Thought experiment-shell theorem

Question

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.

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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.

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Now what's wrong with this explanation?

Answer 1

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).

Answer 2

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.