AP Physics C Mechanics- 2.6 Gravitational Force- Study Notes- New Syllabus

Gravitational Force Due to a Uniform Spherical Distribution of Mass

The gravitational force exerted by a uniform spherical distribution of mass (like a planet or star) on an object can be understood by summing up the effects of all the small differential mass elements that make up the sphere. This principle is described by Newton’s Shell Theorem, which provides powerful simplifications for analyzing spherical bodies.

Gravitational Force as the Sum of Differential Contributions

The total gravitational force exerted on an object by a spherical distribution of mass is the vector sum of the gravitational forces exerted by all infinitesimal mass elements \( \mathrm{dm} \) within the sphere.

\( \mathrm{\vec{F}_{net} = \sum \vec{F}_{dm} = G \int \dfrac{m\,dm}{r^2}\hat{r}} \)

By symmetry, only the radial components of these differential forces add up; tangential components cancel out.

Key Idea: Spherical symmetry allows us to treat the problem as if all mass elements act as one combined gravitational source at the sphere’s center (under certain conditions).

Newton’s Shell Theorem

Newton’s Shell Theorem describes the gravitational effects of spherical shells of mass:

• Inside a Shell: The net gravitational force on an object located anywhere inside a thin spherical shell of uniform mass is zero.
• Outside a Shell: The gravitational force on an object located outside a thin shell is the same as if the shell’s total mass were concentrated at its center.

\( \mathrm{F = G\dfrac{M_{shell} m}{r^2}} \)

Implication: For a planet or star (approximated as a uniform sphere), the gravitational force outside its surface behaves as though all its mass were concentrated at the center.

Force on an Object Inside a Uniform Solid Sphere

If an object is inside a uniform sphere (e.g., underground within Earth), not all of the sphere’s mass contributes to the net gravitational force. Only the mass of the sphere within a radius equal to the object’s distance from the center affects it; the outer layers exert zero net force (from the shell theorem).

 Partial Mass That Contributes to Force Inside a Sphere

The contributing (partial) mass within radius \( \mathrm{r_{partial}} \) can be expressed in terms of the density \( \mathrm{\rho} \):

\( \mathrm{m_{partial} = \dfrac{4}{3}\pi r_{partial}^3 \rho} \)

As the object moves deeper within the sphere, less mass contributes to the net gravitational pull — because the shells outside its radius cancel out their gravitational effects.

Gravitational Force Inside a Uniform Sphere

Using the partial mass expression in Newton’s law of gravitation:

\( \mathrm{F = G\dfrac{m\,m_{partial}}{r_{partial}^2}} \)

Substituting \( \mathrm{m_{partial} = \dfrac{4}{3}\pi r_{partial}^3 \rho} \):

\( \mathrm{F = G\dfrac{m(4/3)\pi r_{partial}^3 \rho}{r_{partial}^2} = \dfrac{4}{3}\pi G\rho m r_{partial}} \)

This shows that the gravitational force inside a uniform sphere increases linearly with the distance from the center:

\( \mathrm{F = -k r_{partial}} \)

where \( \mathrm{k = \dfrac{4}{3}\pi G\rho m} \) is a proportionality constant, and the negative sign indicates the force is directed toward the center.

Key Result: Inside a uniform sphere, \( \mathrm{F \propto r} \), whereas outside the sphere, \( \mathrm{F \propto 1/r^2.} \)