Cannon Ball Paradox: Math Explained

[keode189]

I understand that if you drop a cannonball from a hole drilled through the Earth from pole-to-pole, the ball would reach a speed of 18,000 mph at the center, and then slow to a speed of zero at the other end of the hole. Can anyone give me a mathematical explanation of this?

thanks

 

[chroot]

It's effectively just simple harmonic motion.

The only mass that affects the cannon ball is the mass INSIDE the sphere defined by its radius vector from the center of the earth. The matter outside its radius cancels. In other words, for a cannon ball that's a distance r from the center, the volume inside its radius vector is

$$V = \frac{4}{3} \pi r^3$$

and given some density (which, for a simple approximation, does not depend on depth), the mass inside is

$$M = \frac{4}{3} \rho \pi r^3$$

The gravitational force on the bowling ball from this mass is

$$F = G \frac{ \frac{4}{3} \rho \pi r^3 m }{r^2} = \frac{4}{3} G m \rho \pi r$$

which is linear in r. The rest is just like any spring problem.

- Warren

 

[Julian Solos]

[Allow me to continue if I may . . .]


The period of this simple harmonic motion is

$$T = 2 \pi \sqrt{\frac{m}{k}}$$

where

$$k = \frac{4}{3} G m \rho \pi \mbox{.}$$


Simplifying, we obtain

$$T = 2 \pi \sqrt{\frac{m}{k}} = 2 \pi \sqrt{\frac{3 m}{4 G m \rho \pi }} = \sqrt{\frac{3 \pi}{G \rho }} \mbox{.}$$


Taking

$$\rho = 5.51 \times 10^3 \ \mbox{kg/m}^3$$

and

$$G = 6.67 \times 10^{-11} \ \mbox{N} \cdot \mbox{m}^2\mbox{/kg}^2$$

we obtain

$$T = 5,050 \ \mbox{s} = 84.2 \ \mbox{min.}}$$


For simple harmonic motion,

the maximum speed is $$\omega A$$

where

$$\omega = \frac {2 \pi}{T}$$

and

$$A = \mbox{the maximum displacement} \mbox{.}}$$


Taking $$A = \mbox{the mean radius of the Earth} = 3,960 \ \mbox{mi}$$

we find the maximum speed which occurs at the center of the Earth

$$\omega A = \frac {2 \pi}{T} A = \frac {2 \pi \ 3,960 \ \mbox{mi}} {5,050 \ \mbox{s}} = 17,737 \ \mbox{mi/h} \mbox{.}}$$

 

[Julian Solos]

The last line of my previous post should have been

$$\omega A = \frac {2 \pi}{T} A = \frac {2 \pi \ 3,960 \ \mbox{mi}} {5,050 \ \mbox{s}} = 17,737 \ \mbox{mi/h} \mbox{.}} \simeq 18,000 \ \mbox{mph}.$$

(Actually, I have a question. Should I use $$\approx$$ before 18,000 mph instead? What's the difference between signs $$\simeq$$ and $$\approx$$?)

 

[Julian Solos]

A set of questions:

How is the atmosphere distributed around the Earth when a hole is dug through it?

Will the air be denser inside the hole?

The hole won't be vacuum, will it?

 

[chroot]

Yes, the air will be denser in the hole, and densest right at the center of the earth.

If you want to calculate the pressure at any point in the hole, just consider the weight of the column of air above that point.

- Warren

 

[Julian Solos]

Originally posted by chroot
Yes, the air will be denser in the hole, and densest right at the center of the earth.

Then, the usual premise "neglecting friction" in this sort of a-tunnel-through-the-earth problems is unrealistic, isn't it?

Considering an object dropped into the tunnel will reach its terminal velocity, the object reaching a speed of 18,000 MPH at the center of the Earth and the motion having a period of 84 minutes are both impossible.

 

[LURCH]

Originally posted by Julian Solos
Then, the usual premise "neglecting friction" in this sort of a-tunnel-through-the-earth problems is unrealistic, isn't it?

Only slightly more unrealistc than it is in all the other hypathetical situations proposed in physics. For these kinds of situations, I find it helpfull to re[;ace the phrase "neglect air resistance" with, perform the experiment in a vacuum chamber". If the hole is dug through center of the Earth, and then all the air is pumped out, the model becomes sensible.