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$
地球表面的科里奥利力
令延地轴向北的单位矢量为 $\uvec z$, 质点所在经线与赤道交点的单位矢量为 $\uvec x$, 则 $\uvec y = \uvec z \cross \uvec x$. 若质点运动的方向与所在经线的夹角为 $\phi$ (顺时针为正),运动平面的法向量与赤道平面的夹角为 $\theta$ (若把地球近似看做球形,则 $\theta$ 是质点所在纬度1).这样,运动平面内正北方向的单位矢量为 $\uvec z' = -\sin\theta \,\uvec x + \cos\theta\,\uvec z$, 正东方向的单位矢量为 $\uvec y$, 正上方为 $\uvec x' = \uvec y \cross \uvec z' = \sin\theta \,\uvec z + \cos\theta\,\uvec x$, 速度方向的单位矢量为
\begin{equation}
\uvec v = \sin\phi\,\uvec y + \cos\phi\,\uvec z' = -\cos\phi\sin\theta \,\uvec x + \sin\phi\,\uvec y+\cos\phi\cos\theta\,\uvec z
\end{equation}
现在可以计算科里奥利力
\begin{equation}
\bvec F_{col} = 2mv\omega\,\uvec v\cross\uvec z = 2mv\omega (\cos\phi\sin\theta\,\uvec y + \sin\phi\,\uvec x)
\end{equation}
其向北,向东,向上的分量分别为
\begin{equation}
\bvec F_{col} \vdot \uvec z' = -2mv\omega\sin\theta\sin\phi
\qquad
\bvec F_{col} \vdot \uvec y = 2mv\omega\sin\theta\cos\phi
\end{equation}
\begin{equation}
\bvec F_{col}\vdot\uvec x' = 2mv\omega\sin\phi\cos\theta
\end{equation}
可以证明水平分力可以表示为
\begin{equation}
\bvec F_{col}^{\|} = (\bvec F_{col} \vdot \uvec y)\uvec y + (\bvec F_{col} \vdot \uvec z')\uvec z' = 2m\bvec v\cross\bvec\omega'
\end{equation}
其中 $\bvec\omega' = \omega\sin\theta\,\uvec x'$. 可见科氏力的水平分量始终与速度垂直,且在地球的两极($\theta = \pi/2$)处取最大值 $2mv\omega$, 在赤道处为 0.
要特别注意的是,地球表面的非惯性力除了科里奥利力外还有离心力,但离心力一般被地球的椭球形弥补,可以不计.
例1
假设 30 吨重的高铁车厢在北纬 30 度以 $300\Si{km/h}$ 的速度行驶,其水平方向的科氏力大小为
\begin{equation}
\ali{
F_{col} &= 2\times 30,000\Si{kg}\times \frac{300,000\Si{m}}{3600\Si{s}} \times\frac{2\pi}{24\Si{h}\times 3600\Si{s}}\times\sin\frac{\pi}{6} \\
&= 60.32\Si{N}
}\end{equation}
1. 但严格来说地球由于受离心力,赤道宽,两极窄.
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