Question

Two satellites $S_1$ and $S_2$ revolve around a planet in coplanar circular orbits in the opposite sense. The periods of revolutions are $T$ and $nT$ respectively. Find the angular speed of $S_2$ as observed by an astronaut in $S_1$ , when they are closest to each other.

• A. $\frac{2\pi (n^{\frac{1}{3}}+1)}{T(n^{\frac{1}{3}}-1)}$
  
• B. $\frac{2\pi (n^{-\frac{1}{3}}+1)}{T(n^{\frac{2}{3}}-1)}$
  
• C. $\frac{\pi (n^{\frac{1}{3}}+1)}{T(n^{\frac{2}{3}}-1)}$
• D. $\frac{2\pi (n^{\frac{2}{3}}+1)}{T(n^{\frac{1}{3}}-1)}$

Answer 1

The correct option is B $\frac{2\pi (n^{-\frac{1}{3}}+1)}{T(n^{\frac{2}{3}}-1)}$


According to Kepler’s law of periods, the time period of a planet/satellite and the radius of the circular orbit is related as $$T^2\propto r^3$$

$T\propto r^{3/2}$ or $r\propto T^{2/3}$

$\Rightarrow \frac{r_1}{r_2}={\left(\frac{T_1}{T_2}\right)}^{2/3}$

Substituting the values,

$\Rightarrow \frac{r_1}{r_2}={\left(\frac{T}{nT}\right)}^{2/3}=\frac{1}{n^{2/3}}$

The angular velocity of $S_2$ wrt $S_1$ is given by

${\omega }_{21}=\frac{v_{21}}{l}$,

where, $v_{21}$ is the relative velocity of $S_2$ wrt $S_1$ and $l$ is the distance between them.

Since the satellites are moving in the opposite direction, $v_{21}=v_1+v_2$

And $l=r_2-r_1$

Then,
${\omega }_{21}=\frac{v_2+v_1}{r_2-r_1}$

$\Rightarrow {\omega }_{21}=\frac{\frac{2\pi r_2}{T_2}+\frac{2\pi r_1}{T_1}}{r_2-r_1}=\frac{2\pi (\frac{1}{T_2}+\frac{1}{T_1}.\frac{r_1}{r_2})}{1-\frac{r_1}{r_2}}$

$\Rightarrow {\omega }_{21}=\frac{2\pi (\frac{1}{nT}+\frac{1}{T}.\frac{1}{n^{2/3}})}{1-\frac{1}{n^{2/3}}}=\frac{\frac{2\pi }{T}(n^{-1/3}+1)}{(n^{2/3}-1)}$

Hence, option (b) is correct answer.