Physics Circle home Student Login

Home›Chapters›Gravitation PYQs›Satellites - Orbital Velocity & Period›Question 62

JEE Main 2019 (8 Apr, Shift 1) · Satellites - Orbital Velocity & Period

Satellites - Orbital Velocity & Period question from JEE Main 2019 (8 Apr, Shift 1)

Q62JEE Main 2019 · 8 APR · SHIFT 1Difficult

Satellites - Orbital Velocity & Period

Four identical particles of mass M are located at the corners of a square of side 'a'. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?

Figure

Show answer and solution

Answer: (b)

  1. Circle radius R=a2R = \frac{a}{\sqrt{2}}.
  2. For four masses: v2=GM4R(1+22)v^2 = \frac{GM}{4R}(1+2\sqrt{2}).
  3. v2=2GM4a(3.828)=1.353GMav^2 = \frac{\sqrt{2}GM}{4a}(3.828) = 1.353\frac{GM}{a}.
  4. v=1.16GMav = 1.16\sqrt{\frac{GM}{a}}.
  5. So the answer is (b).
Report a mistake

More questions on Satellites - Orbital Velocity & Period