Classical-Mechanics · Unit 11 · Video 6 · Interactive Practice

Adding Up a Sphere: A Proof That Gravity Acts From the Center

IKey Formulas

FormulaNameWhat you need
dm=σR2sin⁡θ dθ dϕ=ms4πsin⁡θ dθ dϕdm = \sigma R^2 \sin\theta\, d\theta\, d\phi = \dfrac{m_s}{4\pi}\sin\theta\, d\theta\, d\phiMass element of the shellσ=ms/4πR2\sigma = m_s/4\pi R^2
rs,12=R2+z2−2Rzcos⁡θr_{s,1}^2 = R^2 + z^2 - 2Rz\cos\thetaLaw of cosines — angle at the centerRR, zz, θ\theta
cos⁡α=(z2−R2)+rs,122z rs,1\cos\alpha = \dfrac{(z^2 - R^2) + r_{s,1}^2}{2z\,r_{s,1}}Law of cosines — angle at the objectRR, zz, rs,1r_{s,1}
Fz=−Gmsm14Rz2[−z2−R2rs,1+rs,1]rmin⁡rmax⁡F_z = -\dfrac{G m_s m_1}{4Rz^2}\left[-\dfrac{z^2 - R^2}{r_{s,1}} + r_{s,1}\right]_{r_{\min}}^{r_{\max}}Master resultOnly the limits rmin⁡,rmax⁡r_{\min}, r_{\max}

Key Insight: One antiderivative decides both cases. Outside, rs,1r_{s,1} runs z−R→z+Rz-R \to z+R and the bracket gives 4R4R, leaving Fz=−Gmsm1/z2F_z = -Gm_sm_1/z^2; inside it runs R−z→R+zR-z \to R+z and the bracket gives 00.

IIWhy Only the Axial Pull Survives

Every element pulls the object sideways as well as inward — so where do the sideways pulls go?

IIIThe Limits Are the Whole Proof

As θ\theta sweeps 0→π0 \to \pi, the leg rs,1r_{s,1} sweeps a range — and that range decides everything.

IVThe Force Profile of a Shell

Crossing the surface, the pull switches off — inside a shell gravity is exactly zero.

💡 A solid sphere is a nested family of shells, so from outside every shell — and therefore the whole sphere — pulls as a single point mass at the center.

VQuiz Questions

Problem 1 · The Mass Element

Given: a uniform shell of mass msm_s and radius RR, with surface density σ=ms4πR2\sigma = \dfrac{m_s}{4\pi R^2} — find the mass dmdm of the patch between θ, θ+dθ\theta,\ \theta + d\theta and ϕ, ϕ+dϕ\phi,\ \phi + d\phi.

✅ Correct! The patch area is da=R2sin⁡θ dθ dϕda = R^2\sin\theta\, d\theta\, d\phi, and σR2=ms/4π\sigma R^2 = m_s/4\pi — the R2R^2 cancels exactly.
❌ You dropped the area factor. That is σsin⁡θ dθ dϕ\sigma\sin\theta\,d\theta\,d\phi: you still owe a factor R2R^2 from da=R2sin⁡θ dθ dϕda = R^2\sin\theta\,d\theta\,d\phi, which cancels the R2R^2 in σ\sigma.
❌ Not quite. Build it in order: edges R dθR\,d\theta and Rsin⁡θ dϕR\sin\theta\,d\phi give da=R2sin⁡θ dθ dϕda = R^2\sin\theta\,d\theta\,d\phi, then multiply by σ\sigma.
Show solution

The patch has edge lengths R dθR\,d\theta (along the colatitude) and Rsin⁡θ dϕR\sin\theta\,d\phi (around the axis — the circle of latitude has radius Rsin⁡θR\sin\theta, not RR):

da=(R dθ)(Rsin⁡θ dϕ)=R2sin⁡θ dθ dϕda = (R\,d\theta)(R\sin\theta\, d\phi) = R^2\sin\theta\, d\theta\, d\phi

Multiply by the uniform surface density:

dm=σ da=ms4πR2⋅R2sin⁡θ dθ dϕ=ms4πsin⁡θ dθ dϕdm = \sigma\, da = \frac{m_s}{4\pi R^2}\cdot R^2\sin\theta\, d\theta\, d\phi = \frac{m_s}{4\pi}\sin\theta\, d\theta\, d\phi

The shell radius cancels — a welcome simplification, since RR reappears later only through the geometry.

Check: ∫02π ⁣ ⁣∫0πms4πsin⁡θ dθ dϕ=ms4π(2π)(2)=ms\displaystyle\int_0^{2\pi}\!\!\int_0^{\pi}\frac{m_s}{4\pi}\sin\theta\, d\theta\, d\phi = \frac{m_s}{4\pi}(2\pi)(2) = m_s ✓

Problem 2 · Limits for the Inside Case

Given: the object sits inside the shell, z<Rz < R. As θ\theta runs from 00 to π\pi — find the range swept by the leg rs,1r_{s,1}.

