# Correctness of Ball-Dropping Problem Answer

• MHB
• Ackbach
Gold Member
MHB
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?

## Answers and Replies

skeeter
Worked this out long ago and arrived at the same conclusion. I'm used to using $r(t)$ instead of $y(t)$.

$r'' = -\dfrac{GM_e}{R_e^3} \cdot r$

$r'' = -w^2 \cdot r \implies \omega = \sqrt{\dfrac{GM_e}{R_e^3}}$

$r = R_e \cos(\omega t)$

$r' = -R_e \omega \sin(\omega t)$

$r'' = -R_e \omega^2 \cos(\omega t)$

what's interesting is the period of motion ...

$T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{R_e^3}{GM_e}} \implies T^2 = \dfrac{4\pi^2 R_e^3}{GM_e}$ ... Kepler's law of harmony.

The time it takes from ball drop to return from the tunnel is the same as if the ball were orbiting the Earth close to its surface (no air resistance, of course).

Homework Helper
MHB
It means we have a force that is linear with the distance to the center of the earth.
It assumes that the Earth has uniform density, is a perfect ball, and that the ball is free to move without friction.
It's a consequence of the so called Shell Theorem, which Newton first proved.

Such a linear force results in a simple harmonic oscillation (sine).
We can take it a step further and extend it to 3 dimensions with some initial horizontal speed leaving out the movement around the sun.
The beauty of it is that we get an elliptical trajectory, just like a planet around the sun.
The difference is that in this case the center of mass is in the center of the ellipse instead of in one of its focal points.

Gold Member
MHB
Worked this out long ago and arrived at the same conclusion. I'm used to using $r(t)$ instead of $y(t)$.

$r'' = -\dfrac{GM_e}{R_e^3} \cdot r$

$r'' = -w^2 \cdot r \implies \omega = \sqrt{\dfrac{GM_e}{R_e^3}}$

$r = R_e \cos(\omega t)$

$r' = -R_e \omega \sin(\omega t)$

$r'' = -R_e \omega^2 \cos(\omega t)$

what's interesting is the period of motion ...

$T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{R_e^3}{GM_e}} \implies T^2 = \dfrac{4\pi^2 R_e^3}{GM_e}$ ... Kepler's law of harmony.

The time it takes from ball drop to return from the tunnel is the same as if the ball were orbiting the Earth close to its surface (no air resistance, of course).

Nice insight, thanks!

It means we have a force that is linear with the distance to the center of the earth.
It assumes that the Earth has uniform density, is a perfect ball, and that the ball is free to move without friction.
It's a consequence of the so called Shell Theorem, which Newton first proved.

Excellent! Thanks for teasing out those additional assumptions. It's always good to examine your assumptions.

Such a linear force results in a simple harmonic oscillation (sine).
We can take it a step further and extend it to 3 dimensions with some initial horizontal speed leaving out the movement around the sun.
The beauty of it is that we get an elliptical trajectory, just like a planet around the sun.
The difference is that in this case the center of mass is in the center of the ellipse instead of in one of its focal points.

When you say "center of mass" do you mean the reduced center of mass, $\mu?$ Right, we'd get the usual two-body problem if we allow initial horizontal velocity.