Gravitational Force between a Uniform Sphere and a Particle
Historical Context
Newton’s inverse–square law predicts that a spherical body attracts an external particle as if all its mass were concentrated at the center.
This is not obvious: different surface elements pull in different directions. The justification requires a calculus argument, which integrates the contributions of infinitesimal mass elements on a spherical shell.
Gravitational Force of a Uniform Spherical Shell
We show that a thin, uniform spherical shell pulls on an external particle exactly as if the shell’s entire mass were concentrated at its center.
The proof uses symmetry plus a ring-by-ring integration of the gravitational contributions on the shell.
Derivation: Gravitational Force of a Uniform Spherical Shell
Setup and Ring Decomposition
Consider first a thin uniform shell of mass \(M\) and radius \(R\).
Let \(r\) be the distance from the center \(O\) to a test particle \(P\) of mass \(m\). We assume that \(r > R\). We shall divide the shell into circular rings of width \(R \,\Delta\theta\) where, as shown in the figure, the angle \(POQ\) is denoted by \(\theta\), with \(Q\) being a point on the ring.
The circumference of our representative ring element is therefore \(2\pi R \sin\theta\), and its mass \(\Delta M\) is given by
\(\displaystyle \Delta M \;\approx\; \rho \, 2\pi R^{2}\,\sin\theta\,\Delta\theta \tag{6.2.1}\)
where \(\rho\) is the mass per unit area of the shell.
Force from a Ring Element
The gravitational force exerted on \(P\) by a small subelement \(Q\) of the ring is in the direction \(PQ\).
Resolve this force \(\Delta F_{Q}\) into two components: one component along \(PO\), of magnitude \(\Delta F_{Q} \cos\phi\), the other perpendicular to \(PO\), of magnitude \(\Delta F_{Q}\sin\phi\).
Here \(\phi\) is the angle \(OPQ\).
From symmetry, the vector sum of all perpendicular components vanishes.
The force \(\Delta F\) exerted by the entire ring is, therefore, in the direction \(PO\), and its magnitude \(\Delta F\) is obtained by summing the components \(\Delta F_{Q}\cos\phi\).
The result is
\(\displaystyle \Delta F \;=\; G \frac{m \,\Delta M}{s^{2}} \cos\phi \;=\; G \frac{m\,2\pi R^{2}\rho \,\sin\theta \cos\phi}{s^{2}}\,\Delta\theta\)
where \(s\) is the distance \(PQ\) (the distance from the particle \(P\) to the ring) as shown. The magnitude of the force exerted on \(P\) by the whole shell is then obtained by taking the limit of \(\Delta\theta\) and integrating
\(\displaystyle F \;=\; Gm\,2\pi R^{2}\rho \int_{0}^{\pi} \frac{\sin\theta \,\cos\phi}{s^{2}}\, d\theta.\)
Geometric Relations
The integral is most easily evaluated by expressing the integrand in terms of \(s\). From the triangle \(OPQ\) we have, from the law of cosines,
\(\displaystyle r^{2} + R^{2} – 2rR \cos\theta \;=\; s^{2}.\)
Differentiating, because both \(R\) and \(r\) are constant, we have,
\(\displaystyle R \sin\theta \, d\theta \;=\; s \, ds.\)
Also, in the same triangle \(OPQ\), we can write
\(\displaystyle \cos\phi \;=\; \frac{s^{2} + r^{2} – R^{2}}{2rs}.\)
Substitution and Evaluation
On performing the substitutions given by the preceding two equations, and changing the limits of integration from \([0,\pi]\) to \([\,r – R,\; r + R\,]\) we obtain
\(\displaystyle F \;=\; Gm\,2\pi \rho R^{2} \int_{r-R}^{\,r+R} \frac{\,s^{2} + r^{2} – R^{2}\,}{\,2R r\, s^{2}\,}\, ds \;=\; \frac{GmM}{r^{2}},\)
where \(\displaystyle M = 4\pi \rho R^{2}\) is the mass of the shell. We can then write vectorially
\(\displaystyle \mathbf{F} \;=\; -\,G \frac{Mm}{r^{2}}\,\mathbf{e}_{r}.\)
Here \(\mathbf{e}_{r}\) is the unit radial vector from the origin \(O\).
The preceding result means that a uniform spherical shell of matter attracts an external particle as if the whole mass of the shell were concentrated at its center. This is true for every concentric spherical portion of a solid uniform sphere.
A uniform spherical body, therefore, attracts an external particle as if the entire mass of the sphere were located at the center.
