Proving an ellipse fits Kepler's 1st law

  • #1

Homework Statement


(This isn't coursework, just a revision question)
Exercise: An ellipse can be defined as the locus of all points, P, in the plane such that
[tex]PF_1+PF_2=\frac{2p}{(1-ε^2)}[/tex]
where F1 and F2 are two fixed points, and PF1 is the distance from P to F1 (similarly, P F2).
F1 and F2 are known as the foci. By placing F1 at the origin, and F2 at [tex]x = \frac{-2pε}{(1-ε^2)}[/tex]
show that the ellipse satisfies [tex]r = \frac{p}{(1+εcos(θ- θ_0))}[/tex]


Homework Equations


The example is on page 26 of this pdf http://www.nottingham.ac.uk/~ppzap4/PoD.pdf
Hint: Draw a picture and use the cos rule

The Attempt at a Solution


I've assumed that PF1 is r, and therefore PF2 is [tex]\frac{2p}{(1-ε^2)}-r[/tex] and have used the cosine rule to show[tex]r^2 = (\frac{p}{(1+εcos(θ- θ_0))})^2=(\frac{2p}{(1-ε^2)}-r)^2+(\frac{-2pε}{(1-ε^2)})^2+(\frac{2p}{(1-ε^2)}-r)(\frac{-2pε}{(1-ε^2)})cos(θ- θ_0)[/tex]I've tried multiple times but can't seem to rearrange this to prove it is true, can anyone give me a pointer in the right direction.
 

Answers and Replies

  • #2
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I would start with the cosine rule without the relation you want to prove, and then see if solving it for p works.
Alternatively, set up a coordinate system and work with those two coordinates and the first equation and go to r and theta later.

Independently of the approach, you can try to find ##\theta_0## first.
 

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