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ParthKohli

  • one year ago

Another nice physics question.

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  1. ParthKohli
    • one year ago
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    A space vehicle is approaching a planet with a speed \(v\) and is a long way out (treat it as infinity). The trajectory would miss the centre of the planet by distance \(R\) if it continued in a straight line. If the planet has a mass \(m\) and a distance \(r\), what is the smallest value of \(R\) in order that the resulting orbit will just miss the surface?

  2. ParthKohli
    • one year ago
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    So I'll try using conservation of energy here probably.|dw:1441717901150:dw|

  3. ParthKohli
    • one year ago
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    |dw:1441718086223:dw|

  4. Abhisar
    • one year ago
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    Is r the radius of the planet?

  5. ParthKohli
    • one year ago
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    \[\rm{Initial ~energy:}~~\frac{1}{2}mv^2 \]Now the gravitational force does some work to bring it from its original trajectory to the surface. This work is equal to the difference in potential energy.\[= \frac{GMm}{r} - \frac{GMm}{R}\]

  6. ParthKohli
    • one year ago
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    @Abhisar Yes.

  7. ParthKohli
    • one year ago
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    OK, sorry, my OS got completely hung.

  8. ParthKohli
    • one year ago
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    Alright. Ugh. I'll get to the point in that case.\[\text{initial energy + change in potential energy = final energy}\]Now,\[\frac{1}{2}mv^2 + \frac{GMm}{r} - \frac{GMm}{R} = \frac{1}{2}m \left(\sqrt{GM/r}\right)^2 - \frac{GMm}{r}\]

  9. ParthKohli
    • one year ago
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    \[\frac{1}{2}mv^2 + \frac{GMm}{r} - \frac{GMm}{R} = -\frac{GMm}{2r}\]\[\Rightarrow \frac{1}{2}v^2 + \frac{3}{2r}GM = \frac{GM}{R}\]

  10. IrishBoy123
    • one year ago
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    http://openstudy.com/study#/updates/55ec05c1e4b0445dfd1847ef

  11. IrishBoy123
    • one year ago
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    i'd build some DE's to solve this.

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