A satellite is in an elliptical orbit around a planet $$P$$. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is:

For a satellite moving under the central gravitational force of a planet, there is no external torque about the planet. Therefore, the angular momentum of the satellite about the planet remains conserved throughout the orbit. The statement of conservation of angular momentum is:

$$m\,r\,v = \text{constant}$$

Here $$m$$ is the mass of the satellite, $$r$$ is the instantaneous distance from the planet, and $$v$$ is the corresponding linear speed. We compare two special points of the elliptical orbit:

• Closest point (periapsis): distance $$r_c$$ and speed $$v_c$$.
• Farthest point (apoapsis): distance $$r_f$$ and speed $$v_f$$.

Applying the conservation law between these two points, we have

$$m\,r_c\,v_c \;=\; m\,r_f\,v_f.$$

Because the satellite’s mass $$m$$ is common to both terms, it cancels:

$$r_c\,v_c \;=\; r_f\,v_f.$$

The problem states that when the satellite is farthest from the planet, its speed is six times smaller than when it is closest. Expressing this in symbols,

$$v_f \;=\;\dfrac{v_c}{6}.$$

Substituting this value of $$v_f$$ into the conserved‐angular‐momentum equation gives

$$r_c\,v_c \;=\; r_f\left(\dfrac{v_c}{6}\right).$$

Now the speed $$v_c$$ appears on both sides, so it divides out:

$$r_c \;=\;\dfrac{r_f}{6}.$$

Rearranging, the ratio of distances becomes

$$\dfrac{r_c}{r_f} \;=\;\dfrac{1}{6}.$$

Thus, the distance of the satellite from the planet at the closest point is one‐sixth of its distance at the farthest point. Writing this as a ratio closest : farthest, we get

$$r_c : r_f \;=\; 1 : 6.$$

Hence, the correct answer is Option A.