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]

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] (A) (K/π r2m2G) (B) (K/6π r2m2G) (C) (3K/π r2m2G) (D) (K/2π r2m2G)

Tardigrade Admin · Accepted Answer

For a particle rotating in the circular orbit of radius r due to the gravitational attraction of inner cloud of mass M,

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}

∴ M = \frac{v ^{2} r}{G} = \frac{2 m v ^{2} r}{2 G m}

As

K = \frac{1}{2} m v^{2} =

constant, then

M = \frac{2 Kr}{G m}

or

d M = \frac{2 K d r}{G m}

Correspondingly

d M = ρ (r) \times 4 π^{2} d r

∴ ρ (r) \cdot 4 π r^{2} d r = \frac{2 K d r}{G m}

∴ \frac{ρ ( r )}{m} = \frac{K}{2 π G m ^{2} r ^{2}}

$\frac{K}{π r ^{2} m ^{2} G}$

$\frac{K}{6 π r ^{2} m ^{2} G}$

$\frac{3 K}{π r ^{2} m ^{2} G}$

$\frac{K}{2 π r ^{2} m ^{2} G}$

πr2m2GK​

6πr2m2GK​

πr2m2G3K​

2πr2m2GK​

$\frac{K}{π r ^{2} m ^{2} G}$

$\frac{K}{6 π r ^{2} m ^{2} G}$

$\frac{3 K}{π r ^{2} m ^{2} G}$

$\frac{K}{2 π r ^{2} m ^{2} G}$