The same is true also for a nonuniform sphere provided the density depends only on the radial distance \(r\).
The gravitational force on a particle located inside a uniform spherical shell is zero.
Consequences and Extensions
Inside a uniform shell: For \(r<R\), every direction has a counterpart with equal magnitude and opposite component along \(OP\), yielding
\(\displaystyle \mathbf{F}=\mathbf{0}\).
Solid uniform sphere: Decompose into concentric shells; only the mass at radii \(\le r\) contributes, giving \(\displaystyle \|\mathbf{F}\|=G\,\frac{m\,M(r)}{r^{2}}\), where \(M(r)\) is the mass enclosed by radius \(r\). For a uniform density \(\rho_{\mathrm{vol}}\), this reproduces the standard result
\(\displaystyle \|\mathbf{F}\|=G\,\frac{m\,( \frac{4}{3}\pi r^{3}\rho_{\mathrm{vol}} ) }{r^{2}} = \frac{4}{3}\pi G \rho_{\mathrm{vol}}\, m\, r\) (linear in \(r\) inside).
Spherically symmetric but nonuniform bodies: If density depends only on radius, the exterior field still equals that of a point mass \(\displaystyle M\) at the center; the interior field is determined by the enclosed mass
\(\displaystyle M(r)=4\pi\int_{0}^{r}\rho_{\mathrm{vol}}(u)\,u^{2}\,du\).
Takeaways
The result justifies treating planets and stars as point masses in many problems, simplifies orbital mechanics, and underpins potential theory. It also highlights a deep connection: only an inverse–square law produces this elegant cancellation on a sphere, making the center-of-mass equivalence exact for exterior points.
Practice
1. Show that the gravitational force on a test particle inside a thin uniform spherical shell is zero by finding the force directly.
Solution
The derivation of the force is identical to that in the article except here \(r < R\). This means that in the last integral equation, the limits on \(s\) are \(R – r\) to \(R + r\).
\(\displaystyle F \;=\; \frac{G m M}{4 R r^{2}} \int_{R-r}^{R+r} \left( 1 + \frac{r^{2} – R^{2}}{s^{2}} \right) ds \)
\(\displaystyle \phantom{F \;}=\; \frac{G m M}{4 R r^{2}} \Bigg[ R + r – (R – r) \;+\; \frac{R^{2} – r^{2}}{R + r} \;-\; \frac{R^{2} – r^{2}}{R – r} \Bigg] \)
\(\displaystyle \phantom{F \;}=\; \frac{G m M}{4 R r^{2}} \Big[ 2r + R – r – (R + r) \Big] \;=\; 0 \)
2. Find the gravitational attraction between two solid lead spheres of 1 kg mass each if the spheres are almost in contact. Express the answer as a fraction of the weight of either sphere. (The density of lead is 11.35 g/cm3.)
Solution
\(\displaystyle m \;=\; \rho V \;=\; \rho \,\frac{4}{3}\pi r_{s}^{3}\)
\(\displaystyle r_{s} \;=\; \left(\frac{3m}{4\pi \rho}\right)^{\!1/3}\)
\(\displaystyle F \;=\; \frac{G m m}{\left(2 r_{s}\right)^{2}} \;=\; \frac{G m^{2}}{4}\left(\frac{4\pi\rho}{3m}\right)^{\!2/3} \;=\; \frac{G}{4}\left(\frac{4\pi\rho}{3}\right)^{\!2/3} m^{4/3} \)
\(\displaystyle \frac{F}{W} \;=\; \frac{F}{m g} \;=\; \frac{G m^2}{4 g}\left(\frac{4\pi\rho}{3}\right)^{\!2/3}\,m^{1/3} \)
\(\displaystyle \phantom{\frac{F}{W} \;}=\; \frac{6.672\times 10^{-11}\ \mathrm{N\cdot m^{2}\cdot kg^{-2}}}{4\times 9.8\ \mathrm{m\cdot s^{-2}}} \left( \frac{4\pi\times 11.35\ \mathrm{g\cdot cm^{-3}}}{3} \times \frac{1\ \mathrm{kg}}{10^{3}\ \mathrm{g}} \times \frac{10^{6}\ \mathrm{cm^{3}}}{1\ \mathrm{m^{3}}} \right)^{\!2/3} \times \left(1\ \mathrm{kg}\right)^{1/3} \)
\(\displaystyle \phantom{\frac{F}{W} \;}=\; 2.23\times 10^{-9}\)
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