均匀球体对球体外物体的万有引力等效于位于球心处的质点

$\begin{aligned} \cos\alpha &= \frac{D-z}{\sqrt{x^2+y^2+(D-z)^2}} \\ &= \frac{D-r\cos\phi}{\sqrt{r^2\sin^2\phi\cos^2\theta +r^2\sin^2\phi\sin^2\theta+D^2-2rD\cos\phi+r^2\cos^2\phi}} \\ &= \frac{D-r\cos\phi}{\sqrt{r^2+D^2-2rD\cos\phi}} \end{aligned}$ $\begin{aligned} F &= \iiint\frac{Gm\rho\text{d}V}{x^2+y^2+(D-z)^2}\cos\alpha \\ &= Gm\rho\iiint\frac{r^2\sin\phi(D-r\cos\phi)} {(r^2+D^2-2rD\cos\phi)^{\frac{3}{2}}}\text{d}V \end{aligned}$

令

$\begin{aligned} A(\phi) &= r^2\sin\phi(D-r\cos\phi) \\ B(\phi) &= r^2+D^2-2rD\cos\phi \end{aligned}$

则

$\begin{aligned} \frac{\text{d}B(\phi)}{\text{d}\phi} &= 2rD\sin\phi \\ A(\phi) &= r^2\sin\phi(D-r\cos\phi) \\ &= \frac{r(D-r\cos\phi)}{2D}\frac{\text{d}B(\phi)}{\text{d}\phi} \\ &= \frac{r(B(\phi)+D^2-r^2)}{4D^2}\frac{\text{d}B(\phi)}{\text{d}\phi} \\ \end{aligned}$

于是

$\begin{aligned} & \int\frac{B(\phi)+D^2-r^2}{B^{\frac{3}{2}}(\phi)}\text{d}B(\phi) \\ =& -2\int(B(\phi)+D^2-r^2)\text{d}B^{-\frac{1}{2}}(\phi) \\ =& -2\left[\sqrt{B(\phi)}+\frac{D^2-r^2}{\sqrt{B(\phi)}}\right] +2\int B^{-\frac{1}{2}}(\phi)\text{d}B(\phi) \\ =& -2\left[\sqrt{B(\phi)}+\frac{D^2-r^2}{\sqrt{B(\phi)}}\right]+4\sqrt{B(\phi)} \\ =& 2\sqrt{B(\phi)}-2\frac{D^2-r^2}{\sqrt{B(\phi)}} \\ \end{aligned}$

加入上下限

$\begin{aligned} \sqrt{B(0)} =& \sqrt{r^2+D^2-2rD}=D-r \\ \sqrt{B(\pi)} =& \sqrt{r^2+D^2+2rD}=D+r \\ \int_0^{\pi}\frac{B(\phi)+D^2-r^2}{B^{\frac{3}{2}}(\phi)}\text{d}B(\phi) =& 2\left[\sqrt{B(\phi)}-\frac{D^2-r^2}{\sqrt{B(\phi)}}\right]_0^{\pi} \\ =& 2[(D+r)-(D-r)]-2[(D-r)-(D+r)] \\ =& 8r \end{aligned}$

代入$F$

$\begin{aligned} F =& Gm\rho\iiint\frac{r^2\sin\phi(D-r\cos\phi)} {(r^2+D^2-2rD\cos\phi)^{\frac{3}{2}}}\text{d}V \\ =& Gm\rho\int_0^{2\pi}\int_0^{\pi}\int_0^{R}\frac{r^2\sin\phi(D-r\cos\phi)} {(r^2+D^2-2rD\cos\phi)^{\frac{3}{2}}}\text{d}r\text{d}\phi\text{d}\theta \\ =& 2\pi Gm\rho\int_0^{\pi}\int_0^{R}\frac{r^2\sin\phi(D-r\cos\phi)} {(r^2+D^2-2rD\cos\phi)^{\frac{3}{2}}}\text{d}r\text{d}\phi \\ =& 2\pi Gm\rho\int_0^{R}\int_0^{\pi}\int\frac{A(\phi)} {B^{\frac{3}{2}}(\phi)}\text{d}\phi\text{d}r \\ =& 2\pi Gm\rho\int_0^{R}\int_0^{\pi}\frac{r(B(\phi)+D^2-r^2)} {4D^2B^{\frac{3}{2}}(\phi)}\text{d}B(\phi)\text{d}r \\ =& \frac{\pi Gm\rho r}{2D^2}\int_0^{R}r\int_0^{\pi}\frac{B(\phi)+D^2-r^2} {B^{\frac{3}{2}}(\phi)}\text{d}B(\phi)\text{d}r \\ =& \frac{\pi Gm\rho}{2D^2}\int_0^{R}8r^2\text{d}r \\ =& \frac{4\pi Gm\rho}{D^2}\frac{R^3}{3} \\ =& \frac{4\pi GmR^3}{3D^2}\frac{M}{\frac{4}{3}\pi R^3} \\ =& \frac{GMm}{D^2} \end{aligned}$

均匀球壳对球壳内任意一点的万有引力合力为0

积分

$\int_0^{\pi}\frac{B(\phi)+D^2-r^2}{B^{\frac{3}{2}}(\phi)}\text{d}B(\phi)$

中，由于$D\le r$，所以$\sqrt{B(0)}=\sqrt{r^2+D^2-2rD}=r-D$，此时

$\begin{aligned} \int_0^{\pi}\frac{B(\phi)+D^2-r^2}{B^{\frac{3}{2}}(\phi)}\text{d}B(\phi) =& 2\left[\sqrt{B(\phi)}-\frac{D^2-r^2}{\sqrt{B(\phi)}}\right]_0^{\pi} \\ =& 2[(D+r)-(D-r)]-2[(r-D)+(r+D)] \\ =& 0 \end{aligned}$

代入$F$

$\begin{aligned} F =& Gm\rho\int_0^{2\pi}\int_0^{\pi}\int_{R_1}^{R_2}\frac{r^2\sin\phi(D-r\cos\phi)} {(r^2+D^2-2rD\cos\phi)^{\frac{3}{2}}}\text{d}r\text{d}\phi\text{d}\theta \\ =& \frac{\pi Gm\rho r}{2D^2}\int_{R_1}^{R_2}r\int_0^{\pi}\frac{B(\phi)+D^2-r^2} {B^{\frac{3}{2}}(\phi)}\text{d}B(\phi)\text{d}r \\ =& 0 \end{aligned}$