Semi-major axis definition: sum or total of distances

[TrifidBlue]

I hope this the right place to post my question...

should it be, "we can define a as half the sum of distances..."?
please correct and explain if I'm mistaken
thanks

 

[HallsofIvy]

You are mistaken.

When $\theta= \theta'$, $1/r= C(1- \epsilon cos(\theta- \theta')=$$C(1- \epsilon cos(0))=$$C(1- \epsilon)$ so that $r= 1/(C(1-\epsilon)$.

When $\theta= \theta'+ \pi$, $1/r= C(1+\epsilon cos(\theta- \theta- \pi)=$$C(1- \epsilon cos(\pi))=$$C(1+ \epsilon)$ so that $r= 1/(C(1+\epsilon)$.

The total distance between those points is $1/C(1+\epsilon)+$$1/C(1- \epsilon)$$= (1/C)(1/(1+\epsilon)+ 1/(1- \epsilon))$. Getting the common denominator, $(1+\epsilon)(1-\epsilon)= 1-\epsilon^2$, we have
$$(1/C)\frac{1- \epsilon+ 1+ \epsilon}{1- \epsilon^2}= (1/C)\frac{2}{1- \epsilon^2}$$

Half of that is
$$a= \frac{1}{C(1- \epsilon^2)}$$

 

[TrifidBlue]

You are mistaken.

When $\theta= \theta'$, $1/r= C(1- \epsilon cos(\theta- \theta')=$$C(1- \epsilon cos(0))=$$C(1- \epsilon)$ so that $r= 1/(C(1-\epsilon)$.

When $\theta= \theta'+ \pi$, $1/r= C(1+\epsilon cos(\theta- \theta- \pi)=$$C(1- \epsilon cos(\pi))=$$C(1+ \epsilon)$ so that $r= 1/(C(1+\epsilon)$.

The total distance between those points is $1/C(1+\epsilon)+$$1/C(1- \epsilon)$$= (1/C)(1/(1+\epsilon)+ 1/(1- \epsilon))$. Getting the common denominator, $(1+\epsilon)(1-\epsilon)= 1-\epsilon^2$, we have
$$(1/C)\frac{1- \epsilon+ 1+ \epsilon}{1- \epsilon^2}= (1/C)\frac{2}{1- \epsilon^2}$$

Half of that is
$$a= \frac{1}{C(1- \epsilon^2)}$$

and that's what I've said
thank you