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Newton's Law of Universal Gravitation (Differential Equation Question)

  • Thread starter Bazzinga
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  • #1
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I'm having trouble with part a) of this question...

[PLAIN]http://img69.imageshack.us/img69/5815/98157006.png [Broken]


So I started off by solving the DE above a), and I've gotten it down to:

[tex]\frac{1}{2} m v^{2} = \frac{mgR^{2}}{(x + R)} + C[/tex]

I can tell I'm getting close, but I'm a little confused where h is coming from, and how to get rid of the [tex]R^{2}[/tex]. Also, what do I do with C??

I know it says x = x(t) which is the height, so does that mean that x(t) = h? I've never taken physics (and this was in my Calculus homework) so this is all new to me!
 
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Answers and Replies

  • #2
hunt_mat
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Good so far, to find C set v=0 when x=h, then to find v_0 set x=0.
 
  • #3
dextercioby
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Hi, you should use a definite integration

[tex] \int_{v_0}^{0} mv \ dv = \int_{0}^{h} -\frac{mgR^2}{(R+x)^2} dx [/tex]

That way you'll get the formula in the book.

As for the limit, I think it's trivial, right ?
 
  • #4
hunt_mat
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or, he could have done what I told him to to find the constant C, both methods are equivalent.
 
  • #5
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So C = (-mgR^2) / (h+R) ?
 
  • #6
hunt_mat
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Correct. From there you can easily derive the required answer.
 
  • #7
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Thanks a lot guys! So what about b)? will V0 approach infinity because the top approaches infinity faster than the bottom?
 
  • #8
hunt_mat
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divide the top and bottom of the fraction and you will have on the demoninator a R/h which is the only place h appears, take the limit as h tends to infinity and that yerm dissapears.
 

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