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chris215

  • one year ago

Va is the orbital speed of a satellite at a distance x from the centre of the Earth. If the satellite is at a distance 2x from the centre of the Earth and with the speed of Vb, which of the following statements is correct?

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  1. chris215
    • one year ago
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    va = vb^2?

  2. Astrophysics
    • one year ago
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    |dw:1435735326495:dw| we need to use the following formula \[\huge v_{escape}^2 = \frac{ 2GM }{ r }\] so we have \[\large v_a^2 = \frac{ 2GM }{ x }\] and \[\large v_b^2 = \frac{ 2GM }{ 2x } = \frac{ GM }{ x }\]

  3. Astrophysics
    • one year ago
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    So now we have \[\large v_a = \sqrt{\frac{ 2GM }{ x }} \implies \sqrt{2}\sqrt{\frac{ GM }{ x }}\] and we have \[\large v_b = \sqrt{\frac{ GM }{ x }}\] therefore we can say \[\huge v_a = \sqrt{2}v_b\]

  4. Michele_Laino
    • one year ago
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    for mechanical stability, we can write: \[\Large \begin{gathered} m\frac{{v_a^2}}{x} = \frac{{GMm}}{{{x^2}}} \hfill \\ \hfill \\ m\frac{{v_b^2}}{{2x}} = \frac{{GMm}}{{{{\left( {2x} \right)}^2}}} \hfill \\ \end{gathered} \] where m is the mass of our satellite, and M is the mass of Earth. Now simplifying both equations, I get: \[\Large {v_a} = \sqrt {\frac{{GM}}{x}} ,\quad {v_b} = \frac{1}{{\sqrt 2 }}\sqrt {\frac{{GM}}{x}} \] So, what can you conclude?

  5. Astrophysics
    • one year ago
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    Why did you do that @Michele_Laino ?

  6. Michele_Laino
    • one year ago
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    It is another way to solve that question @Astrophysics

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