No.
This is a simple concept but I am finding it difficult to put into words. I think the problem is that it is not intuitive; we cannot normally see the Coriolis effect and any attempt to simulate it using optics or mechanical means fails.
Let's try this for an example: Say there are two lighthouses, one mile apart. And to make it easier, they are on a line of longitude, meaning one is directly north of the other, no east / west difference at all. And one of them is located at the equator and the other one is one- mile away from the equator, on a perfect north / south line.
Now, you are at the top of the southern lighthouse and you train a telescope on the north lighthouse and measure the angle of the telescope (relative to the Earth's axis, or the North Pole) and find it to be zero (of course). If the telescope is not moved, it will point north and look at a specific spot on the other lighthouse forever.
Then we set up a rifle next to the telescope and also point it perfectly, absolutely north. It is sighted directly on the other lighthouse, in a very specific point. We will ignore vertical drop in this example to make it simple but again, the rifle is pointed directly at the other lighthouse, as is the telescope next to the rifle. We put a mark on the north lighthouse exactly where the rifle is aimed.
In this example, there is no wind, there is no inaccuracy in the rifle and the projectile will follow a path absolutely straight left / right path, due north, on a line that intersects the north lighthouse where we have marked that lighthouse.
We fire the rifle and wait for the projectile to impact the other lighthouse, directly in- line with the mark placed there. There will be some drop of course but the projectile WILL hit the lighthouse some distance exactly below our previously marked point. It cannot do otherwise. There are no tricks or gimmicks or hidden wrinkles in this example.
But the bullet impacts the north lighthouse SOME DISTANCE to the SIDE of the mark by some inches or feet. We check the rifle, and it is still pointed directly at the mark on the other lighthouse so in our confusion, we fire it again. And again it misses our mark but it impacts in exactly the same point the previous projectile did.
Now, just to make absolutely sure there is nothing wrong here, we now string a cable between the two lighthouses, exactly on the line between the rifle and the point on the other lighthouse that the rifle is aimed toward. The cable is pulled tight and we can verify that the two points on the two lighthouses are indeed directly in- line as shown by the cable (we do not care about any vertical sag in the cable, only the left- right relationship) and perfetly aligned north / south as we knew they had to be.
So the cable is straight. The rifle sends the projectile directly toward a point on the other lighthouse and it travels straight. But the impact point is off to one side, every-time it is tested.
The difference between the impact points and the line- of- sight of the rifle is due to the Coriolis effect. It appears that both lighthouses are fixed and not moving but that is incorrect they are indeed moving, because they are rotating around the axis of the Earth at ~1,000 MPH. The key to this is the fact that both lighthouses (as well as the telescope, the rifle, you and everything else in sight) is fixed to the Earth and being pushed sideways at 1,000 MPH. Because EVERYTHING around you is also moving at this same speed, the movement is not apparent but it is this sideways motion, or rotation, that causes the sun to appear and disappear each day. But in our example, the instant the projectile leaves the barrel, it is NO LONGER being pushed sideways but is now free to travel in a truly straight path, which it does. And when it arrives one mile north, the northern lighthouse has MOVED SIDEWAYS since the projectile was fired and so it misses the point of aim by a substantial amount.
The amount of side movement is dependent on the time it takes the projectile to travel from one lighthouse to the other.
The Coriolis effect is only applicable when an object is sent on a path that will lead toward an object on the Earth's surface but not attached to the Earth's surface. It is a pretty simple math problem to figure out the deceleration rate of the projectile (it starts off moving 1,000 MPH sideways but begins to slow when it leaves the barrel), and the calculate the flight time to yield a specific distance.
And again, per the Paris gun, given the projectile's flight time of three minutes, the 'target' rotated away from the point of aim by almost exactly one kilometer.
Now that is the simplest example. The next thing is to think that this effect reduces as the angle changes toward one more in- line with the equator until it is no longer applicable but because a projectile will always travel in a arc, and the projectile must be launched at some upward angle to hit a distant target, the Coriolis effect still 'bites' the shooter, but instead of a left / right error, we now get a closer / farther error where the projectile lands in front of or behind the point where the arc (called a ballistic path in our case and not a true arc but close enough for this example) would have it land. Again, this if fairly simple to calculate. Of course in the real world, it is almost certain a firearm, an artillery piece, a rocket, a golf ball sent a long way or any other example would ever be on a perfect north / south or east / west line so both parts of the Coriolis effect must be calculated and then added together to get the left / right and too short / too far corrections in the launch azimuth and attitude (the angle of the gun, in rotation around in a circle and the angle the barrel is elevated to before firing).
That is the simplest example I can think of.
'Everything should be made as simple as possible, but not simpler.'
-an axiom attributed to A. Einstein
What he meant by this is to reduce everything down to the simplest possible explanation or example but not so much that the example or explanation becomes wrong. In my example above, the temptation is to simply say that one seemingly fixed object moves in relationship to another seemingly fixed object but that is too simple and incorrect. So in my view, the simplest explanation is that both objects attached to the Earth are being moved sideways all of the time, and the projectile is NOT being moved sideways for the duration of its flight. That is as simple as I believe it can be made and still be correct and not some bad or wrong example.
The Coriolis effect is also why water swirls in a bowl as it drains out the hole in the bottom (round sink, toilets and so forth), and while the reason is absolutely identical to the above example, the explanation is more complex. So that can be an optional, no added cost, homework assignment.
The rocket traveling straight up is not sufficient to show the Coriolis effect but if the rocket did have sufficient power or fuel to leave the Earth's gravitation and fell straight back down (straight up, stopped, then straight back down) it would NOT hit the point from which it was launched. That would be an example of the Coriolis effect.
Next weeks' class will be on the Doppler shift (or Doppler effect.... to maintain a steady course curriculum) and how it is used to tell if a distant star is moving toward us, away from us or neither. And if it is moving toward or away from us, exactly how fast. That is a much longer, more detailed and more complex thing than the Coriolis effect is though.
And still on- topic: math answers.
Brian
When a rocket takes off and does not go straight up, would that be the same thing?