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ParthKohli
 one year ago
Another nice physics question.
ParthKohli
 one year ago
Another nice physics question.

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0A 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
 one year ago
Best ResponseYou've already chosen the best response.0So I'll try using conservation of energy here probably.dw:1441717901150:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441718086223:dw

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Is r the radius of the planet?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\rm{Initial ~energy:}~~\frac{1}{2}mv^2 \]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
 one year ago
Best ResponseYou've already chosen the best response.0OK, sorry, my OS got completely hung.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Alright. 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}mv^2 + \frac{GMm}{r}  \frac{GMm}{R} = \frac{GMm}{2r}\]\[\Rightarrow \frac{1}{2}v^2 + \frac{3}{2r}GM = \frac{GM}{R}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/55ec05c1e4b0445dfd1847ef

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0i'd build some DE's to solve this.
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