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Step-by-Step Solution
Step 1: Understand the Problem
A satellite orbits a planet along an elliptical path. The velocity of the satellite at its farthest point (aphelion) is given to be one-sixth the velocity at its closest point (perihelion). We need to find the ratio of the distances of the satellite from the planet at these two points.
Step 2: Apply Conservation of Angular Momentum
For an elliptical orbit, the angular momentum of the satellite about the planet remains constant. Let:
$ r_{\text{closest}} $ = distance of satellite from the planet at the closest point
$ r_{\text{farthest}} $ = distance of satellite from the planet at the farthest point
$ v_{\text{closest}} $ = corresponding orbital speed at the closest point
$ v_{\text{farthest}} $ = corresponding orbital speed at the farthest point
By the law of conservation of angular momentum:
$ m \, r_{\text{closest}} \, v_{\text{closest}} \;=\; m \, r_{\text{farthest}} \, v_{\text{farthest}} $
(m is the mass of the satellite, which cancels out on both sides.)
Step 3: Relate Velocities and Distances
Given:
$ v_{\text{farthest}} = \frac{v_{\text{closest}}}{6} \quad \Longrightarrow \quad v_{\text{farthest}} : v_{\text{closest}} = 1 : 6. $
Substitute this into the angular momentum equation:
$ r_{\text{closest}} \, v_{\text{closest}} = r_{\text{farthest}} \, \bigl(\tfrac{v_{\text{closest}}}{6}\bigr). $
Simplify:
$ r_{\text{closest}} \, v_{\text{closest}} = \tfrac{r_{\text{farthest}} \, v_{\text{closest}}}{6} $
Cancelling $v_{\text{closest}}$ on both sides:
$ r_{\text{closest}} = \tfrac{r_{\text{farthest}}}{6}. $
Rearranging,
$ \frac{r_{\text{closest}}}{r_{\text{farthest}}} = \frac{1}{6}. $
Step 4: Conclude the Distance Ratio
Thus, the ratio of the distance of the satellite from the planet (closest : farthest) is:
$ 1 : 6. $
Which matches the correct answer.
Reference Diagram
