I shot an arrow into the air,
It fell to earth, I knew not where;
For, so swiftly it flew, the sight
Could not follow it in its flight.
– Henry Wadsworth Longfellow
This is what happens when you don’t pay attention in freshman physics. You don’t need to see where the arrow goes to find out where it lands, you just need the initial speed and angle. Two simple accelerated motion equations later, you have the range. The method isn’t perfect since it fails to account for air resistance and other factors, but for finding an arrow it’s a lot better than shrugging and giving up once you lose sight of it.
In the early days of physics the question of refining the predictions of projectile motion was not merely one of academic or poetic interest. If you’re commander Wellington of the noble British army wishing to lob cannon shells at the nefarious French (or vice versa), these other more delicate physics corrections to the projectile equations can mean the difference between life and death – or at least which side experiences which.
One of these corrections is due to the fact that the earth is not standing still. When a projectile leaves the ground, the earth keeps on rotating right under it. While this is a tiny effect for (say) a line drive in a baseball stadium, it can be substantial for artillery firing over longer ranges. The effect is called the Coriolis force, with the usual caveat that it’s not a force as such but rather a “fictitious” force due to the fact that the earth is an accelerated coordinate system.
The acceleration of the projectile from the path it would have taken on a non-rotating earth is to first order:
Where v is the velocity in the inertial frame. For simplicity, let’s assume we’re just throwing the projectile straight up into the air. On a stationary earth, it would just land exactly where it started. On the actual moving earth, it’ll have that additional acceleration deflecting it to the side as a function of its velocity. Plug in the velocity in the above equation:
Now omega is along the earth’s axis, and z is the upward direction at the point of the throw. Thus the cross product will be in the y direction, defining y as east:
Integrate that twice to get the position:
Though there’s not much to be gained by going through the algebra directly, it’s not so difficult to find t and the initial velocity in terms of h. Doing this, we get a total deflection of:
in the westerly direction. Theta is the angle between omega and your local z – in other words, your colatitude. Plugging in a value of h = 100 meters, the earth’s rotational angular frequency, and my colatitude of roughly 60 degrees, we get a total deflection of 7.6 centimeters. Very small considering the total distance of travel, but not microscopic either. It would be especially difficult to test due to the difficulty of throwing a projectile exactly straight upward.
The experiment has been done with the much more simple method of dropping a projectile off of a tall building, therefore automatically causing it to fall exactly straight. Bizarrely, if you work out the calculation the total deflection is exactly 1/8th of what we get for the up-and-down case – in the opposite direction (east). Nobody said being a modern major general would be easy.
(This problem more-or-less happens to be Fetter and Walecka 2.5. Great book, btw.)