User:Skeptical scientist/scratchwork From Wikipedia, the free encyclopedia 1 z = 1 x + i y = 1 x + i y ⋅ x − i y x − i y = x − i y x 2 + y 2 {\displaystyle {\frac {1}{z}}={\frac {1}{x+iy}}={\frac {1}{x+iy}}\cdot {\frac {x-iy}{x-iy}}={\frac {x-iy}{x^{2}+y^{2}}}} ∭ C G ρ ( r → − ( R − d , 0 , 0 ) ) | | r → − ( R − d , 0 , 0 ) | | 3 d 3 V ( r → ) {\displaystyle \iiint _{C}{\frac {G\rho ({\vec {r}}-(R-d,0,0))}{||{\vec {r}}-(R-d,0,0)||^{3}}}d^{3}V({\vec {r}})} ∫ 0 ∞ 1 m 2 ⋅ ρ ( r ) d r = ∫ 0 ∞ 1 m 2 ⋅ ρ 0 e − k g r d r = [ − ρ 0 ⋅ 1 m 2 k g e − k g r ] 0 ∞ = ρ 0 ⋅ 1 m 2 k g , {\displaystyle \int _{0}^{\infty }1m^{2}\cdot \rho (r)\,dr=\int _{0}^{\infty }1m^{2}\cdot \rho _{0}e^{-kgr}\,dr=\left[-{\frac {\rho _{0}\cdot 1m^{2}}{kg}}e^{-kgr}\right]_{0}^{\infty }={\frac {\rho _{0}\cdot 1m^{2}}{kg}},} g ∫ h ∞ ρ ( r ) d r = g ∫ h ∞ ρ 0 e − k g r d r = g [ − ρ 0 k g e − k g r ] h ∞ = ρ 0 e − k g h k = ρ ( h ) k = P ( h ) . {\displaystyle g\int _{h}^{\infty }\rho (r)\,dr=g\int _{h}^{\infty }\rho _{0}e^{-kgr}\,dr=g\left[-{\frac {\rho _{0}}{kg}}e^{-kgr}\right]_{h}^{\infty }={\frac {\rho _{0}e^{-kgh}}{k}}={\frac {\rho (h)}{k}}=P(h).} ∫ 0 2 π ∫ 0 ∞ G m d h 2 + r 2 h r 2 + h 2 r d r d θ {\displaystyle \int _{0}^{2\pi }\int _{0}^{\infty }{\frac {Gmd}{h^{2}+r^{2}}}{\frac {h}{\sqrt {r^{2}+h^{2}}}}rdrd\theta } = π ∫ 0 ∞ G m d h ( h 2 + r 2 ) 3 / 2 2 r d r {\displaystyle =\pi \int _{0}^{\infty }{\frac {Gmdh}{(h^{2}+r^{2})^{3/2}}}2rdr} = π ∫ h 2 ∞ G m d h u 3 / 2 d u {\displaystyle =\pi \int _{h^{2}}^{\infty }{\frac {Gmdh}{u^{3/2}}}du} = π [ − 2 G m d h u ] h 2 ∞ {\displaystyle =\pi \left[-2{\frac {Gmdh}{\sqrt {u}}}\right]_{h^{2}}^{\infty }} Related Articles
![{\displaystyle \int _{0}^{\infty }1m^{2}\cdot \rho (r)\,dr=\int _{0}^{\infty }1m^{2}\cdot \rho _{0}e^{-kgr}\,dr=\left[-{\frac {\rho _{0}\cdot 1m^{2}}{kg}}e^{-kgr}\right]_{0}^{\infty }={\frac {\rho _{0}\cdot 1m^{2}}{kg}},}](//wikimedia.org/api/rest_v1/media/math/render/svg/5f1fd5bb8170af59a1914c42a3e8556a45f584f0)
![{\displaystyle g\int _{h}^{\infty }\rho (r)\,dr=g\int _{h}^{\infty }\rho _{0}e^{-kgr}\,dr=g\left[-{\frac {\rho _{0}}{kg}}e^{-kgr}\right]_{h}^{\infty }={\frac {\rho _{0}e^{-kgh}}{k}}={\frac {\rho (h)}{k}}=P(h).}](//wikimedia.org/api/rest_v1/media/math/render/svg/206f74882b4eb65c3fe574bd0f00f2fcbe36d053)
![{\displaystyle =\pi \left[-2{\frac {Gmdh}{\sqrt {u}}}\right]_{h^{2}}^{\infty }}](//wikimedia.org/api/rest_v1/media/math/render/svg/415dd8d4610c9ac3bba7dc2d1b79a1e8e8e8f4da)
![{\displaystyle \int _{0}^{\infty }1m^{2}\cdot \rho (r)\,dr=\int _{0}^{\infty }1m^{2}\cdot \rho _{0}e^{-kgr}\,dr=\left[-{\frac {\rho _{0}\cdot 1m^{2}}{kg}}e^{-kgr}\right]_{0}^{\infty }={\frac {\rho _{0}\cdot 1m^{2}}{kg}},}](http://wikimedia.org/api/rest_v1/media/math/render/svg/5f1fd5bb8170af59a1914c42a3e8556a45f584f0)
![{\displaystyle g\int _{h}^{\infty }\rho (r)\,dr=g\int _{h}^{\infty }\rho _{0}e^{-kgr}\,dr=g\left[-{\frac {\rho _{0}}{kg}}e^{-kgr}\right]_{h}^{\infty }={\frac {\rho _{0}e^{-kgh}}{k}}={\frac {\rho (h)}{k}}=P(h).}](http://wikimedia.org/api/rest_v1/media/math/render/svg/206f74882b4eb65c3fe574bd0f00f2fcbe36d053)
![{\displaystyle =\pi \left[-2{\frac {Gmdh}{\sqrt {u}}}\right]_{h^{2}}^{\infty }}](http://wikimedia.org/api/rest_v1/media/math/render/svg/415dd8d4610c9ac3bba7dc2d1b79a1e8e8e8f4da)