2012: Problem 25

Topic: Gravity
Concepts: Circular orbits, Conservation of angular momentum, Elliptical orbits


Solution:

Conserving angular momentum between the apogee and perigee of the ellipse, we have

mvbr1=mvar2mv_br_1=mv_ar_2
vavb=r1r2\frac{v_a}{v_b}=\frac{r_1}{r_2}

Since 2Rr13R2R \leq r_1 \leq 3R and R/3r2R/2R/3 \leq r_2 \leq R/2, we can bound

4=2RR/2vavb3RR/3=94=\frac{2R}{R/2} \leq \frac{v_a}{v_b} \leq \frac{3R}{R/3}=9
4vbva9vb(1)4v_b \leq v_a \leq 9v_b \qquad (1)

Recall the velocity of a circular orbit of radius rr is given by

vcirc=GMrv_{\text{circ}}=\sqrt{\frac{GM}{r}}

We can relate vbv_b to vc=GM/Rv_c=\sqrt{GM/R} by considering a circular orbit of radius r1r_1. Then

vb<GMr1GM2R=vc2v_b < \sqrt{\frac{GM}{r_1}} \leq \sqrt{\frac{GM}{2R}}=\frac{v_c}{\sqrt{2}}

since the ellipse is contained in the circle and r12Rr_1 \geq 2R. We can also relate vav_a to vcv_c by considering a circular orbit of radius r2r_2. Then

va>GMr2GMR/2=2vcv_a>\sqrt{\frac{GM}{r_2}} \geq \sqrt{\frac{GM}{R/2}}=\sqrt{2}v_c

since the circle is contained in the ellipse and r2R/2r_2 \leq R/2. Combining these two equations,

vb<vc2<vc<2vc<va(2)v_b<\frac{v_c}{\sqrt{2}}<v_c<\sqrt{2}v_c<v_a \qquad (2)

We now go through the possible choices:

  • A) vb>vc>2vav_b>v_c>2v_a
    Not correct because vb<vcv_b<v_c from equation 2.
  • B) 2vc>vb>va2v_c>v_b>v_a
    Not correct because vb<vav_b<v_a from equation 2.
  • C) 10vb>va>vc10v_b>v_a>v_c
    Correct because va9vb<10vbv_a \leq 9v_b < 10v_b from equation 1 and vc<vav_c<v_a from equation 2.
  • D) vc>va>4vbv_c>v_a>4v_b
    Not correct because vc<vav_c<v_a from equation 2.
  • E) 2va>2vb>vc2v_a>\sqrt{2}v_b>v_c
    Not correct because 2vb<vc\sqrt{2}v_b<v_c from equation 2.

Thus, the answer is C.

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