Physics Circle home Student Login

Home›Chapters›Gravitation PYQs›Satellites - Orbital Velocity & Period›Question 54

MHT CET 2021 (24 Sep, Shift 2) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from MHT CET 2021 (24 Sep, Shift 2)

Q54MHT CET 2021 · 24 SEP · SHIFT 2Difficult

Satellites - Orbital Velocity & Period

Three point masses, each of mass 'm' are kept at the corners of an equilateral triangle of side 'L'. The system rotates about the centre of the triangle without any change in the separation of masses during rotation. The period of rotation is directly proportional to (cos⁡30∘\left(\cos 30^\circ\right. =32)\left.= \frac{\sqrt{3}}{2}\right)

Show answer and solution

Answer: (c)

  1. Net force on each mass towards centre: F=2Gm2L2cos⁡30∘=3Gm2L2F = 2\frac{Gm^2}{L^2}\cos 30^\circ = \frac{\sqrt{3}Gm^2}{L^2}.
  2. Radius of rotation r=L3r = \frac{L}{\sqrt{3}}.
  3. mω2L3=3Gm2L2⇒ω2=3GmL3m\omega^2 \frac{L}{\sqrt{3}} = \frac{\sqrt{3}Gm^2}{L^2} \Rightarrow \omega^2 = \frac{3Gm}{L^3}.
  4. T=2πω∝L3/2T = \frac{2\pi}{\omega} \propto L^{3/2}.
  5. So the answer is (c) L3/2L^{3/2}.
Report a mistake

More questions on Satellites - Orbital Velocity & Period