最近30从www.hoelymoley.com 2023 - 07 - 09 - t00:04:22z //www.hoelymoley.com/feeds/question/24928 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/24928 4 木 //www.hoelymoley.com/users/11611 2023 - 02年- 18 - t17:28:37z 2023 - 02 - 24 t12:35:01z

< p >地球的平均温度是什么当你考虑所有的层,不仅表面吗?一切我发现到目前为止只关注表面或单独每一层。但是如果你把所有层的平均(加权)?和它是如何计算的?这个答案< a href = " https://astronomy.stackexchange.com/a/27260 " > < / >似乎表明权重应该热容:< / p > < p > <跨类= " math-container " > $ $ T_{地球}= \识别裂缝分析{\ iiint_{地球}C \ \, dV} {\ iiint_{地球}C \, dV} = \压裂{\ int_0 ^ 4 \πR ^ 2 \ C (R) \ T (R) \博士}{\ int_0 ^ 4 \πR ^ 2 \ C (R) \,博士}$ $ < / span > < / p > < p >(这是一个重复的一个封闭的< a href = " https://astronomy.stackexchange.com/questions/27252/what-is-earths-average-temperature-as-a-whole " > < / >问题张贴在天文学堆栈交换)< / p >江南电子竞技平台

//www.hoelymoley.com/questions/24928/-/24933 # 24933 7 木 //www.hoelymoley.com/users/11611 2023 - 02年- 20 - t05:42:03z 2023 - 02 - 24 t12:35:01z

< p > <强> TL;博士:< / >强约<跨类= " math-container " > 2700美元\ \文字K美元< / span >(<跨类= " math-container " > \文本{°C} 2400美元< / span >, < span class = " math-container " > \文本{°F} 4400美元< / span >) < / p > <人力资源/ > < p >假设权重的确是单位体积热容<跨类= " math-container " >美元加元< / span >, < span class = " math-container " > C \大约\ρC_P美元< / span >,在< span class = " math-container " > \ρ< / span >是美元的密度和<跨类= " math-container " > C_P < / span >是美元在恒压比热容,我们可以计算平均温度的公式:< span class = " math-container " > $ $ \{对齐}开始T_{地球},识别= \压裂{\ iiint_{地球}C \ \, dV} {\ iiint_{地球}C \, dV} = \压裂{\ int_0 ^ 4 \πR ^ 2 \ C (R) \ T (R) \博士}{\ int_0 ^ 4 \πR ^ 2 \ C (R) \,博士}\ \,= \压裂{\ int_0 ^ R R ^ 2 \ C (R) \ T (R) \博士}{\ int_0 ^ R R ^ 2 \ C (R) \,博士}\结束{对齐}$ $ < / span > < a href = " https://www.cambridge.org/core/books/physics-of-the-earth/05CC5483F1DD8FCC2604C550946AC2E6 " rel = " nofollow noreferrer " >地球物理学< / >,斯泰西,戴维斯[1]我们发现表<跨类= " math-container " > \ρ< / span >美元(表F.1,页469 - 471),和一个表类< span = " math-container " > $ T $ < / span >和<跨类= " math-container " > C_P < / span >美元(表G.1,页472 - 473)。< / p > < p >我们近似类< span = " math-container " > C (r)美元< / span >和<跨类= " math-container " > T (r) < / span >美元通过分段仿射函数,每个间隔<跨类= " math-container " >的$ r \ [r_n, r_ {n + 1}] < / span >给美元以下近似:< span class = " math-container " > $ $ C (r) \大约C_n + \压裂{r - r_n} {r_ {n + 1} - r_n} \ cdot (C_ {n + 1} - C_n) \ \ T (r) \大约T_n + \压裂{r - r_n} {r_ {n + 1} - r_n} \ cdot (T_ {n + 1} - T_n识别)$ $ < / span >所以可以近似积分分子(根据< a href = " https://www.wolframalpha.com/input?我从% 20 = % 20积分r1 % 20 % 20 r2 % 20 % 20 R % 5的e2 * % 28 c1 % 20% 2 b % 20% 28 r-r1 % 29 * % 28路径c1 % 2 f % 28 r2-r1 % 29% 29% 29 * % 28 t1 % 20% 2 b % 20% 28 r-r1 28 t2-t1 % 29 * % % 2 f % 28 r2-r1 % 29% 29% 29 *博士nofollow noreferrer“rel = > WolframAlpha < / >): < span class = " math-container " > $ $ \ int_0 ^ R R ^ 2 \ C (R) \ T (R) \ \ \ \博士约\ sum_ {n = 1} ^ {n} \ int_ {r_n} ^ {r_ {n + 1}} R ^ 2 \离开(C_n + \压裂{R - r_n} {r_ {n + 1} - r_n} \ cdot (C_ {n + 1} - C_n) \) \离开(T_n + \压裂{R - r_n} {r_ {n + 1} - r_n} \ cdot (T_ {n + 1} - T_n识别)\)\,博士\ \ = \ sum_ {n = 1} ^ {n} \压裂{1}{60}(r_ {n + 1} -r_n) \左\ {C_n \离开[3 r_n ^ 2 (4 T_n + T_ {n + 1})识别+ r_n r_ {n + 1} (6 T_n + 4 T_ {n + 1})识别+ r_ {n + 1} ^ 2 (2 T_n + 3 T_ {n + 1})识别正确\]+ C_左{n + 1} \ [r_n ^ 2 (3 T_n + 2 T_ {n + 1})识别+ r_n r_ {n + 1} (4 T_n + 6 T_ {n + 1})识别+ 3 r_ {n + 1} ^ 2 (T_n + 4 T_ {n + 1})识别正确\]正确\ \}$ $ < / span >和分母变得(根据< a href = " https://www.wolframalpha.com/input?我从% 20 = % 20积分r1 % 20 % 20 r2 % 20 % 20 R % 5的e2 * % 28 c1 % 20% 2 b % 20% 28 r-r1 % 29 * % 28路径c1 % 2 F % 28 r2-r1 % 29% 29% 29 *博士nofollow noreferrer“rel = > WolframAlpha < / >): < span class = " math-container " > $ $ \ int_0 ^ R R ^ 2 \ C (R) \ \ \ \博士约\ sum_ {n = 1} ^ {n} \ int_ {r_n} ^ {r_ {n + 1}} R ^ 2 \离开(C_n + \压裂{R - r_n} {r_ {n + 1} - r_n} \ cdot (C_ {n + 1} - C_n) \) \,博士\ \ = \ sum_ {n = 1} ^ {n} \压裂{1}{12}(r_ {n + 1} - r_n)左\ [r_n ^ 2 (3 C_n + C_ {n + 1}) + 2 r_n r_ {n + 1} (C_n + C_ {n + 1}) + r_ {n + 1} ^ 2 (C_n + 3 C_ {n + 1}) \右]$ $ < / span > < / p > < p >现在我们可以写这些表达式在< a href = " https://docs.google.com/spreadsheets/d/1wGtwVJboFD6Tay_arlQmh_pub3EVGubz/ " rel = " nofollow noreferrer " >电子表格的> < /值[1],然后和他们划分结果得到最终答案:< / p > < p > <跨类= " math-container " > $ $ \{对齐}开始T_{地球}和识别\大约2692 \ \文字K \ \, = 2419 {°C} \ \ \文本,文本{°F} = 4386 \ \结束{对齐}$ $ < / span > < / p > < p >[1]斯泰西,F。,,戴维斯,p (2008)。地球物理(第四版)。剑桥:剑桥大学出版社。doi: 10.1017 / CBO9780511812910 < / p >

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