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Since the earth is spinning, why does a cannon ball shot straight up land exactly where it was shot from?

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closed as off-topic by AccidentalFourierTransform, ZeroTheHero, Kyle Kanos, John Duffield, Jon Custer Apr 8 '18 at 17:40

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It doesn't; a cannonball shot straight up will be deflected due to the Coriolis force. The effect is small in most everyday-life scenarios, though.

To make a rough calculation, let's imagine dropping something off a skyscraper at the equator. Imagine the skyscraper is 200m tall. That means the top of the skyscraper is going a little faster than the ground beneath it. The ground and the skyscraper top both go in a big circle once per day, but the skyscraper top goes in a circle slightly bigger - with a radius 200m greater. That means the top of the skyscraper goes $200m * 2\pi \approx 1200m$ further every day.

If you were to drop something off the skyscraper, maybe it would take about 10s to fall, about $10^{-4}$ days. So it should travel about 0.1 m, or 10cm, more in the horizontal direction than the spot on the ground directly underneath it did while it was falling. We'd only see about half that show up in deflection, though (because the object is, on average, half way up the tower while it falls), so about 5 cm deflection.

That's enough to matter if you're being super-precise, but for the most part it won't be noticeable. The cannonball is similar, but I thought it was a little easier to work out this example. The main point is that there is deflection, but it's small because the Earth rotates slowly.

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  • $\begingroup$ Can this argument about the differential distance be adapted to a qualitative discussion of the Foucault pendulum? I'm thinking here this would have to be with the length of the pendulum. Can dimensional analysis help? $\endgroup$ – ZeroTheHero Apr 14 '18 at 17:11
  • $\begingroup$ Interesting; I'm not quite sure how to do that, but it's certainly the same effect, so it seems plausible you could work something out! $\endgroup$ – Mark Eichenlaub Apr 14 '18 at 17:13
  • $\begingroup$ Yeah I wouldn't know how to do it either. Time to put the thinking cap on... $\endgroup$ – ZeroTheHero Apr 14 '18 at 17:17
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The other answers have addressed how the Earth’s rotation makes it come down in a slightly different place, but it sounds like this is asking about why it doesn’t come down way over there as the Earth rotates underneath it.

The next time you’re on a bus, in a car or on a plane, toss a coin or pen up and down. When the motion is smooth and unchanging, it will go straight up and down. Only when the vehicles speed is changing, faster or slower, will is miss its launching spot.

Your intuition is based on changing, jumpy motion. That’s what people do. But the Earth has a huge amount of inertia. It moves smoothly, so smoothly that only tiny effects are visible.

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It doesn't land exactly where it was originally launched. Here's an example that illustrates what I'm talking about.

If we shot a cannonball weighing 10 kg at an initial velocity of 200 m/s (still subsonic, but definitely a respectable speed for a cannonball) straight up at the equator, the magnitude of the Coriolis force on the cannonball would be

$$F=2mv\Omega$$

where $\Omega$ is the rate of rotation of the Earth. Our cannonball, under the influence of gravity, and assuming no atmosphere, has time-dependent velocity

$$v(t)=v_0-gt$$

where $v_0$ is the initial velocity of the cannonball and $g$ is gravitational acceleration (assumed constant because our cannonball will not go very high). So our cannonball experiences a time-dependent force

$$F(t)=2m\Omega(v_0-gt)$$

which means we can apply Newton's Second Law to get its acceleration due to the Coriolis effect

$$a(t)=\frac{F(t)}{m}=2\Omega(v_0-gt)$$

This acceleration is directed along the equator, to the west, so it will push the cannonball westward before it lands. How far westward? Well, we can integrate the acceleration to get the velocity:

$$v(t)=\int a(t)=2\Omega\left(v_0t-\frac{1}{2}gt^2\right)$$

And we can integrate the velocity to get the object's position along the equator $x(t)$:

$$x(t)=\int v(t)=2\Omega\left(\frac{1}{2}v_0t^2-\frac{1}{6}gt^3\right)$$

where we have set both the initial position and initial velocity of the cannonball along the Equator to be zero. The time of flight of the cannonball is equal to the time at which its velocity will be $-v_0$, so plugging that into $v(t)$ gives us a flight time of

$$\tau=\frac{2v_0}{g}$$

Plugging this back into our equation for $x$, we find that, when the cannonball lands, it will land a distance $x(\tau)$ to the west, where

$$x(\tau)=\Omega\left(\frac{4v_0^3}{g^2}-\frac{8v_0^3}{3g^2}\right)=\frac{4v_0^3\Omega}{3g^2}$$

Letting $v_0=200$ m/s, $\Omega=7.272\times 10^{-5}$ radians/s, and $g=9.81$ m/s^2, we have that the cannonball lands $x(\tau)=8.06$ m to the west. This is definitely a measurable difference, but it's not likely to be easily confirmed for several reasons. First, we have neglected air resistance, and that tends to be both very important and very complicated for objects moving at speeds approaching the speed of sound. As such, the cannonball's velocity will decrease much faster than the above formulas suggest, and so the observed deflection will be substantially less than 8 m. Second, the atmosphere is not totally still. It has high-altitude winds and turbulence which, if you're unlucky, can cause significant deviations in the cannonball's path. But there still should be a measurable deflection to the west in the average cannonball's landing location, which is exactly what the Coriolis force would produce. This is indeed what we see in real life; everyone from artillery field manuals to recreational shooters (https://loadoutroom.com/thearmsguide/external-ballistics-the-coriolis-effect-6-theory-section/) take into account the deviation caused by the Coriolis effect.

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  • $\begingroup$ Interesting opening sentence, but I must disagree! $\endgroup$ – Tom B. Apr 7 '18 at 23:28
  • $\begingroup$ @TomB. My bad. Obvious contradiction is now fixed. $\endgroup$ – probably_someone Apr 8 '18 at 0:05
  • $\begingroup$ It seems to me another calculation approach is possible, expressing the actual physics taking place exactly. During its flight the trajectory of the projectile is orbital motion (be it orbital motion at such a low altitude that it impacts the Earth again in seconds). As the projectile ascends to its highest point its angular velocity relative to the Earth center will decrease. As it descends the angular velocity will increase again. Overall during its flight a lag relative to the Earth will accumulate. It would be interesting to see a calculation using that conservation of angular momentum. $\endgroup$ – Cleonis Apr 8 '18 at 20:02

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