- #1
Ackbach
Gold Member
MHB
- 4,155
- 93
Mary decided to drop the ball drastically. She drilled a hole all the way through the earth, and dropped the ball from the Earth's surface. What is the motion of the ball?
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?
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?