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Fragen 11
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 #4
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 Integration- kann jemand helfen?

 

\(\text{$10 a$) in eine geostationäre Bahn ($ h_2 = 4,22\cdot 10^4\ km$ ) zu bringen;}\)

\(\begin{array}{|rcll|} \hline W = \displaystyle \int \limits_{h_1}^{h_2} F(s) ds &=& \displaystyle \int \limits_{h_1}^{h_2} \gamma \dfrac{mM}{s^2} ds \\\\ &=& \displaystyle \gamma mM \int \limits_{h_1}^{h_2} \dfrac{1}{s^2} ds \\\\ &=& \displaystyle \gamma mM \int \limits_{h_1}^{h_2} s^{-2} ds \\\\ &=& \gamma mM \left[ \dfrac{s^{-2+1}}{-2+1} \right]_{h_1}^{h_2} \\\\ &=& \gamma mM \left[ \dfrac{s^{-1}}{-1} \right]_{h_1}^{h_2} \\\\ &=& \gamma mM \left[ -s^{-1} \right]_{h_1}^{h_2} \\\\ &=& -\gamma mM \left[ s^{-1} \right]_{h_1}^{h_2} \\\\ &=& -\gamma mM \left[ \dfrac{1}{s} \right]_{h_1}^{h_2} \\\\ &=& -\gamma mM \left( \dfrac{1}{h_2} - \dfrac{1}{h_1} \right) \\\\ &=& -\gamma mM \left( \dfrac{h_1-h_2}{h_1h_2} \right) \\\\ \mathbf{W} & \mathbf{=} & \mathbf{ \gamma mM \left( \dfrac{h_2-h_1}{h_1h_2} \right) } \\ \hline \end{array}\)

 

\(\text{$W=\ ?$} \)

\(\begin{array}{|rcll|} \hline W &=& 6,67\cdot 10^{-11} \dfrac{m^3}{kg s^2} \cdot 10^3\ kg \cdot 5,97\cdot 10^{24}\ kg \left( \dfrac{4,22-0,6370}{4,22\cdot 0,6730} \right)\cdot \dfrac{1}{10^7\ m} \\\\ &=& 6,67\cdot 5,97 \cdot \left( \dfrac{4,22-0,6370}{4,22\cdot 0,6730} \right)\cdot 10^{-11}\cdot 10^3\cdot 10^{24}\cdot 10^{-7} \dfrac{m^3}{kg s^2}\ kg^2 \dfrac{1}{m} \\\\ &=& 39,8199 \cdot\left( \dfrac{3,5830}{2,68814} \right)\cdot 10^{9} \dfrac{kg\cdot m^2}{s^2} \\\\ &=& 39,8199 \cdot 1.33289188807\cdot 10^{9} \ Nm \\\\ &=& 53,0756216938\cdot 10^{9} \ Nm \\\\ \mathbf{W} & \mathbf{=} & \mathbf{5,30756216938\cdot 10^{10} \ Nm} \\ \hline \end{array}\)

 

 

\(\text{$10 b$) aus dem Anziehungsbereich der Erde "hinauszubefördern" ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{W} & \mathbf{=} & \mathbf{-\gamma mM \left( \dfrac{1}{h_2} - \dfrac{1}{h_1} \right) } \\\\ & = & \gamma mM \left( \dfrac{1}{h_1} - \dfrac{1}{h_2} \right) \quad & | \quad \dfrac{1}{h_2} = \dfrac{1}{\infty} = 0 \\\\ \mathbf{W} & \mathbf{=} & \mathbf{ \gamma mM \left( \dfrac{1}{h_1} \right) } \\ \hline \end{array}\)

 

\(\text{$W=\ ?$}\)

\(\begin{array}{|rcll|} \hline W &=& 6,67\cdot 10^{-11} \dfrac{m^3}{kg s^2} \cdot 10^3\ kg \cdot 5,97\cdot 10^{24}\ kg \left( \dfrac{1}{0,6730} \right)\cdot \dfrac{1}{10^7\ m} \\\\ &=& 6,67\cdot 5,97 \cdot \left( \dfrac{1}{0,6730} \right)\cdot 10^{-11}\cdot 10^3\cdot 10^{24}\cdot 10^{-7} \dfrac{m^3}{kg s^2}\ kg^2 \dfrac{1}{m} \\\\ &=& 39,8199 \cdot \left( \dfrac{1}{0,6730} \right)\cdot 10^{9} \dfrac{kg\cdot m^2}{s^2} \\\\ &=& \left( \dfrac{39,8199}{0,6730} \right)\cdot 10^{9} \ Nm \\\\ &=& 59,1677563150\cdot 10^{9} \ Nm \\\\ \mathbf{W} & \mathbf{=} & \mathbf{5,91677563150\cdot 10^{10} \ Nm} \\ \hline \end{array}\)

 

 

laugh

heureka 11.04.2018