1. Introduction

In many gravitational problems, objects are treated as point masses.

However, real objects such as:

• planets
• stars
• moons
• spheres

have mass distributed throughout a volume.

An extended body is an object whose size cannot be neglected.

To calculate its gravitational attraction, we must consider how the mass is distributed.

Fortunately, several important simplifications exist for spherically symmetric bodies.

2. Point Mass Review

Newton’s Law of Gravitation for two point masses is:

FG=Gm1m2r2F_G = \frac{Gm_1m_2}{r^2}FG​=r2Gm1​m2​​

where:

• $G$ = universal gravitational constant
• m1m_1m1​ and m2m_2m2​ are masses
• $r$ is the distance between their centers

For extended objects, this equation is not always directly applicable.

3. Extended Bodies

Definition

An extended body is an object whose mass is distributed over a region of space rather than concentrated at a single point.

Examples include:

• Earth
• the Moon
• the Sun
• solid spheres
• planets

To determine the gravitational force, contributions from every small piece of mass must be added together.

4. Mass Elements

An extended body can be divided into many tiny mass elements:

$$dm$$

Each mass element exerts a small gravitational force:

dF=Gm dmr2dF = \frac{Gm\,dm}{r^2}dF=r2Gmdm​

The total force is obtained by integration:

$$F=\int dF$$

This calculus approach is the foundation of gravitational field calculations.

5. Spherically Symmetric Bodies

Many astronomical objects are approximately spherical.

Examples:

• Earth
• the Sun
• planets
• stars

For these objects, a remarkable theorem greatly simplifies calculations.

6. The Shell Theorem

Statement

For a spherically symmetric body:

• the gravitational force outside the body is identical to the force produced if all of its mass were concentrated at its center.

This result is known as the Shell Theorem.

It was first demonstrated by Isaac Newton.

7. Gravitational Force Outside a Sphere

For a spherical object of mass $M$:

FG=GMmr2F_G = \frac{GMm}{r^2}FG​=r2GMm​

where:

• $r$ is the distance from the center of the sphere

This equation is exactly the same as the point-mass equation.

Important Result

Outside a spherical body:

$$\text{Sphere}⟹\text{Point Mass at Center}$$

This simplification is used extensively in AP Physics C.

8. Gravitational Field Outside a Sphere

Because the sphere behaves like a point mass:

g=GMr2g = \frac{GM}{r^2}g=r2GM​

where:

• $M$ = mass of the sphere
• $r$ = distance from the center

This equation applies to:

• planets
• stars
• moons
• satellites orbiting Earth

9. Gravitational Force at Earth’s Surface

At Earth’s surface:

r=REr=R_Er=RE​

Thus:

g=GMERE2g = \frac{GM_E}{R_E^2}g=RE2​GME​​

Substituting Earth’s values gives:

g≈9.8 m/s2g \approx 9.8\,m/s^2g≈9.8m/s2

10. Gravity Above Earth’s Surface

At altitude $h$:

r=RE+hr=R_E+hr=RE​+h

Therefore:

g=GME(RE+h)2g = \frac{GM_E}{(R_E+h)^2}g=(RE​+h)2GME​​

As altitude increases:

$$g\downarrow$$

Gravity becomes weaker.

11. Gravitational Force Inside a Hollow Sphere

The Shell Theorem provides another remarkable result.

For a hollow spherical shell:

Inside the Shell

FG=0F_G=0FG​=0

Every gravitational contribution is exactly canceled by contributions from other parts of the shell.

Consequence

An object placed anywhere inside a hollow spherical shell experiences no net gravitational force.

12. Gravitational Field Inside a Hollow Shell

Inside a hollow shell:

$$g=0$$

This result is independent of:

• position inside the shell
• shell size
• shell mass

As long as the shell is spherically symmetric.

13. Inside a Uniform Solid Sphere

A uniform solid sphere behaves differently.

Only the mass enclosed within radius $r$ contributes to the gravitational force.

The mass outside radius $r$ produces no net force.

14. Enclosed Mass

For a sphere of radius $R$ and total mass $M$:

Mr=Mr3R3M_r = M\frac{r^3}{R^3}Mr​=MR3r3​

where:

• MrM_rMr​ = mass enclosed within radius $r$

15. Gravitational Field Inside a Uniform Sphere

Substituting the enclosed mass into Newton’s Law gives:

g=GMrr2g = \frac{GM_r}{r^2}g=r2GMr​​

Using:

Mr=Mr3R3M_r = M\frac{r^3}{R^3}Mr​=MR3r3​

yields:

g=GMrR3g = \frac{GMr}{R^3}g=R3GMr​

Important Result

Inside a uniform sphere:

$$g\propto r$$

The gravitational field increases linearly with distance from the center.

16. Gravitational Field at the Center

At the center:

$$r=0$$

Therefore:

$$g=0$$

The gravitational field vanishes at the exact center of a uniform sphere.

17. Graph of Gravitational Field

Inside the Sphere

$$g\propto r$$

A straight-line increase.

Outside the Sphere

g∝1r2g\propto \frac1{r^2}g∝r21​

An inverse-square decrease.

18. Applications

Planets

Treat Earth as a point mass when studying satellites.

Stars

Model stellar gravity using the Shell Theorem.

Planetary Orbits

Apply:

FG=GMmr2F_G = \frac{GMm}{r^2}FG​=r2GMm​

using the planet’s center as the source.

Astrophysics

The Shell Theorem is fundamental in studying stars and galaxies.

19. Example Problem

Earth’s mass:

ME=5.97×1024 kgM_E = 5.97\times10^{24}\,kgME​=5.97×1024kg

A satellite is located at:

r=2REr = 2R_Er=2RE​

Find the gravitational field strength.

Solution

Using:

g=GME(2RE)2g = \frac{GM_E}{(2R_E)^2}g=(2RE​)2GME​​ g=14(GMERE2)g = \frac14 \left( \frac{GM_E}{R_E^2} \right)g=41​(RE2​GME​​) g=14(9.8)g = \frac14(9.8)g=41​(9.8) g=2.45 m/s2g = 2.45\,m/s^2g=2.45m/s2

The field strength is one-fourth of its surface value.

20. Common AP Physics C Mistakes

Mistake 1

Using the surface radius when the object is above Earth.

Always measure distance from the center.

Mistake 2

Assuming gravity is zero outside Earth.

Gravity extends infinitely far from Earth.

Mistake 3

Assuming all of a sphere’s mass contributes inside the sphere.

Only the enclosed mass contributes.

Summary

Outside a spherical body:

FG=GMmr2F_G = \frac{GMm}{r^2}FG​=r2GMm​ g=GMr2g = \frac{GM}{r^2}g=r2GM​

Inside a hollow shell:

FG=0F_G=0FG​=0 $$g=0$$

Inside a uniform solid sphere:

g=GMrR3g = \frac{GMr}{R^3}g=R3GMr​

Key ideas:

• spherical bodies can be treated as point masses when viewed from outside
• the Shell Theorem greatly simplifies gravitational calculations
• the gravitational field inside a hollow shell is zero
• inside a uniform sphere, gravity increases linearly with distance from the center

The study of gravitational attraction due to extended bodies provides the foundation for understanding planetary structure, satellite motion, stellar physics, and many advanced topics in AP Physics C Mechanics.