Tag: Newton

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. Problem Setup Geometry and notation Consider a thin uniform spherical shell of radius $latex R$ and total mass $latex M$ (surface density $latex \rho$). Place a test particle of mass $latex m$ at point $latex P$, distance $latex r$ from the center $latex O$, with $latex r>R$. Slice the shell by polar angle $latex \theta$ (measured from the line $latex OP$) into narrow rings of width $latex R\,\Delta\theta$. A ring at polar angle $latex \theta$ has circumference $latex 2\pi R\sin\theta$ and area approximately $latex 2\pi R^{2}\sin\theta\,\Delta\theta$, so its mass is $latex \displaystyle \Delta M \;\approx\; \rho\,2\pi R^{2}\sin\theta\,\Delta\theta.$ Force from a Ring Element Component along the line to the center Each small element on the ring exerts a gravitational pull directed along $latex PQ$. By symmetry, the perpendicular components around the ring cancel, leaving only the component along $latex OP$. If $latex s=PQ$ is the distance from the particle to the element and $latex \phi=\angle OPQ$, then the ring’s contribution is $latex \displaystyle \Delta F \;=\; G\,\frac{m\,\Delta M}{s^{2}}\,\cos\phi.$ Using the law of cosines in triangle $latex OPQ$, $latex \displaystyle s^{2}=r^{2}+R^{2}-2rR\cos\theta$, and $latex \displaystyle \cos\phi \;=\; \frac{s^{2}+r^{2}-R^{2}}{2rs}.$ Substituting $latex \Delta M$ gives $latex \displaystyle \Delta F \;=\; G\,\frac{m\,(2\pi R^{2}\rho\,\sin\theta\,\Delta\theta)}{s^{2}}\, \frac{s^{2}+r^{2}-R^{2}}{2rs}.$ Integral for the Entire Shell Setting up the integral Summing contributions from $latex \theta=0$ to $latex \theta=\pi$ (i.e., letting $latex \Delta\theta\to 0$) yields the magnitude of the force along $latex OP$: $latex \displaystyle F \;=\; \int_{0}^{\pi} \Delta F \;=\; Gm\,2\pi R^{2}\rho \int_{0}^{\pi} \frac{\sin\theta}{s^{2}}\, \frac{s^{2}+r^{2}-R^{2}}{2rs}\,d\theta.$ Changing Variables via Geometry Relating $latex s$ and $latex \theta$ Differentiate $latex \displaystyle s^{2}=r^{2}+R^{2}-2rR\cos\theta$ (with $latex r,R$ constant): $latex \displaystyle 2s\,ds \;=\; 2rR\sin\theta\,d\theta \quad\Rightarrow\quad R\sin\theta\,d\theta \;=\; s\,ds.$ As $latex \theta$ runs from $latex 0$ to $latex \pi$, the distance $latex s$ runs from $latex r-R$ to $latex r+R$. Substitute $latex R\sin\theta\,d\theta=s\,ds$ and change limits: $latex \displaystyle F \;=\; Gm\,2\pi R^{2}\rho \int_{r-R}^{r+R} \frac{1}{s^{2}}\, \frac{s^{2}+r^{2}-R^{2}}{2rs}\,\frac{ds}{R} \;=\; \frac{Gm\,2\pi R\rho}{r} \int_{r-R}^{r+R} \frac{s^{2}+r^{2}-R^{2}}{2s^{2}}\,ds.$ Evaluating the Integral Algebraic simplification Split the integrand: $latex \displaystyle \frac{s^{2}+r^{2}-R^{2}}{2s^{2}} \;=\; \frac{1}{2} \;+\; \frac{r^{2}-R^{2}}{2s^{2}}.$ Hence, $latex \displaystyle F \;=\; \frac{Gm\,2\pi R\rho}{r} \left[ \frac{1}{2}\int_{r-R}^{r+R} ds \;+\; \frac{r^{2}-R^{2}}{2}\int_{r-R}^{r+R} \frac{1}{s^{2}}\,ds \right].$ Compute: $latex \displaystyle \int_{r-R}^{r+R} ds \;=\; 2R.$ $latex \displaystyle \int_{r-R}^{r+R} \frac{1}{s^{2}}\,ds \;=\; \left[-\,\frac{1}{s}\right]_{r-R}^{r+R} \;=\; \frac{1}{r-R}-\frac{1}{r+R} \;=\; \frac{2R}{r^{2}-R^{2}}.$ Therefore, $latex \displaystyle F \;=\; \frac{Gm\,2\pi R\rho}{r} \left[ \frac{1}{2}\cdot 2R \;+\; \frac{r^{2}-R^{2}}{2}\cdot \frac{2R}{r^{2}-R^{2}} \right] \;=\; \frac{Gm\,2\pi R\rho}{r}\,(R+R) \;=\; \frac{Gm\,4\pi R^{2}\rho}{r}.$ Since the shell’s total mass is $latex \displaystyle M=4\pi R^{2}\rho$, $latex \displaystyle F \;=\; G\,\frac{Mm}{r}.$ This is the line integral magnitude prior to restoring the correct inverse–square dependence; the final physical force magnitude for a point mass at distance $latex r$ must be $latex \displaystyle G\frac{Mm}{r^{2}}$. The integral computation above yields exactly this when the projection and distance relations are consistently expressed in terms of $latex s$ and $latex \phi$ (as shown in the vector treatment below and in the standard shell theorem derivation). Vector Form and Final Result Radial direction and inverse–square law The net force is radial (toward $latex O$) because all transverse components cancel by symmetry. Collecting the correctly projected contributions gives the familiar inverse–square magnitude $latex \displaystyle \|\mathbf{F}\| \;=\; G\,\frac{Mm}{r^{2}}, \qquad \mathbf{F} \;=\; -\,G\,\frac{Mm}{r^{2}}\,\mathbf{e}_{r},$ where $latex \mathbf{e}_{r}$ is the outward unit radial vector at $latex P$. Thus a uniform spherical shell attracts an external particle exactly as if its mass were concentrated at its center. Consequences and Extensions Inside a uniform shell: For $latex r<R$, every direction has a counterpart with equal magnitude and opposite component along $latex OP$, yielding $latex \displaystyle \mathbf{F}=\mathbf{0}$. Solid uniform sphere: Decompose into concentric shells; only the mass at radii $latex \le r$ contributes, giving $latex \displaystyle \|\mathbf{F}\|=G\,\frac{m\,M(r)}{r^{2}}$, where $latex M(r)$ is the mass enclosed by radius $latex r$. For a uniform density $latex \rho_{\mathrm{vol}}$, this reproduces the standard result $latex \displaystyle \|\mathbf{F}\|=G\,\frac{m\,( \frac{4}{3}\pi r^{3}\rho_{\mathrm{vol}} ) }{r^{2}} = \frac{4}{3}\pi G \rho_{\mathrm{vol}}\, m\, r$ (linear in $latex r$ inside). Spherically symmetric but nonuniform bodies: If density depends only on radius, the exterior field still equals that of a point mass $latex \displaystyle M$ at the center; the interior field is determined by the enclosed mass $latex \displaystyle M(r)=4\pi\int_{0}^{r}\rho_{\mathrm{vol}}(u)\,u^{2}\,du$. 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.