For large aa (a gg bab), what is the approximate difference delta e equiv e' - eeee in terms of a, ba,b, and ee? In

 For large aa (a gg ba≫b), what is the approximate difference delta e equiv e' - eδe≡e′−e in terms of a, ba,b, and ee? In other words, Taylor expand delta eδe to first order in b/ab/a. How well does this formula work for the numerical values you calculated in the previous part?

delta e equiv e' - e equivδe≡e′−e≡__________

Hint (Added on 6/18/2017): One can do this problem with the following steps:

1. For the Newtonian case, express the energy EE of the orbit in terms of the closet and furthest distance (and relevant constant like k=GMmk=GMm). You should did this already in part (a).
2. Then, along with r = frac{b}{1+ecos(theta)}r=1+ecos(θ)b one can then express the eccentricity ee in terms of the energy of the orbit and bb, i.e. e =e (E, b)e=e(E,b).
3. Now for the Yukawa case, Taylor expand the exponential to the first order. Comparing to the energy of the Newtonian case, you will find that the energy of the Yukawa case E'E′ is just a constant shift comparing to the Newtonian case, i.e. E' = E + text{constant}E′=E+constant, and the constant depends oaa.
4. As aa is large, the constant shift you found in the previous step is actually small. We therefore assume that the relation between the eccentricity and energy you found in step 2. for the Newtonian case also applies to the Yukawa case, but just replacing ee with e'e′ and EE with E'E′. So now you should have the relation e' = e'(E', b) = e'(E+text{constant}, b)e′=e′(E′,b)=e′(E+constant,b). You can then express e'e′ in terms of ee, aa and bb. Taylor expand this expression and you will be able to find delta eδe.