Does the diameter of the earth decrease because of soil subsiding due to irrigation and rainfall over the years? - Earth Science Stack Exchange most recent 30 from earthscience.stackexchange.com 2019-09-18T05:14:33Z https://earthscience.stackexchange.com/feeds/question/11898 https://creativecommons.org/licenses/by-sa/4.0/rdf https://earthscience.stackexchange.com/q/11898 1 Does the diameter of the earth decrease because of soil subsiding due to irrigation and rainfall over the years? Krishna https://earthscience.stackexchange.com/users/8520 2017-07-23T04:06:27Z 2017-07-23T19:57:00Z <p>I am taking a course on coursera where the instructor talks of a <strong>soil subsidence concrete post</strong> that was put there in <strong>Everglades southern Florida in 1920s</strong>. He goes on to describe the level of earth that has decreased over the years due to <strong>soil subsidence</strong>.</p> <p>Now there have been rains on the earth for years and in past couple of centuries, the area for agriculture has increased significantly which in my opinion leads to increased soil subsidence due to irrigation.</p> <p><em>Doesn't this continuous rainfall and irrigation lead to higher subsidence and hence decrease in the diameter of the earth?</em></p> https://earthscience.stackexchange.com/questions/11898/-/11903#11903 3 Answer by Z W for Does the diameter of the earth decrease because of soil subsiding due to irrigation and rainfall over the years? Z W https://earthscience.stackexchange.com/users/6221 2017-07-23T19:46:06Z 2017-07-23T19:57:00Z <p>Disclaimer: this is probably TLWR, but you can absolutely apply some physical reasoning to arrive at the answer.</p> <p>Changes in surface height of the earth are governed by the conservation of mass, i.e. that if mass ($M$) is removed in one location (everglades soil erosion), then an equal amount of mass must be gained at another location (perhaps accretion on the seafloor).The total change in the earths mass is zero $$\Delta M = M_1-M_2=0$$ $$-M_1=M_2$$. </p> <p>Mass may be broken up into the volume ($V=\Delta x \Delta y \Delta z$) multiplied by density ($\rho$), so $M=\rho V$. In terms of the conservation of mass equation from above, then: $$-\rho_{1} \Delta x_1 \Delta y_1 \Delta z_1 = \rho_{2} \Delta x_2 \Delta y_2 \Delta z_{2}$$</p> <p>The place where erosion occurs is described by the left side (with subscript 1). The place where accretion occurs is described by the right side (with subscript 2). This equation says the total mass lost by area 1, must equal the total mass gained by area 2. Now consider the following two situations.</p> <p><strong>1) The transported sediment <em>does not undergo</em> any change in density</strong>. If the density does not change, you have $\rho_1=\rho_2$. Let's also say the eroding area ($\Delta x_1 \Delta y_1$) equals the accreting area ($\Delta x_2 \Delta y_2$) for simplicity. Then all equal terms in the conservation of mass cancel out and you're left with $$-\Delta z_{1} = \Delta z_{2}$$</p> <p><strong>2) The transported sediment <em>does undergo</em> a change in density</strong>. There's a host of reasons why density <em>can</em> change. 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).</p> <p><strong>Putting it all together to answer the question</strong>, the average earth radius (r) is the average of all heights (z) with respect to the center of the earth $$r=\frac{z_1+z_2+...+z_n}{n}$$</p> <p><strong>Try this all out with arbitrary numbers to convince yourself</strong>. 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 $$r_{after}=\frac{z_1+z_2+z_3 + (\Delta z_1+\Delta z_2)}{3}$$</p> <p><strong>Case 1</strong> 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.</p> <p><strong>Case 2</strong> 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 $r_{before}$.</p>