Consider a spherical gaseous cloud of mass density in free space where is the radial distance from its center. [2019]

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 centre 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] [2019]

Smart Achievers · Accepted Answer

(2)

Gravitational pull of the mass $' M'$ present in the sphere of radius $' r'$ provides the required centripetal force of particle of mass $' m'$ to revolve in a circular path.

$\frac{m v^{2}}{r} = \frac{G M m}{r^{2}} \Rightarrow \frac{1}{2} m v^{2} = \frac{G M m}{2 r} \Rightarrow K = \frac{G M m}{2 r}$

$∴$ $M = \frac{2 K r}{G m}$

$Differentiating the above equation w.r.t. ' r' we get$

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

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

$∴$ $4 π r^{2} ρ d r = \frac{2 K}{G m} d r \Rightarrow ρ = \frac{K}{2 π r^{2} m G}$

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

Solution

1 $\frac{3 K}{π r^{2} m^{2} G}$

2 $\frac{K}{2 π r^{2} m^{2} G}$

3 $\frac{K}{π r^{2} m^{2} G}$

4 $\frac{K}{6 π r^{2} m^{2} G}$

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Solution

1 3Kπr2m2G

2 K2πr2m2G

3 Kπr2m2G

4 K6πr2m2G

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1 $\frac{3 K}{π r^{2} m^{2} G}$

2 $\frac{K}{2 π r^{2} m^{2} G}$

3 $\frac{K}{π r^{2} m^{2} G}$

4 $\frac{K}{6 π r^{2} m^{2} G}$