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(B) The gravitational force on a particle of mass $m$ at a distance $r$ from the center inside the core of density $\rho_c = 3\rho$ is given by:$F = -

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$\frac{1}{2}\sqrt{\frac{\pi}{\rho G}}$

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(B) The gravitational force on a particle of mass $m$ at a distance $r$ from the center inside the core of density $\rho_c = 3\rho$ is given by:$F = -\frac{G M_{enc} m}{r^2} = -\frac{G (\frac{4}{3} \pi r^3 \cdot 3\rho) m}{r^2}$$F = -4\pi G \rho m r$Since $F = ma$,the acceleration is:$a = -4\pi G \rho r$This is the equation of simple harmonic motion $a = -\omega^2 r$,where $\omega^2 = 4\pi G \rho$.Thus,the angular frequency is $\omega = \sqrt{4\pi G \rho}$.The time period of a full oscillation is $T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{4\pi G \rho}} = \sqrt{\frac{\pi}{G \rho}}$.The particle travels from $A$ to $B$ in half the time period:$t = \frac{T}{2} = \frac{1}{2}\sqrt{\frac{\pi}{G \rho}}$.

$A$ planet has a core of density $3\rho$ and an outer crust of density $\rho$. There is a small tunnel through the core. $A$ small particle of mass $m$ is released from end $A$. What is the time required to reach end $B$?

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