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Help with proving this Kepler's laws based theorem

  • Thread starter Dvsdvs
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  • #1
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the proposition is that if r(t) has a constant length (//r(t)// is constant), then prove that r(t) is perpendicular to dr/dt at all t.


I was thinking that if r(t) is just a function of the radius and its magnitude //r// is constant, then its basically talking about a circle? and by saying prove that r is perpendicular to dr/dt is basically asking to prove that the radius vector is always orthogonal to the velocity v(t) vector? I need a mathematical proof of this proposition using vectors. Thank, you
 

Answers and Replies

  • #2
181
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What is d/dt(r.r)?

BTW Kepler hasn't got much to do with it.
Any wiggly path on the surface of a sphere has constant r.

David
 
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  • #3
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I'm not sure what your asking. this is a general proof so i dont know what d/dt(r.r) is. all the information i have is posted.
 
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  • #4
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i got an idea... would the //x(t)// be equal to the length of the curve of x(t) and therefore //x(t)//=integral(x(t).dx/dt) and since //x(t)// is constant the indefinite integral can only=c if the x(t).dx/dt=0 and so they are perpendicular. Can someone plz tell me if all this makes sense?
 
  • #5
1,838
7
I'm not sure what your asking. this is a general proof so i dont know what d/dt(r.r) is. all the information i have is posted.
r^2 is constant:


0 = d/dt r^2 = 2 r dot dr/dt
 
  • #6
24
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0 = d/dt r^2 = 2 r dot dr/dt
oh ok so you're saying that since //r(t)// is constant then //r(t)//^2 is constant which means (r.r) is constant and that derivative which is a dot product is 0. yeah this was easier than i thought my integral solution also works i think
 

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