SOLUTION

$\frac{1}{r} = \frac{1+ e \cos(1 + \epsilon)\theta}{q}$

$\frac{d}{d\theta} \Big{(} \frac{1}{r}\Big{)} = \frac{- e (1+\epsilon) \sin (1+\epsilon)\theta}{q}$

$\frac{d^2}{d\theta^2} \Big{(} \frac{1}{r}\Big{)} = \frac{- e (1+\epsilon)^2 \cos (1+\epsilon)\theta}{q} \simeq \frac{-e (1 + 2 \epsilon) \cos (1+\epsilon) \theta}{q},$ car $\epsilon <<1$ .

Donc $\frac{d^2}{d\theta^2} \Big{(} \frac{1}{r}\Big{)} + \frac{1}{r} = \frac{1}{q} - \frac{2 \epsilon e \cos (1+\epsilon)\theta}{q}.$

$K_1 + \frac{K_2}{r^2} = K_1 + \frac{K_2}{q^2} (1 + e \cos (1+\epsilon) \theta)^2 \simeq K_1 + \frac{K_2}{q^2} + \frac{2K_2}{q^2} e \cos (1+\epsilon)\theta$ , car e<<1 .

Donc $K_1 + \frac{K_2}{r^2} \simeq K_1 + \frac{2K_2}{q^2} e \cos (1+\epsilon)\theta$ , car $\frac{K_2}{r^2} \simeq \frac{K_2}{q^2} <<K_1$ .

Donc par identification $K_1 = \frac{1}{q}$ et $\epsilon = \frac{-K_2}{q}$ .

Si $\epsilon \longrightarrow 0$ , $\displaystyle r=\frac{a(1-e^2)}{1 + e \cos \theta}$ (solution du problème keplerien) donc q = a (1 - e^2) .

$\epsilon = \frac{-K_2}{q} = - \frac{3 \alpha R^2 K_1}{a(1- e^2)}= - 3 \alpha \frac{R^2}{a^2 (1 - e^2)^2}$ donc $\epsilon \simeq - 3 \alpha \frac{R^2}{a^2}$ .