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LESSON 11 / 22 · TOPIC 2.6

Gravity inside a sphere and a shell

You will be able to: Apply the shell theorem to a uniform sphere and compare interior and exterior field laws.

Calculus-based dynamicsFree study resourceReview editionTeacher review pending

Does gravity keep increasing all the way to a planet’s center?

In an ideal planet of uniform density, gravitational field is zero at the center, grows linearly to the surface and decreases as 1/r² outside. The interior cannot be treated as all mass sitting at the center.

A useful starting point: What does an elevator scale measure? →

Words and symbols before equations

Uniform density ρ
Constant mass per volume, in kg/m³.
Enclosed mass M(r)
Mass within radius r of the center.
Shell theorem
A uniform thin spherical shell gives zero net field inside and the point-mass field outside.
Radial direction
Along a line from the sphere’s center; gravity points inward.
Uniform solid sphereProbe radius = 0.5Rg / g_surface = 0.5Enclosed mass / M = 0.125
Read this model snapshot. At r/R = 0.5, field magnitude / surface field = 0.5, enclosed mass / M = 0.125. Direction is inward wherever field is nonzero.
What this picture assumes

Uniform solid sphere or uniform ideal thin shell with equal total mass and radius. Field is normalized to GM/R². At the thin shell surface, the exterior limiting value is displayed; the ideal surface field is discontinuous.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. At r/R = 0.5, field magnitude / surface field = 0.5, enclosed mass / M = 0.125. Direction is inward wherever field is nonzero.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the physics

Inside a uniform sphere of radius R and mass M, outer shells cancel their contributions. Only enclosed mass contributes to the net field. From dm = ρ4πs²ds, M(r) = ∫₀ʳρ4πs²ds = 4πρr³/3 = M(r/R)³.

For r > 0 inside, g(r) = GM(r)/r² = GMr/R³. The limit at r = 0 is zero by symmetry. At the surface this joins GM/R² continuously; outside, use GM/r². The inward vector direction is separate from these nonnegative magnitudes.

A hollow thin shell instead gives zero field everywhere strictly inside it. At its idealized material surface the limiting field is discontinuous; do not confuse that shell with a filled solid sphere. The theorem is applied here, not derived, matching the course boundary.

A worked example, step by step

A uniform solid sphere has surface field 12 N/kg. Find the field at r = R/2 and r = 2R.

  1. At R/2 the enclosed mass is M/8.
  2. The interior law gives g = g_surface(r/R) = 12/2 = 6 N/kg inward.
  3. At 2R use the exterior inverse square: g = 12/4 = 3 N/kg inward.
  4. At the center the field is zero; extrapolating the exterior formula inward would be wrong.
Common mix-up

GM/r² with the whole mass M is not valid inside a filled sphere; use the enclosed mass.

CHECK THE IDEA

Why does a uniform sphere’s interior field scale as r rather than 1/r²?

Compare with an explanation

Enclosed mass grows as r³; dividing by r² leaves a factor r.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Move the probe from the center through the surface and outward. Predict where field magnitude peaks. Switch between solid sphere and thin shell and explain the interior difference.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Uniform solid sphereProbe radius = 0.5Rg / g_surface = 0.5Enclosed mass / M = 0.125

At r/R = 0.5, field magnitude / surface field = 0.5, enclosed mass / M = 0.125. Direction is inward wherever field is nonzero.

Field magnitude inside and outsideg / (GM/R²)r/R000.750.31.50.62.250.931.2

Uniform solid sphere or uniform ideal thin shell with equal total mass and radius. Field is normalized to GM/R². At the thin shell surface, the exterior limiting value is displayed; the ideal surface field is discontinuous.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus relationship to justify your prediction.

Apply the idea to a fresh problem Practice →

Show what you understand.

Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.

1. Inside a uniform thin spherical shell, net field is…

Show answer and reasoning

Zero. Contributions from the complete shell cancel.

2. At r = R/3 inside a uniform solid sphere, enclosed mass is…

Show answer and reasoning

M/27. Volume and enclosed mass scale as r³.

Original written challenge

4 points · self-check · not an official AP question

A uniform sphere has mass M and radius R. Derive M(r), derive its interior g(r), compare g(R/4) to surface field and explain the center limit.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: M(r) = ρ(4πr³/3) = M(r/R)³.
  2. 1 point: g(r) = GM(r)/r² = GMr/R³.
  3. 1 point: g(R/4) = g_surface/4.
  4. 1 point: The field tends to zero at r = 0; symmetry leaves no preferred inward direction at the center.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1Which mass enters the interior law?

Only mass enclosed within the probe radius.

RECALL 2Where is field largest for a uniform solid sphere?

At its surface.

RECALL 3Must this unit prove the shell theorem?

No. It applies the theorem to the stated symmetric distributions.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Gravity inside a sphere and a shell

  • Uniform sphere: M(r) = M(r/R)³ for r < R.
  • Inside: g = GMr/R³. Outside: g = GM/r².
  • Uniform thin shell: g = 0 strictly inside.

Remember: GM/r² with the whole mass M is not valid inside a filled sphere; use the enclosed mass.

Conditions: Uniform solid sphere or uniform ideal thin shell with equal total mass and radius. Field is normalized to GM/R². At the thin shell surface, the exterior limiting value is displayed; the ideal surface field is discontinuous.

Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.6.E · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 2.6, objectives 2.6.E. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.

Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.

Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.

Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.

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