Question

A planet is in a highly eccentric orbit about a star. The distance of its closest approach is 300 times smaller than its farthest distance from the star. If the corresponding speeds are \( v_c \) and \( v_f \), then \( \frac{v_c}{v_f} \) is:

• A. \( \frac{1}{300} \)
• B. \( \frac{1}{\sqrt{300}} \)
• C. \( \sqrt{300} \)
• D. 300

Hint:

In orbital motion, the product \( v^2 r \) remains approximately constant for highly eccentric orbits.

Answer 1

The Correct Option is D

Step 1: Use conservation of angular momentum.
For an elliptical orbit, angular momentum is conserved: \[ m v_c r_c = m v_f r_f \] \[ \Rightarrow \frac{v_c}{v_f} = \frac{r_f}{r_c}. \]
Step 2: Apply the given condition.
It is given that \( r_f = 300 r_c \). Hence: \[ \frac{v_c}{v_f} = 300. \] However, this would violate conservation of energy if used alone. Considering energy conservation, \[ v_c^2 r_c = v_f^2 r_f \Rightarrow \frac{v_c}{v_f} = \sqrt{\frac{r_f}{r_c}} = \sqrt{300}. \]
Step 3: Final Answer.
The ratio \( \frac{v_c}{v_f} = \sqrt{300}. \)