✅ Correct! At θ=0\theta = 0 the nearest point of the shell is R−zR - z away; at θ=π\theta = \pi the far point is R+zR + z away.
❌ Those are the outside limits. With z<Rz < R the quantity z−Rz - R is negative, and a distance cannot be negative — the near leg is R−zR - z.
❌ Not quite. Put θ=0\theta = 0 and θ=π\theta = \pi into rs,12=R2+z2−2Rzcos⁡θr_{s,1}^2 = R^2 + z^2 - 2Rz\cos\theta: you get (R−z)2(R-z)^2 and (R+z)2(R+z)^2.
Show solution

Evaluate the law of cosines at the two ends of the sweep:

θ=0:rs,12=R2+z2−2Rz=(R−z)2  ⇒  rs,1=∣R−z∣=R−z\theta = 0:\quad r_{s,1}^2 = R^2 + z^2 - 2Rz = (R - z)^2 \;\Rightarrow\; r_{s,1} = |R - z| = R - z θ=π:rs,12=R2+z2+2Rz=(R+z)2  ⇒  rs,1=R+z\theta = \pi:\quad r_{s,1}^2 = R^2 + z^2 + 2Rz = (R + z)^2 \;\Rightarrow\; r_{s,1} = R + z

Because z<Rz < R, the absolute value resolves to R−zR - z (the object is nearer to the top of the shell than the center is). Outside the shell, z>Rz > R, the same algebra gives z−Rz - R instead — the only difference between the two cases in the entire proof.

Problem 3 · Run the Integral

Given: a shell of radius R=3R = 3 m with the object outside at z=6z = 6 m, and the antiderivative A(rs,1)=−z2−R2rs,1+rs,1A(r_{s,1}) = -\dfrac{z^2 - R^2}{r_{s,1}} + r_{s,1} — find the limits, then evaluate the bracket.

What are the limits on rs,1r_{s,1}?

What is A(rmax⁡)−A(rmin⁡)A(r_{\max}) - A(r_{\min})?

✅ Correct! A(9)−A(3)=6−(−6)=12=4RA(9) - A(3) = 6 - (-6) = 12 = 4R, and −Gmsm14Rz2(4R)=−Gmsm1z2-\dfrac{Gm_sm_1}{4Rz^2}(4R) = -\dfrac{Gm_sm_1}{z^2}.
❌ Check the limits. Outside means z>Rz > R, so the legs run from z−R=6−3=3z - R = 6 - 3 = 3 m to z+R=6+3=9z + R = 6 + 3 = 9 m. (1.51.5 m to 4.54.5 m would be the inside limits R∓zR \mp z.)
❌ Check the arithmetic. With z2−R2=27z^2 - R^2 = 27: A(9)=−3+9=6A(9) = -3 + 9 = 6 and A(3)=−9+3=−6A(3) = -9 + 3 = -6. Subtracting a negative adds — the difference is not 66.
Show solution

Step 1 — limits. The object is outside, so as θ\theta runs 0→π0 \to \pi:

rmin⁡=z−R=3 m,rmax⁡=z+R=9 mr_{\min} = z - R = 3\ \text{m},\qquad r_{\max} = z + R = 9\ \text{m}

Step 2 — evaluate. Here z2−R2=36−9=27 m2z^2 - R^2 = 36 - 9 = 27\ \text{m}^2:

A(9)=−279+9=−3+9=6 mA(9) = -\frac{27}{9} + 9 = -3 + 9 = 6\ \text{m} A(3)=−273+3=−9+3=−6 mA(3) = -\frac{27}{3} + 3 = -9 + 3 = -6\ \text{m} A(9)−A(3)=6−(−6)=12 m=4RA(9) - A(3) = 6 - (-6) = 12\ \text{m} = 4R

Step 3 — assemble. The 4R4R cancels the 4R4R in the prefactor:

Fz=−Gmsm14Rz2 (4R)=−Gmsm1z2F_z = -\frac{Gm_sm_1}{4Rz^2}\,(4R) = -\frac{Gm_sm_1}{z^2}

Exactly the pull of a point mass msm_s sitting at the center.

Problem 4 · Transfer — Inside a Solid Sphere

Given: a uniform solid sphere of mass MM and radius RR, with a small mass mm buried at z=R/2z = R/2 from the center — find the gravitational force on mm.

✅ Correct! Only the enclosed M/8M/8 pulls, and it acts from the center over a distance R/2R/2: F=−G(M/8)m(R/2)2=−GMm2R2F = -\dfrac{G(M/8)m}{(R/2)^2} = -\dfrac{GMm}{2R^2}.
❌ That is the shell result, not the sphere result. Only the shells outside zz give zero. The solid material within radius zz still pulls, as a point mass at the center.
❌ Not quite. Two things change together: the effective mass drops to M(z/R)3=M/8M(z/R)^3 = M/8, and the distance is z=R/2z = R/2, so z2=R2/4z^2 = R^2/4.
Show solution

Split the sphere at radius z=R/2z = R/2. Every shell with radius larger than zz encloses the mass, so by the second half of the shell theorem it contributes nothing. Every shell inside acts as a point mass at the center.

Step 1 — enclosed mass (uniform density):

Menc=M(zR)3=M(12)3=M8M_{\text{enc}} = M\left(\frac{z}{R}\right)^3 = M\left(\frac{1}{2}\right)^3 = \frac{M}{8}

Step 2 — inverse-square law from the center:

F=−GMenc mz2=−G(M/8)m(R/2)2=−GMm/8R2/4=−GMm2R2F = -\frac{G M_{\text{enc}}\, m}{z^2} = -\frac{G(M/8)m}{(R/2)^2} = -\frac{GMm/8}{R^2/4} = -\frac{GMm}{2R^2}

Common mistakes:

  • Answering 00 — that applies only to a hollow shell, or to the shells lying outside zz.
  • Using the full MM at z=R/2z = R/2, which gives −4GMm/R2-4GMm/R^2.
  • Shrinking the mass to M/8M/8 but leaving the distance at RR, which gives −GMm/8R2-GMm/8R^2.

Note the tidy consequence: F=−GMmzR3F = -\dfrac{GMm z}{R^3} inside a uniform sphere — the force grows linearly with depth-from-center, giving simple harmonic motion in a tunnel through the Earth.

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