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MHT CET 2025 (19 Apr, Shift 1) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from MHT CET 2025 (19 Apr, Shift 1)

Q14MHT CET 2025 · 19 APR · SHIFT 1Medium

Satellites - Orbital Velocity & Period

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 ( G=G = Universal gravitational constant)

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Answer: (a)

mmFd = 2r
  1. The two particles are at opposite ends of a diameter, so their separation is 2r2r.
  2. Mutual attraction gives the centripetal force: Gm2(2r)2=mv2r\frac{Gm^2}{(2r)^2} = \frac{mv^2}{r}
  3. v2=Gm4rv^2 = \frac{Gm}{4r}
  4. So the answer is (a) Gm4r\sqrt{\frac{Gm}{4r}}.
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