本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
Consider the two orbits around the sun sho *g. dtal*3if kbi y9ny9reo; ;w fsv(3j b7h: cvn below. Orbit P is circular with radiu s:h3(cb v vj7fs $R$, orbit Q is elliptical such that the farthest point $b$ is between $2R$ and $3R$, and the nearest point $a$ is between $R/3$ and $R/2$. Consider the magnitudes of the velocity of the circular orbit $v_{c}$, the velocity of the comet in the elliptical orbit at the farthest point $v_{b}$, and the velocity of the comet in the elliptical orbit at the nearest point $v_{a}$. Which of the following rankings is correct? 
A. $v_{b} > v_{c} > 2v_{a}$
B. $2v_{c} > v_{b} > v_{a}$
C. $10v_{b} > v_{a} > v_{c}$
D. $v_{c} > v_{a} > 4v_{b}$
E. $2v_{a} > \sqrt{2}v_{b} > v_{c}$
参考答案: C
本题详细解析:
1. **Compare $v_a$ and $v_c$**: The speed for a circular orbit is $v_{circ}(r) = \sqrt{GM/r}$. The speed at periapsis $v_a$ of an ellipse is *always* greater than the circular orbit speed at that same radius $r_a$. $v_a > v_{circ}(r_a)$. We are given $r_a < R$. Since $r_a < R$, $v_{circ}(r_a) = \sqrt{GM/r_a} > \sqrt{GM/R} = v_c$. Therefore, $v_a > v_{circ}(r_a) > v_c$, which means **$v_a > v_c$**. This eliminates options A, B, D, and E (since $v_b > v_c$ is false).
2. **Analyze Option C**: This option claims $10v_b > v_a > v_c$. We already proved $v_a > v_c$.
3. **Check $10v_b > v_a$**: By conservation of angular momentum: $m v_a r_a = m v_b r_b$, so $v_a / v_b = r_b / r_a$.
We are given $R/3 < r_a < R/2$ and $2R < r_b < 3R$.
Let's check the extremes for the ratio $r_b / r_a$:
* Max ratio: $r_b = 3R$, $r_a = R/3$. Ratio = $3R / (R/3) = 9$. So $v_a = 9v_b$.
* Min ratio: $r_b = 2R$, $r_a = R/2$. Ratio = $2R / (R/2) = 4$. So $v_a = 4v_b$.
In all cases, $v_a$ is between $4v_b$ and $9v_b$. The inequality $10v_b > v_a$ is always true, since $10v_b > 9v_b \ge v_a$.
Since both $10v_b > v_a$ and $v_a > v_c$ are correct, option C is the correct ranking.
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