What happens at the end of Coriolis Deflection

Question

Consider we launch a cannonball due south from a point at 45 degrees latitude in the Northern Hemisphere (e.g the point defined with the co-ordinate system on this diagram). The cannonball travels for 5 minutes (through vacuum - ignore air resistance), experiences Coriolis deflection and drifts 'right/west' ending up at 40 degrees latitude.

A few nanoseconds before it hits the ground, the rotational/tangential speed of the cannonball is slower than the tangential speed of the ground below. In order for the cannonball to stick/stay at the point of impact, its tangential speed must match the speed of the ground.

What effect does this difference in speed have on the cannonball on landing/impact? Does it experience a sudden west-east acceleration due to friction? If yes, would this effect be even more pronounced in this case where the project is fired from North Pole?

Answer 1

There is no significant difference between what happens in the east-west direction and what happens in the north-south one. Upon impact it will experience a pretty sudden acceleration due to the impact, and depending on the trajectory and the nature of where it hits the ground, it may not stick, but rebound, skid, or whatever...

Answer 2

Yes, I'm THAT GUY who answers a 6 year old question, LOL.

Your intuition is correct. When it left the cannon, the cannonball had an eastward velocity of 1,074.9 feet per second, which is how fast the cannon was moving east. (Assume earth radius = 3959 statute miles).

When it arrives, the ground is moving east at 1,164.5 feet per second because of the lower latitude. How significant is that 90 FPS difference? Let's see.

In order to cover 5 degrees of latitude in 5 minutes, the cannonball is covering 345.5 statute miles in 5 minutes. That's 1.82 million feet in 300 seconds. That's about 6,080 feet per second.

The 90 FPS east-west delta on landing is 1.5% of the southerly velocity of 6,080. So yes, it will impart a small clockwise rotation (from shooter's POV) and deflect the cannonball a tiny, but measurable distance to the east. The clockwise spin imparted by the first few bounces will actually tend to make future bounces deflect the ball back west as well.

And yes, this delta would be larger, but not astronomically larger at the poles. Instead of 90 FPS, the delta between the cannonball's eastward vector (now zero because it's fired from the pole) would land on ground moving 132 FPS east faster than the ball, or 2.2% of it's southerly velocity.