what if you dug a hole and jumped inside? you would have some velocity from gravitational acceleration while you were above the surface, so would you continue at a constant velocity inside the sphere? also, could air be pulled inside the sphere?

Well that would be petty cool, and gravity doesn’t need to be so small as Dave said depending on how you design this world. Make the diameter 1/4 the size of Earth - we would need only 1/16 the mass of Earth to create a1 G gravity. (Likely already to dense to be real)
Now if we concentrate that mass into a 10% thick shell (An even denser configuration) with a familiar surface. Plus a few openings with no leaks below a certain altitude so the surface could hold a couple oceans of water. Now just add enough air till gravity brings it up to a pressure that we like. It would take a lot of air as it would of course leak into the inside where it would experience no significant gravity increase thus the pressure inside would rise to a uniform level matching the surface with a small gravitation bias to the center due to the mass of the air.

All the making for a breathable relatively weightless environment.
Could be fun - But what of the weather inside?
Interior rain would collect at the center for a low pressure / gravity blob of an ocean, could get very weird.
I expect possible only within the imagination of a Sci-Fi man made planet.

---------

Also related to the OP, as to “checking” the inside and outside gravity mathematically you can also do an approximate manual calculation in a flat circle. Use four points each with a mass of 1/4 that of Earth in a circular orbit to replace the earth. The orbit of the Moon would still be the same. Calculations at and near the center would be zero. But if you move out to close to the inside edge of the circle, or measure the outside force to near the circle would have some geometry aberrations biased to a nearby point. That distortion would be reduced by increasing the mass points to 16 or more but then the number of calculations start to get out of hand.

 

[matt.o]

whomever you heard that from is not to be trusted for accurate knowledge of physics.

Hi rbj. To put this in context, this is what I posted in answer to a post by Chaos;

Chaos - you need to remember that the gravitational field within a spherical shell of uniformly distributed matter is zero everywhere inside that shell.