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(B) From the conservation of energy,the total energy at the perihelion and aphelion is equal:$\frac{1}{2}mv_1^2 - \frac{GM_Sm}{r_1} = \frac{1}{2}mv_2^

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$\sqrt{\frac{2GM_Sm^2r_1r_2}{r_1 + r_2}}$

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(B) From the conservation of energy,the total energy at the perihelion and aphelion is equal:$\frac{1}{2}mv_1^2 - \frac{GM_Sm}{r_1} = \frac{1}{2}mv_2^2 - \frac{GM_Sm}{r_2}$Angular momentum is conserved,so $L = mv_1r_1 = mv_2r_2$,which implies $v_2 = v_1\frac{r_1}{r_2}$.Substituting $v_2$ into the energy equation:$\frac{1}{2}mv_1^2 - \frac{GM_Sm}{r_1} = \frac{1}{2}m\left(v_1\frac{r_1}{r_2}\right)^2 - \frac{GM_Sm}{r_2}$Solving for $v_1$:$v_1^2 \left(1 - \frac{r_1^2}{r_2^2}\right) = 2GM_S \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$v_1^2 \left(\frac{r_2^2 - r_1^2}{r_2^2}\right) = 2GM_S \left(\frac{r_2 - r_1}{r_1r_2}\right)$$v_1^2 \left(\frac{(r_2 - r_1)(r_2 + r_1)}{r_2^2}\right) = 2GM_S \left(\frac{r_2 - r_1}{r_1r_2}\right)$$v_1 = \sqrt{\frac{2GM_Sr_2}{r_1(r_1 + r_2)}}$Finally,the angular momentum $L = mv_1r_1 = m \sqrt{\frac{2GM_Sr_2}{r_1(r_1 + r_2)}} \cdot r_1 = \sqrt{\frac{2GM_Sm^2r_1r_2}{r_1 + r_2}}$.

$A$ planet of mass $m$ moves in an ellipse around the sun of mass $M_S$ such that its maximum and minimum distances are $r_1$ and $r_2$ respectively. The angular momentum of the planet relative to the center of the sun is

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