Particle falling inside a sphere due to gravity

[Robin04]

Homework Statement

We drill a hole from the surface of the Earth to the other side through the center, drop a test particle and measure the time it takes to come back. How does this time depend on the radius and density of the Earth?

Homework Equations

The Attempt at a Solution

The gravitation field inside a sphere is $g=\frac{4\pi G\rho}{3}r$, where $G$ is the gravitational constant, $\rho$ is the density, $r$ is the distance from the center.
Now I have to find the function $r(t)$. I already found $v(r)=r\sqrt{4\pi G \rho}$ (the initial velocity is zero). How can I bring time into the equation?

 

[Arman777]

Homework Statement

We drill a hole from the surface of the Earth to the other side through the center, drop a test particle and measure the time it takes to come back. How does this time depend on the radius and density of the Earth?

Homework Equations

The Attempt at a Solution

The gravitation field inside a sphere is $g=\frac{4\pi G\rho}{3}r$, where $G$ is the gravitational constant, $\rho$ is the density, $r$ is the distance from the center.
Now I have to find the function $r(t)$. I already found $v(r)=r\sqrt{4\pi G \rho}$ (the initial velocity is zero). How can I bring time into the equation?

Since the motion is oscillatory motion, you can think a form of Second order DE, such that

$$d^2r/dt^2+ω^2r=0$$ where $ω$ will be the anguler frequency or $w=2π/T$

 

[Robin04]

Since the motion is oscillatory motion, you can think a form of Second order DE, such that

$$d^2r/dt^2+ω^2r=0$$ where $ω$ will be the anguler frequency or $w=2π/T$

So I could say $g+\omega^2r=0$, from which $T=\sqrt{\frac{3\pi}{G\rho}}$ Is this correct? It doesn't depend on the radius of the sphere.

 

[Arman777]

So I could say $g+\omega^2r=0$, from which $T=\sqrt{\frac{3\pi}{G\rho}}$ Is this correct? It doesn't depend on the radius of the sphere.

It's true yes. I don't think we can get the density and the radius in the same equation. For the radius dependence put the $ρ$ value in the equation that you find and you ll see that it depends also on radius in some other aspect.

 

[Robin04]

It's true yes. I don't think we can get the density and the radius in the same equation. For the radius dependence put the $ρ$ value in the equation that you find and you ll see that it depends also on radius in some other aspect.

Thank you! :)

 

[gneill]

A bit of trivia: the period for your particle is the same as for a (theoretical) satellite orbiting the Earth at its surface.