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(D) Let $M$ be the total mass enclosed within a sphere of radius $r$.For a particle of mass $m$ moving in a circular orbit of radius $r$,the gravitati

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$\frac{K}{2 \pi r^2 m^2 G}$

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(D) Let $M$ be the total mass enclosed within a sphere of radius $r$.For a particle of mass $m$ moving in a circular orbit of radius $r$,the gravitational force provides the necessary centripetal force:$\frac{GMm}{r^2} = \frac{mv^2}{r}$Since the kinetic energy $K = \frac{1}{2}mv^2$,we have $mv^2 = 2K$. Substituting this into the force equation:$\frac{GMm}{r^2} = \frac{2K}{r} \Rightarrow M = \frac{2Kr}{Gm}$Differentiating both sides with respect to $r$ to find the mass $dM$ in a shell of thickness $dr$:$dM = \frac{2K}{Gm} dr$Also,the mass of the shell is $dM = \rho(r) \cdot 4 \pi r^2 dr$. Equating the two expressions for $dM$:$4 \pi r^2 \rho(r) dr = \frac{2K}{Gm} dr$Solving for density $\rho(r)$:$\rho(r) = \frac{2K}{4 \pi r^2 Gm} = \frac{K}{2 \pi r^2 Gm}$The particle number density $n(r)$ is given by $\rho(r) / m$:$n(r) = \frac{K}{2 \pi r^2 m^2 G}$

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Take $AG = BG = CG = 1 \; m$.

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