Spherical gas cloud

JEE Advanced 2019 Paper 1, Question 1

Consider a spherical gas cloud of mass density \rho(r) in free space where r is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If \rho(r) is constant in time, the particle number density n(r) = \rho(r)/m is [G is the universal gravitational constant]

  1. \frac{K}{2 \pi r^2 m^2 G}
  2. \frac{K}{\pi r^2 m^2 G}
  3. \frac{3 K}{\pi r^2 m^2 G}
  4. \frac{K}{6 \pi r^2 m^2 G}

Solution

Consider a small parcel of gas in the cloud with mass \Delta m at a distance r from the center of the cloud (see figure). …

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