MHT CET · Physics · Gravitation
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 ( \(\mathrm{G}=\) Universal gravitational constant)
- A \(\sqrt{\frac{G m}{4 r}}\)
- B \(\sqrt{\frac{G m}{r}}\)
- C \(\sqrt{\frac{G m}{2 r}}\)
- D \(\sqrt{\frac{4 \mathrm{Gm}}{\mathrm{r}}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{G m}{4 r}}\)
Step-by-step Solution
Detailed explanation
\(G \frac{m \cdot m}{(2r)^2} = \frac{mv^2}{r}\) \(G \frac{m^2}{4r^2} = \frac{mv^2}{r}\)
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