土壤是地球直径的减少因为下沉由于灌溉和降雨多年来吗?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换 最近30从www.hoelymoley.com 2023 - 07 - 10 - t21:26:55z //www.hoelymoley.com/feeds/question/11898 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/11898 2 土壤是地球直径的减少因为下沉由于灌溉和降雨多年来吗? 克利须那神 //www.hoelymoley.com/users/8520 2017 - 07 - 23 - t04:06:27z 2017 - 07 - 23 - t19:57:00z < p >我上coursera教练会谈的<强>土壤沉降具体文章< /强>是在<强>佛罗里达南部大沼泽1920年代< /强>。他继续描述地球的水平多年来减少由于土壤沉降< /强> <强>。< / p > < p >现在地球上有雨水很多年了,在过去的几个世纪,农业面积显著增加,在我看来会导致增加由于灌溉土壤沉降。< / p > < p > < em >这连续降雨和灌溉导致更高的沉降,因此减少地球的直径? < / em > < / p > //www.hoelymoley.com/questions/11898/-/11903 # 11903 4 回答的Z W的地球直径的减少因为土壤沉降由于灌溉和降雨多年来吗? Z W //www.hoelymoley.com/users/6221 2017 - 07 - 23 - t19:46:06z 2017 - 07 - 23 - t19:57:00z < p >免责声明:这可能是TLWR,但你完全可以运用一些物理推理得出答案。< / p > < p >地球表面高度的变化都是由质量守恒定律,即,如果质量(M美元)删除一个位置(湿地土壤侵蚀),那么必须获得等量的质量在另一个位置(可能在海底吸积)。地球质量为零的总变化δM = $ $ \ M_1-M_2 = 0 $ $ $ $ 1 = M_2 $ $。< / p > < p >质量可以分为体积(V =美元\δx \δy \δz $)乘以密度(ρ\美元),所以$ M = \ρV $。从上面的质量守恒方程,然后:$ $ - \ rho_{1} \δδy_1 x_1 \ \δz_1 = \ rho_{2} \δδy_2 x_2 \ \δz_ {2} $ $ < / p > < p >侵蚀发生的地方是所描述的左边(下标1)。吸积发生的地方是所描述的右边(下标2)。这个方程表示的总质量损失区域1,必须等于所获得的总质量区域2。现在考虑以下两种情况。< / p > < p > <强> 1)运输沉积物< em >不接受< / em > < / >强密度的变化。如果不改变密度,\ rho_1 = \ rho_2美元。让我们也说,侵蚀面积(\ x_1 \δy_1美元)=增生区(\ x_2 \δy_2美元)简单。那么平等的质量守恒消掉了,剩下$ $ -δz_ \{1} =δz_ \ {2} $ $ < / p > < p > <强> 2)运输沉积物< em >并接受< / em > < / >强密度的变化。有许多原因密度< em > < / em >可以改变。 If the density changes then we can't let the density terms cancel in the conservation of mass. Lets still say the eroding area and accreting area are equal. The conservation of mass becomes $$-\rho_1 \Delta z_1 = \rho_2 \Delta z_2$$ or conversely $$-\Delta z_1=\frac{\rho_2}{\rho_1}\Delta z_2$$ This says that the amount of change in height at location 1 is not equal to the amount of change at location 2 (sure they're proportional, but not definitely not exactly equal).

Putting it all together to answer the question, the average earth radius (r) is the average of all heights (z) with respect to the center of the earth $$r=\frac{z_1+z_2+...+z_n}{n}$$

Try this all out with arbitrary numbers to convince yourself. Add the changes $\Delta z_1$ to $z_1$, and $\Delta z_2$ to $z_2$ $$r_{before}=\frac{z_1 + z_2 + z_3}{3}$$ $$r_{after}=\frac{(z_1+\Delta z_1)+(z_2+\Delta z_2)+z_3}{3}$$ Rearrange $r_{after}$ to be $$r_{after}=\frac{z_1+z_2+z_3 + (\Delta z_1+\Delta z_2)}{3}$$

Case 1 If density doesn't change (i.e. $-\Delta z_1=\Delta z_2$), then $$r_{after}=\frac{z_1+z_2+z_3 + (\Delta z_1-\Delta z_1)}{3}$$ The changes cancel out and $r_{before}=r_{after}$. The earths radius does not at all change.

Case 2 If the density changes, $r_{after}$ becomes $$r_{after}=\frac{z_1+z_2+z_3 + (\Delta z_1-\frac{\rho_2}{\rho_1}\Delta z_1)}{3}$$ and you can see that $r_{after}$ does not equal $r_{before}$.

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