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QJEEM23Ph2a55: JEE Main 2023 Physics - Class 11 JEEM23Ph2a55 Solution: Two particles of equal mass ' ' move in a circle of radius ' '…

Two particles of equal mass ' mm ' move in a circle of radius ' rr ' under the action of their mutual gravitational attraction. The speed of each particle will be :

Multiple Choice Options

  • A Gm4r\sqrt{\frac{G m}{4 r}}
  • B Gm2r\sqrt{\frac{G m}{2 r}}
  • C Gmr\sqrt{\frac{G m}{r}}
  • D 4Gmr\sqrt{\frac{4 G m}{r}}

Step-by-Step Verified Solution

Gm24r2=mv2r\frac{\mathrm{Gm}^{2}}{4 \mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}

JEE Main 2023 (Online) 29th January Morning Shift Physics - Gravitation Question 62 English Explanation
v=Gm4rv=\sqrt{\frac{G m}{4 r}}

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