Question

Two satellites ${S_1}$ and ${S_2}$ are revolving around a planet in the opposite sense in coplanar circular concentric orbits. At time t = 0, the satellites are farthest apart. The period of revolution of ${S_1}$ and ${S_2}$ are 3h and 24h respectively. The radius of the orbit of ${S_1}$ is $3 \times {10^4}km$. Then the orbital speed of ${S_2}$ as observed from
$\eqalign{
  & {\text{A}}{\text{. The planet is }}4\pi \times {10^4}km{h^{ - 1}}{\text{ when }}{S_2}{\text{ is closest from }}{S_1} \cr
  & {\text{B}}{\text{. The planet is }}2\pi \times {10^4}km{h^{ - 1}}{\text{ when }}{S_2}{\text{ is farthest from }}{S_1} \cr
  & {\text{C}}{\text{. }}{S_1}{\text{ is }}\pi \times {\text{1}}{{\text{0}}^4}km{h^{ - 1}},{\text{ when }}{S_2}{\text{ is closest from }}{S_1} \cr
  & {\text{D}}{\text{. }}{S_1}{\text{ is 3}}\pi \times {\text{1}}{{\text{0}}^4}km{h^{ - 1}},{\text{ when }}{S_2}{\text{ is closest to }}{S_1} \cr} $

Answer 1

Hint: According to Keplers’s third law, the square of the time period of revolution of a planet is directly proportional to the cube of its radius of revolution. Here, we have the time periods of revolution and the radius of one planet. Thus, we can find the radius of the other planet. By find the radius, we can find the velocity of the planets. This will help us to find the relative velocity of ${S_2}$ with respect to ${S_1}$.

Formula used:
If the time period of revolution of a planet is ‘T’ and its radius of curvature is ‘R’, then,
${T^2} \propto {R^3}$
If the velocity of${S_1}$ is ${v_1}$and velocity of${S_2}$ is ${v_2}$, then the relative velocity of ${S_2}$ with respect to ${S_1}$ is,
$v = {v_2} + {v_1}$
If the radius of a planet is ‘R’ and the time of its revolution is ‘T’, then its velocity,
$v = \dfrac{{2\pi R}}{T}$

Complete step by step answer:
According to Keplers’s third law, the square of the time period of revolution of a planet is directly proportional to the cube of its radius of revolution. If the time period of revolution of a planet is ‘T’ and its radius of curvature is ‘R’, then,
${T^2} \propto {R^3}$
Let ${T_1}$ and ${R_1}$ be the time period and radius of curvature of ${S_1}$ and ${T_2}$ and ${R_2}$ be the time period and radius of curvature of ${S_2}$. So we have:
$\dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{R_1^3}}{{R_2^3}} \cdots \cdots \cdots \cdots \cdots \left( 1 \right)$
Given:
The time period of revolution of ${S_1}$, ${T_1} = {\text{ }}3h$
The time period of revolution of ${S_2}$, ${T_2} = {\text{ 24}}h$
The radius of the orbit of ${S_1}$, ${R_1} = 3 \times {10^4}km$
Substituting the given values in equation (1), we get:
$\eqalign{
  & \dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{R_1^3}}{{R_2^3}} \cr
  & \Rightarrow {\left( {\dfrac{{3h}}{{24h}}} \right)^2} = \dfrac{{{{\left( {3 \times {{10}^4}} \right)}^3}}}{{R_2^3}} \cr
  & \Rightarrow {\left( {\dfrac{1}{8}} \right)^2} = {\left( {\dfrac{{3 \times {{10}^4}}}{{{R_2}}}} \right)^3} \cr
  & \Rightarrow {\left( {\dfrac{1}{8}} \right)^{\dfrac{2}{3}}} = \left( {\dfrac{{3 \times {{10}^4}}}{{{R_2}}}} \right) \cr
  & \Rightarrow {\left( {\dfrac{1}{{64}}} \right)^{\dfrac{1}{3}}} = \left( {\dfrac{{3 \times {{10}^4}}}{{{R_2}}}} \right) \cr
  & \Rightarrow \dfrac{1}{4} = \left( {\dfrac{{3 \times {{10}^4}}}{{{R_2}}}} \right) \cr
  & \Rightarrow {R_2} = 3 \times {10^4} \times 4 \cr
  & \therefore {R_2} = 12 \times {10^4}km \cr} $
Let ${v_1}$ be the velocity of planet ${S_1}$ and ${v_2}$ be the velocity of planet ${S_2}$, so substituting values we get:
$\eqalign{
  & {v_1} = \dfrac{{2\pi {R_1}}}{{{T_1}}} \cr
  & {v_1} = \dfrac{{2\pi \times 3 \times {{10}^4}km}}{3} = 2\pi \times {10^4}km{h^{ - 1}} \cr
  & {v_2} = \dfrac{{2\pi {R_2}}}{{{T_2}}} \cr
  & {v_2} = \dfrac{{2\pi \times 12 \times {{10}^4}km}}{{24}} = \pi \times {10^4}km{h^{ - 1}} \cr} $
Now, the relative velocity of ${S_2}$ with respect to ${S_1}$ is,
$\eqalign{
  & v = {v_2} + {v_1} \cr
  & v = \left( {\pi \times {{10}^4} + 2\pi \times {{10}^4}} \right)km{h^{ - 1}} \cr
  & \therefore v = 3\pi \times {10^4}km{h^{ - 1}} \cr} $
Therefore, the correct option D, i.e., the orbital speed of ${S_2}$ as observed from ${S_1}$ is $3\pi \times {10^4}km{h^{ - 1}}$, when ${S_2}$ is closest to ${S_1}$.

Note:
The planetary motion in the universe is such that the planetary bodies are moving in opposite directions to each other. It is important to keep that in mind for the calculation of relative velocity because if we have taken it otherwise, we will subtract the velocities instead of adding them. This would result in an incorrect answer.