# Thought experiment-shell theorem

Consider a uniformly dense perfectly spherical planet. At any point inside the planet, the gravitational attraction is only due to the mass below the point. (red circle) according to shell theorem.

Now onto the thought experiment, as per Newtons law of gravitation, every particle attracts every other particle with $\frac{G m_1 m_2}{r^2}$

By symmetry, the particles in the blue region should nullify their(net) attraction at the point. Similarly, all particles in the green region would have a component of the force along the -ve X-axis. Even the particles in the green region but not inside the red circle would have a small component of attraction along -ve X-axis at the point.(contradicting shell theorem) Again, their Y-axis components would nullify each other by symmetry. The same would hold true by spinning this cross-section along the X-axis to a 3d planet.

Now what's wrong with this explanation?

• Shell theorem is valid for a complete shell. So no you can't apply shell theorem solely for the green mass outside the red circle, so it doesn't contradict shell theorem – udiboy1209 Aug 20 '13 at 17:29
• I'm saying that shell theorem proves that only the mass inside the red circle would attract, but this thought experiment concludes that a mass outside of the red circle would also attract. – user80551 Aug 20 '13 at 17:52
• Your blue region mixes bits of mass inside and outside the red circle, which makes seeing the shell theorem from this breakdown very difficult. You haven't contradicted it, but you've made it very difficult to see intuitively. – Kyle Oman Aug 20 '13 at 17:55

The shell theorem states that a test particle in a spherically symmetric system experiences no net force from an entire shell of mass at a radius greater than the radius of the test particle. So for some shell larger than your red circle, the contributions from that entire shell needs to cancel. However, you've split such a shell into pieces; some of the shell is in the green region and some is in the blue region, and then you've invoked some contribution from mass interior to the red circle to cancel out a contribution from outside. This is where you went wrong.

You need to cancel each piece of mass at a constant radius with other pieces of mass at the same radius. This is not obvious from a geometrical analysis alone, you need some calculus to verify it. A small amount of mass on the nearby side of an exterior shell will be compensated by a larger amount of mass on the distant side.

Also note that your analysis gives a qualitatively correct result; the net force on the test point must be directed radially inward, because it only feels a contribution from mass interior to its position (and the y-component in your diagram is trivially zero).

• Oh, so does shell theorem actually state that the entire mass outside the red circle would result in no net acceleration implying that only the mass inside the red circle would result in acceleration? – user80551 Aug 20 '13 at 18:31
• Yes. Of course there is acceleration due to all mass, but the part outside causes no net acceleration. I think you get the idea now. – Kyle Oman Aug 20 '13 at 20:51
• Yeah, I was confused as my textbook stated that only the part inside causes acceleration instead of the part outside does not result in net acceleration – user80551 Aug 21 '13 at 4:32

As I interpret your diagram, this is a circular cross-section through the sphere. In the diagram, the blue region is clearly symmetric, as you state. So if we kept the circle itself as a 2D object by itself, or extruded it into a cylinder, then indeed the blue region would cancel, and your logic would validly show there is a net gravitational force (for points not at the very centre).

But the issue is rotating the circle to form a sphere, because the points are not counted equally. Label the circle's radius $r$, and label a little piece of area from the circle as $dA$. Then for a bit of blue region near the outside, its volume becomes $\approx 2\pi r\,dA$. But for a same-sized bit of blue region nearer the middle, its volume might become $\approx 2\pi(\frac{2}{3}r)\,dA$. So they have different volumes, so different masses, and the symmetry in your 2D diagram is lost.