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MCQMediumJEE 2023Satellites & Orbital Velocity

JEE Physics 2023 Question with Solution

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:

  • A

    GM2r\sqrt{\frac{GM}{2r}}

  • B

    4GMr\sqrt{\frac{4GM}{r}}

  • C

    GMr\sqrt{\frac{GM}{r}}

  • D

    GM4r\sqrt{\frac{GM}{4r}}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two particles of equal mass mm move in circular paths due to their mutual gravitational attraction, each with orbital radius rr.

Find: The speed of each particle.

The solution uses gravitational force as the centripetal force.

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

Solving for vv,

v2=Gm4rv^2 = \frac{Gm}{4r}v=Gm4rv = \sqrt{\frac{Gm}{4r}}

Substituting mm as the planet's mass,

v=GM4rv = \sqrt{\frac{GM}{4r}}

Therefore, the speed of each particle is GM4r\sqrt{\frac{GM}{4r}}. The solution states that the correct option is B, although this value matches option D in the listed options.

Common mistakes

  • Using the separation between the two particles as rr instead of 2r2r. This makes the gravitational force incorrect. The force must be calculated using distance 2r2r, so the denominator becomes 4r24r^2.

  • Equating gravitational force to mv22r\frac{mv^2}{2r} for one particle. This is wrong because each particle moves in a circle of radius rr, so the required centripetal force is mv2r\frac{mv^2}{r}.

  • Trusting the answer key key without checking the derived expression against the options. Here the worked result is GM4r\sqrt{\frac{GM}{4r}}, which matches option D, not the solution label B.

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