📂Classical Mechanics

Uniform Sphere Shell and the Gravity of a Separated Particle

Uniform Spherical Shell and a Displaced Particle’s Gravity¹

Let’s assume there is a uniform spherical shell with a total mass of $M$ and a radius of $R$. And there exists a particle with a mass of $m$, displaced at a distance of $r$ from the center $O$ of the spherical shell. In this case, $R<r$ holds. Let’s first calculate the force exerted by a portion of the spherical shell on the particle.

Consider a band of the shell as shown in the figure above. The radius of the shell band is $R\sin \theta$. Therefore, the circumference of the shell band is $2\pi R \sin \theta$. The width can be approximated as $R\Delta \theta$. Hence, the mass of the shell band is as follows.

$$\Delta M = \rho 2\pi R^{2} \sin \theta \Delta \theta$$

Here, $\rho=\frac{M}{4\pi R^{2}}$ denotes the mass per unit area of the spherical shell. Now, let’s denote the gravity exerted by a point Q on the shell band on the particle $P$ as $\Delta \mathbf{F}_{Q}$. Splitting this into the horizontal component $\Delta \mathbf{F}_{Q}\cos \phi$ and the vertical component $\Delta \mathbf{F}_{Q}\sin \phi$, it can be seen that the vertical component is canceled out by the vertical component of the force exerted by a mass point on the opposite side, due to $Q$. So, it can be realized that the direction of the total force $\Delta \mathbf{F}$ exerted by the spherical shell band on the particle is in the direction from the particle $P$ towards the center of the shell $O$. And the magnitude is given by the law of universal gravitation as follows.

$$\Delta F=G\frac{m\Delta M}{s^{2}}\cos \phi =G\frac{ m2\pi \rho R^{2} \sin \theta \cos \phi}{s^{2}}\Delta \theta$$

Now, by integrating over $\theta$, we can find the magnitude of the total force that the spherical shell exerts on the particle.

$$\begin{equation} F = Gm2\pi \rho R^{2}\int_{0}^{\pi}\frac{\sin \theta \cos \phi}{s^{2}}d\theta \end{equation}$$

When the integration variable $\theta$ is converted to $s$, integration becomes simpler. First, by applying the cosine rule to triangle $OPQ$, we have the following.

$$r^{2} +R^{2} -2rR\cos\theta=s^{2}$$

Since $r$ and $R$ are constants, differentiating both sides gives us the following equation.

$$\begin{align} &&2rR\sin \theta d \theta &= 2s ds \nonumber \\ \implies && \sin \theta d \theta &= \frac{s }{rR}ds \end{align}$$

Similarly, by applying the cosine rule for angle $\phi$, we get the following equation.

$$\begin{equation} \cos \phi = \frac{s^{2}+r^{2}-R^{2}}{2rs} \end{equation}$$

Now, substituting $(2)$ and $(3)$ into $(1)$ and calculating the integration yields the following.

$$\begin{align*} F &= Gm2\pi\rho R^{2}\int_{r-R}^{r+R} \frac{s^{2}+r^{2}-R^{2}}{2Rr^{2}s^{2}}ds \\ &= \frac{ Gm2\pi R^{2}M}{4\pi R^{2}}\frac{1}{2Rr^{2}}\int _{r-R}^{r+R}\frac{s^{2}+r^{2}-R^{2}}{s^{2}}ds \\ &= \frac{ GmM}{4R r^{2}}\int _{r-R}^{r+R}\frac{s^{2}+r^{2}-R^{2}}{s^{2}}ds \\ &= \frac{ GmM}{4R r^{2}}\int _{r-R}^{r+R}\left( 1+\frac{r^{2}-R^{2}}{s^{2}}\right)ds \\ &= \frac{ GmM}{4R r^{2}}\left[ s-\frac{(r-R)(r+R)}{s} \right]_{r-R}^{r+R} \\ &= \frac{ GmM}{4R r^{2}}\Big[ (r+R)-(r-R)-(r-R)+(r+R)\Big] \\ &= \frac{GmM}{r^{2}} \end{align*}$$

Adding the direction to express it as a vector gives the following.

$$\mathbf{F} = -G\frac{mM}{r^{2}}\mathbf{e}_{r}$$

Here, $\mathbf{e}_{r}$ is the unit vector in the radial direction when $O$ is the origin.