ParthKohli
  • ParthKohli
Another nice physics question.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
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?
ParthKohli
  • ParthKohli
So I'll try using conservation of energy here probably.|dw:1441717901150:dw|
ParthKohli
  • ParthKohli
|dw:1441718086223:dw|

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Abhisar
  • Abhisar
Is r the radius of the planet?
ParthKohli
  • ParthKohli
\[\rm{Initial ~energy:}~~\frac{1}{2}mv^2 \]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.\[= \frac{GMm}{r} - \frac{GMm}{R}\]
ParthKohli
  • ParthKohli
@Abhisar Yes.
ParthKohli
  • ParthKohli
OK, sorry, my OS got completely hung.
ParthKohli
  • ParthKohli
Alright. Ugh. I'll get to the point in that case.\[\text{initial energy + change in potential energy = final energy}\]Now,\[\frac{1}{2}mv^2 + \frac{GMm}{r} - \frac{GMm}{R} = \frac{1}{2}m \left(\sqrt{GM/r}\right)^2 - \frac{GMm}{r}\]
ParthKohli
  • ParthKohli
\[\frac{1}{2}mv^2 + \frac{GMm}{r} - \frac{GMm}{R} = -\frac{GMm}{2r}\]\[\Rightarrow \frac{1}{2}v^2 + \frac{3}{2r}GM = \frac{GM}{R}\]
IrishBoy123
  • IrishBoy123
i'd build some DE's to solve this.

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