JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 1)
Two particles of equal mass '$$m$$' move in a circle of radius '$$r$$' under the action of their mutual gravitational attraction. The speed of each particle will be :
$$\sqrt{\frac{G m}{4 r}}$$
$$\sqrt{\frac{G m}{2 r}}$$
$$\sqrt{\frac{G m}{r}}$$
$$\sqrt{\frac{4 G m}{r}}$$
Explanation
$\frac{\mathrm{Gm}^{2}}{4 \mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$
_29th_January_Morning_Shift_en_1_1.png)
$$ v=\sqrt{\frac{G m}{4 r}} $$
$$ v=\sqrt{\frac{G m}{4 r}} $$
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