Consider a spherical gaseous cloud of mass density ρ 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 ab

Question

Consider a spherical gaseous cloud of mass density ρ 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 ρ r is constant in time, the particle number density n r = ρ r / m is: [ G is universal gravitational constant]

TestHub · Accepted Answer

Given that all the particles of gaseous cloud has mass $‘ m ’$ and are moving due to mutual attraction with same kinetic energy $‘ K ’$ .
Now in that cloud, let us consider a sphere of radius $‘ r ’$ containing a group of particles adding to a total mass $‘ M ’$ .
Now, another particle of mass $‘ m ’$ is moving with speed $‘ V ’$ due to gravitational attraction of $M,$ in a circle of radius $‘ r ’$ .
$∴$ Centripetal force is here given by $‘ M ’$
$\Rightarrow \frac{G M \cdot m}{r^{2}} = \frac{m v^{2}}{r}$ ....(1)
But, kinetic energy, $K = \frac{1}{2} m v^{2}$
$\Rightarrow m v^{2} = 2 K$
Put it in $(1)$
$\Rightarrow \frac{G M m}{r^{2}} = \frac{2 K}{r}$
$\Rightarrow M = \frac{2 K r}{G m}$
Differentiating it we get,
$d M = \frac{2 K}{G m} \cdot d r$ .....(2)
Now, if density of cloud is $' ρ^{'},$
Then elementary mass,
$d M = (4 π r^{2} d r) \cdot ρ = \frac{2 K}{G m} \cdot d r$
$\Rightarrow ρ = \frac{K}{2 π G m r^{2}}$

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