- #1

Ackbach

Gold Member

MHB

- 4,155

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So the tricky part here is that the force of gravity is non-constant as you drill through the earth. It is near-zero at the center, and climbs back up to $g$ at the surface.

Definitions: let

\begin{align*}

M_e&=\text{total mass of the earth}\\

y&=\text{position of the ball; }y=0\text{ at the center of the earth, and positive where Mary is}\\

M_y&=\text{mass of the Earth enclosed by a sphere of radius $y$ centered at the Earth's center} \\

m&=\text{mass of the ball} \\

F&=\text{force on the ball exerted by Earth's gravity} \\

R_e&=\text{radius of the earth, assumed spherical}\\

G&=\text{gravitational constant}\\

\rho&=\text{mass density of the earth, defined as } \frac{3M_e}{4\pi R_e^3}\\

t&=\text{time, with the clock starting at the drop: } t=0.

\end{align*}

Now, my big assumption here is that

$$F=-\frac{GM_y\, m}{y^2}. $$

I believe I have seen elsewhere that the sort of "annulus" of mass outside the sphere of radius $y$ centered at the origin cancels out. So we need to calculate $M_y$ in terms of $y$. I do the following:

\begin{align*}

M_y&=\rho\,\frac{4\pi y^3}{3} \\

&=\frac{3M_e}{4\pi R_e^3}\,\frac{4\pi y^3}{3}\\

&=\frac{M_e y^3}{R_e^3}.

\end{align*}

It follows that

$$F=-\frac{Gm}{y^2}\cdot \frac{M_e y^3}{R_e^3}=-\frac{GM_e\,my}{R_e^3}. $$

Now we just do Newton's Second Law and solve the resulting DE:

\begin{align*}

-\frac{GM_e\,m}{R_e^3}\,y&=m\ddot{y}\\

-\frac{GM_e}{R_e^3}\,y&=\ddot{y}.

\end{align*}

This is the regular harmonic oscillator, with solution

\begin{align*}

y&=A \sin\left(\sqrt{\frac{GM_e}{R_e^3}}\,t\right)+B\cos\left(\sqrt{\frac{GM_e}{R_e^3}}\,t\right)\\

\dot{y}&=A\sqrt{\frac{GM_e}{R_e^3}} \cos\left(\sqrt{\frac{GM_e}{R_e^3}}\,t\right)-B\sqrt{\frac{GM_e}{R_e^3}}\sin\left(\sqrt{\frac{GM_e}{R_e^3}}\,t\right).

\end{align*}

We use the initial conditions $y(0)=R_e$ and $\dot{y}(0)=0$ to obtain $A=0$ and $B=R_e,$ for a final solution of

$$y(t)=R_e\cos\left(\sqrt{\frac{GM_e}{R_e^3}}\,t\right). $$

Is this analysis correct? Obviously, I'm ignoring some factors such as Earth's rotation, travel about the sun, and the fact that the Earth is an oblate spheroid, not a sphere. But ignoring all that, is this correct?