# Correctness of Ball-Dropping Problem Answer

• MHB
• Ackbach
In summary, Mary decided to drop the ball drastically and drilled a hole all the way through the earth, dropping the ball from the Earth's surface. The tricky part is that the force of gravity is non-constant as you drill through the earth, but this can be calculated using the total mass of the earth, the position of the ball, and other relevant factors. This leads to a regular harmonic oscillator with a solution of $y(t)=R_e\cos\left(\sqrt{\frac{GM_e}{R_e^3}}\,t\right)$. This analysis assumes uniform density of the earth, a perfect spherical shape, and no friction or other external forces. It is a consequence of the Shell Theorem, first proved
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?

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).

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.

skeeter said:
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!

Klaas van Aarsen said:
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.

Klaas van Aarsen said:
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.

## 1. What is the "Ball-Dropping Problem" and why is its correctness important?

The Ball-Dropping Problem is a thought experiment used to demonstrate the concept of probability. It involves dropping a ball into a grid of pegs, each with an equal chance of the ball landing on either side. The correctness of this problem is important because it helps us understand the concept of probability and its application in real-world scenarios.

## 2. How do we know if the answer to the Ball-Dropping Problem is correct?

The answer to the Ball-Dropping Problem is considered correct if it follows the principles of probability, specifically the law of large numbers. This means that the more times the ball is dropped, the closer the results will be to the expected probability.

## 3. Is there a specific formula or method for solving the Ball-Dropping Problem?

There is no specific formula or method for solving the Ball-Dropping Problem, as it is a thought experiment and not a mathematical problem. However, the concept of probability and the law of large numbers can be applied to determine the expected outcome.

## 4. Can the Ball-Dropping Problem be applied to real-world scenarios?

Yes, the Ball-Dropping Problem can be applied to real-world scenarios where there are multiple possible outcomes with equal chances of occurring. For example, it can be used to understand the probability of winning a game of chance or the likelihood of a certain event happening.

## 5. Are there any limitations to the Ball-Dropping Problem and its answer?

The Ball-Dropping Problem is a simplified thought experiment and does not take into account external factors that may affect the outcome. Additionally, the answer to the problem is based on theoretical probabilities and may not always reflect the actual results in a real-world scenario.

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