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User:GabrielVelasquez

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This is an old revision of this page, as edited by GabrielVelasquez (talk | contribs) at 16:34, 7 September 2009. The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

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This user is a beginner archer.
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This user does weight training.
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This user supports the right of all women to go without a top at the beach, the local pool, or in public.
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This user supports Humans United Against Robots.
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Oxford EncyclopediasThis user is interested in ].




T p = ( L ( 1 A ) 16 π σ D 2 ) 1 4 {\displaystyle T_{\mathrm {p} }=\left({\frac {L(1-A)}{16\pi \sigma D^{2}}}\right)^{\tfrac {1}{4}}}
T p = T [ ( R d ) 2 1 A p 4 ] 1 4 {\displaystyle T_{\mathrm {p} }=T_{\ast }\left^{\frac {1}{4}}}
σ T p 4 = σ T 4 4 π d 2 4 π R 2 4 π R p 2 π R p 2 ( 1 A p ) {\displaystyle \sigma \cdot T_{\mathrm {p} }^{4}={\frac {\sigma \cdot T_{\ast }^{4}}{4\pi \cdot d^{2}}}\cdot {\frac {4\pi \cdot R_{\ast }^{2}}{4\pi \cdot R_{\mathrm {p} }^{2}}}\cdot \pi \cdot R_{\mathrm {p} }^{2}(1-A_{\mathrm {p} })}

code for Earth at Perihelion: f p = ( ( 6.955 × 10 8 ) 2 ) × ( 5.67051 × 10 8 ) × ( 5778 4 ) ( ( 1 ( 1 × 0.016710219 ) ) × 149597876600 ) 2 = 1 , 412.903   W / m 2 {\displaystyle f_{p}={\frac {((6.955\times 10^{8})^{2})\times (5.67051\times 10^{-8})\times (5778^{4})}{((1-(1\times 0.016710219))\times 149597876600)^{2}}}=1,412.903\ W/m^{2}}

code for Gliese 581 c at Periastron: f p = ( ( 0.38 × 6.955 × 10 8 ) 2 ) × ( 5.67051 × 10 8 ) × ( 3480 4 ) ( ( 0.073 ( 0.073 × 0.16 ) ) × 149597876600 ) 2 = 6.9 × 10 3   W / m 2 {\displaystyle f_{p}={\frac {((0.38\times 6.955\times 10^{8})^{2})\times (5.67051\times 10^{-8})\times (3480^{4})}{((0.073-(0.073\times 0.16))\times 149597876600)^{2}}}=6.9\times 10^{3}\ W/m^{2}}

G l   581   c   P e r i a s t r o n   I n s o l a t i o n E a r t h s   S o l a r   C o n s t a n t = {\displaystyle {\frac {Gl\ 581\ c\ Periastron\ Insolation}{Earth's\ Solar\ Constant}}=} 6.9 × 10 3   W / m 2 1 , 366.079   W / m 2 = 505 % {\displaystyle {\frac {6.9\times 10^{3}\ W/m^{2}}{1,366.079\ W/m^{2}}}=505\%}

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