Q81 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
Two identical particles each of mass ' $m$ ' go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be:
- A$\sqrt{\frac{G m}{2 a^3}}$
- B$\sqrt{\frac{\mathrm{Gm}}{8 \mathrm{a}^3}}$
- Correct answerC$\sqrt{\frac{G m}{4 a^3}}$
- D$\sqrt{\frac{G m}{a^3}}$
Solution
Radius of circle, $\mathrm{r}=\mathrm{a}$
Centrifugal force, $F=m \omega^2 r$
$\begin{aligned}<br />& \Rightarrow \frac{\mathrm{Gmm}}{(2 \mathrm{a})^2}=\mathrm{m} \omega^2 \mathrm{a} \\<br />& \left(\because \mathrm{F}=\frac{\mathrm{G} \mathrm{M} \mathrm{M}_2}{\mathrm{~d}}\right)<br />\end{aligned}$
Here $M_1=m ; M_2=m$ and $d=2 a$
$\Rightarrow$ angular speed, $\omega=\sqrt{\frac{G m}{4 a^3}}$
Centrifugal force, $F=m \omega^2 r$
$\begin{aligned}<br />& \Rightarrow \frac{\mathrm{Gmm}}{(2 \mathrm{a})^2}=\mathrm{m} \omega^2 \mathrm{a} \\<br />& \left(\because \mathrm{F}=\frac{\mathrm{G} \mathrm{M} \mathrm{M}_2}{\mathrm{~d}}\right)<br />\end{aligned}$
Here $M_1=m ; M_2=m$ and $d=2 a$
$\Rightarrow$ angular speed, $\omega=\sqrt{\frac{G m}{4 a^3}}$
Hence, the correct answer is option: 'C'.