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…

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]

Aicharya · 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}}

Consider a spherical gaseous cloud of mass density ρ r in free space where r is the radial distance from its…

Solution