万有引力的几个结论证明

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 α & = \frac{D - z}{\sqrt{x^{2} + y^{2} + (D - z)^{2}}} \\ = \frac{D - r \cos ϕ}{\sqrt{r^{2} \sin^{2} ϕ \cos^{2} θ + r^{2} \sin^{2} ϕ \sin^{2} θ + D^{2} - 2 r D \cos ϕ + r^{2} \cos^{2} ϕ}} \\ = \frac{D - r \cos ϕ}{\sqrt{r^{2} + D^{2} - 2 r D \cos ϕ}} \end{aligned}

cos α = \frac{D - z}{x ^{2} + y ^{2} + ( D - z ) ^{2}} = \frac{D - r cos ϕ}{r ^{2} sin ^{2} ϕ cos ^{2} θ + r ^{2} sin ^{2} ϕ sin ^{2} θ + D ^{2} - 2 rD cos ϕ + r ^{2} cos ^{2} ϕ} = \frac{D - r cos ϕ}{r ^{2} + D ^{2} - 2 rD cos ϕ}

\begin{aligned} F & = ∭ \frac{G m ρ d V}{x^{2} + y^{2} + (D - z)^{2}} \cos α \\ = G m ρ ∭ \frac{r^{2} \sin ϕ (D - r \cos ϕ)}{(r^{2} + D^{2} - 2 r D \cos ϕ)^{\frac{3}{2}}} d V \end{aligned}

令

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

则

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

于是

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

加入上下限

\begin{aligned} \sqrt{B (0)} = & \sqrt{r^{2} + D^{2} - 2 r D} = D - r \\ \sqrt{B (π)} = & \sqrt{r^{2} + D^{2} + 2 r D} = D + r \\ \int_{0}^{π} \frac{B (ϕ) + D^{2} - r^{2}}{B^{\frac{3}{2}} (ϕ)} d B (ϕ) = & 2 {[\sqrt{B (ϕ)} - \frac{D^{2} - r^{2}}{\sqrt{B (ϕ)}}]}_{0}^{π} \\ = & 2 [(D + r) - (D - r)] - 2 [(D - r) - (D + r)] \\ = & 8 r \end{aligned}

代入

F

\begin{aligned} F = & G m ρ ∭ \frac{r^{2} \sin ϕ (D - r \cos ϕ)}{(r^{2} + D^{2} - 2 r D \cos ϕ)^{\frac{3}{2}}} d V \\ = & G m ρ \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{R} \frac{r^{2} \sin ϕ (D - r \cos ϕ)}{(r^{2} + D^{2} - 2 r D \cos ϕ)^{\frac{3}{2}}} d r d ϕ d θ \\ = & 2 π G m ρ \int_{0}^{π} \int_{0}^{R} \frac{r^{2} \sin ϕ (D - r \cos ϕ)}{(r^{2} + D^{2} - 2 r D \cos ϕ)^{\frac{3}{2}}} d r d ϕ \\ = & 2 π G m ρ \int_{0}^{R} \int_{0}^{π} \int \frac{A (ϕ)}{B^{\frac{3}{2}} (ϕ)} d ϕ d r \\ = & 2 π G m ρ \int_{0}^{R} \int_{0}^{π} \frac{r (B (ϕ) + D^{2} - r^{2})}{4 D^{2} B^{\frac{3}{2}} (ϕ)} d B (ϕ) d r \\ = & \frac{π G m ρ r}{2 D^{2}} \int_{0}^{R} r \int_{0}^{π} \frac{B (ϕ) + D^{2} - r^{2}}{B^{\frac{3}{2}} (ϕ)} d B (ϕ) d r \\ = & \frac{π G m ρ}{2 D^{2}} \int_{0}^{R} 8 r^{2} d r \\ = & \frac{4 π G m ρ}{D^{2}} \frac{R^{3}}{3} \\ = & \frac{4 π G m R^{3}}{3 D^{2}} \frac{M}{\frac{4}{3} π R^{3}} \\ = & \frac{G M m}{D^{2}} \end{aligned}

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}^{π} \frac{B (ϕ) + D^{2} - r^{2}}{B^{\frac{3}{2}} (ϕ)} d B (ϕ) = & 2 {[\sqrt{B (ϕ)} - \frac{D^{2} - r^{2}}{\sqrt{B (ϕ)}}]}_{0}^{π} \\ = & 2 [(D + r) - (D - r)] - 2 [(r - D) + (r + D)] \\ = & 0 \end{aligned}

\int_{0}^{π} \frac{B ( ϕ ) + D ^{2} - r ^{2}}{B ^{\frac{3}{2}} ( ϕ )} d B (ϕ) = = = 2 [B (ϕ) - \frac{D ^{2} - r ^{2}}{B ( ϕ )}]_{0}^{π} 2 [(D + r) - (D - r)] - 2 [(r - D) + (r + D)] 0

代入

F

\begin{aligned} F = & G m ρ \int_{0}^{2 π} \int_{0}^{π} \int_{R_{1}}^{R_{2}} \frac{r^{2} \sin ϕ (D - r \cos ϕ)}{(r^{2} + D^{2} - 2 r D \cos ϕ)^{\frac{3}{2}}} d r d ϕ d θ \\ = & \frac{π G m ρ r}{2 D^{2}} \int_{R_{1}}^{R_{2}} r \int_{0}^{π} \frac{B (ϕ) + D^{2} - r^{2}}{B^{\frac{3}{2}} (ϕ)} d B (ϕ) d r \\ = & 0 \end{aligned}

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原文地址：https://blog.csdn.net/qq_34288751/article/details/126247837