Question

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 the closest and farthest points is:

• A. $1:2$
• B. $1:3$
• C. $1:6$
• D. $3:4$

Answer 1

The correct option is C

$1:6$

Step 1. Given data and the Diagram:

Let the velocity at the farthest point be $v_{far}$ and the velocity at the nearest point be $v_{near}$, then

The ratio of velocities for the satellite is $\frac{v_{far}}{v_{near}}=\frac{1}{6}$

Step 2. Find the ratio for the distances:

If $m$ is the mass of the planet and $r_1$and $r_2$are the radius of orbits at the nearest and the farthest ends respectively, then

By the conservation of angular momentum we have,

${\left(mvr\right)}_{initial}={\left(mvr\right)}_{final}$

$\Rightarrow$ $m{v}_{far}{r}_2=m{v}_{near}{r}_1$

$\Rightarrow$ $\frac{r_1}{r_2}=\frac{v_{far}}{v_{near}}$

$\Rightarrow$ $\frac{r_1}{r_2}=\frac{1}{6}$

Hence, option (C) is correct.