Irodov 1.214: Shell Theorem Inside and Outside – Amit & Deepali Sinha Foundation

Problem Irodov 1.214

A sphere of uniform density has mass $M$ and radius $R$ . Find how the gravitational field strength $\vect G$ and the gravitational potential $\varphi$ depend on the distance $r$ from the sphere’s center, giving separate expressions for $r\le R$ (inside the sphere) and $r\ge R$ (outside it), and sketch the approximate shapes of the two functions $G(r)$ and $\varphi(r)$ .

The uniform sphere of mass M and radius R, with a field point a distance r from its center. — The uniform sphere of mass $M$ and radius $R$ , with a field point a distance $r$ from its center.

Given

Find

G(r)

and

\varphi(r)

for

r\le R

and

r\ge R

; their approximate plots

Solution

Step 1Gauss’s theorem reduces the field to the enclosed mass

The concentric Gaussian sphere of radius r (dashed) over which the flux is taken, drawn inside the body of radius R. At the external field point the field G is radial, directed toward the center. — The concentric Gaussian sphere of radius $r$ (dashed) over which the flux is taken, drawn inside the body of radius $R$ . At the external field point the field $\vect G$ is radial, directed toward the center.

By symmetry $\vect G$ at any point is radial and depends only on $r$ . Applying the flux form of the law of gravitation over a concentric sphere of radius $r$ , and using that $G_n$ is uniform over it, relates the field there to the mass $m(r)$ enclosed within that sphere.

\begin{align*}\oint_S G_n\,dS &= -4\pi\gamma\, m(r) \\ G(r)\cdot 4\pi r^2 &= -4\pi\gamma\, m(r)\end{align*}

G(r) = -\frac{\gamma\, m(r)}{r^2} \tag{1}

Step 2Field and potential outside the sphere (

r\ge R

)

For $r\ge R$ the whole mass lies inside the Gaussian sphere, so $m(r)=M$ in (1). Integrating $G=-d\varphi/dr$ from infinity, where $\varphi\to0$ , down to $r$ then gives the potential there.

G(r) = -\frac{\gamma M}{r^2}, \qquad r\ge R.

\begin{align*}\varphi(r) &= -\int_\infty^r G(r')\,dr' \\ &= -\int_\infty^r \left(-\frac{\gamma M}{r'^2}\right)dr' \\ &= \gamma M\left[-\frac{1}{r'}\right]_\infty^r\end{align*}

\varphi(r) = -\frac{\gamma M}{r}, \qquad r\ge R. \tag{2}

Step 3Field and potential inside the sphere (

r\le R

)

The enclosed mass is the total mass times the volume ratio.

m(r) = M\,\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}

m(r) = M\,\frac{r^3}{R^3}

(3)

Substituting into (1) gives the interior field; since $G$ is finite at $r=R$ , $\varphi$ is continuous there, so (2) evaluated at $r=R$ fixes the constant of integration for the inward integral.

\begin{align*}G(r) &= -\frac{\gamma}{r^2}\cdot M\frac{r^3}{R^3} = -\frac{\gamma M r}{R^3}, \qquad r\le R.\end{align*}

\begin{align*}\varphi(r) &= \varphi(R) - \int_R^r G(r')\,dr' \\ &= -\frac{\gamma M}{R} - \int_R^r \left(-\frac{\gamma M r'}{R^3}\right)dr' \\ &= -\frac{\gamma M}{R} + \frac{\gamma M}{R^3}\left[\frac{r'^2}{2}\right]_R^r \\ &= -\frac{\gamma M}{R} + \frac{\gamma M(r^2-R^2)}{2R^3} \\ &= -\frac{3\gamma M}{2R} + \frac{\gamma M r^2}{2R^3}, \qquad r\le R.\end{align*}

Step 4Collecting the results

The field's signed radial component falls from zero at the center to a minimum of -γ M/R^2 at the surface, then rises back toward the asymptote G=0 as 1/r^2 outside. — The field’s signed radial component falls from zero at the center to a minimum of $-\gamma M/R^2$ at the surface, then rises back toward the asymptote $G=0$ as $1/r^2$ outside.

The potential is a parabola opening upward from its minimum -3γ M/2R at the center, joins smoothly to -γ M/r outside, and approaches the asymptote φ=0 as r->∞. — The potential is a parabola opening upward from its minimum $-3\gamma M/2R$ at the center, joins smoothly to $-\gamma M/r$ outside, and approaches the asymptote $\varphi=0$ as $r\to\infty$ .

The two regions above combine into one continuous piecewise function for each quantity, plotted above. Each region’s scalar $G(r)$ is the radial component of a vector pointing along $\vect r$ , the position vector from the center, so the vector form each piece asks for is

\vect G(r) = G(r)\,\frac{\vect r}{r}.

\vect G(r) = \begin{cases} -\dfrac{\gamma M}{R^3}\,\vect r, & r\le R \\[6pt] -\dfrac{\gamma M}{r^3}\,\vect r, & r\ge R \end{cases}

(4)

\varphi(r) = \begin{cases} -\dfrac{3\gamma M}{2R}+\dfrac{\gamma M r^2}{2R^3}, & r\le R \\[6pt] -\dfrac{\gamma M}{r}, & r\ge R \end{cases}

(5)

Check

Both pieces agree at $r=R$ : the interior field gives $-\gamma M/R^2$ and the exterior gives the same; the interior potential gives $-3\gamma M/2R+\gamma M/2R=-\gamma M/R$ , matching (2) at $r=R$ . Dimensionally $[\gamma][M]/[R^2]$ is $\mathrm{m/s^2}$ , a field, and $[\gamma][M]/[R]$ is $\mathrm{J/kg}$ , a potential, as required, and $\varphi\to0$ as $r\to\infty$ , matching the outside plot’s asymptote.

Remark

Inside the sphere $\vect G=-(\gamma M/R^3)\vect r$ is an isotropic linear restoring field, exactly a three dimensional Hooke’s law with $k/m_{\text{test}}=\gamma M/R^3$ . Its component along any fixed straight line is proportional to the coordinate measured along that line from the line’s closest approach to the center, since the projection of $\vect r$ onto a fixed direction is linear in that coordinate. So a frictionless straight tunnel bored through a uniform planet, along any chord and not only a diameter, would let a dropped object execute simple harmonic motion with the same angular frequency $\sqrt{\gamma M/R^3}$ , independent of which chord is drilled.