# Spherical Mass Distributions

We have stated without proof that the gravitational interaction between two spherically symmetric mass distributions is the same as though all the mass of each were concentrated at its center. Now we’re ready to prove this statement. Newton searched for a proof for several years, and he delayed publication of the law of gravitation until he found one.

Rather than starting with two spherically symmetric masses, we’ll tackle the simpler problem of a point mass m interacting with a thin spherical shell with total mass M. We’ll show that when m is outside the sphere, the potential energy associated with this gravitational interaction is the same as though M were concentrated in a point at the center of the sphere. We learned in Section 7.4 that the force is the negative derivative of the potential energy, so the force on m is also the same as for a point mass M. Our result will also hold for any spherically symmetric mass distribution, which we can think of as being made of many concentric spherical shells.

**Frequently Asked Questions**

## Recommended Posts:

- Nature of physics
- Solving Physics Problems
- Standards and Units
- Using and Converting Units
- Uncertainty and significant figures
- Estimates and order of magnitudes
- Vectors and vector addition
- Equilibrium and Elasticity
- Conditions for equilibrium
- Center of gravity
- finding and using the Center of gravity
- solving rigid-body equilibrium problems
- SOLVED EXAMPLES ON EQUILIBRIUM
- stress, strain, and elastic moduLi
- tensile and Compressive stress and strain

**4/5**