Bruntilda, the “centripedal force” is an imaginary force introduced to make the equations work again, and effectively turning Earth into an inertial frame.
Centrifugal force is the imaginary force that accounts for you being pushed to the sides of a spinning object. Centripetal force is quite real, and in this case is accounted for by earth’s gravity.
I’m far from being able to follow your discussion, but if I get this part right, it seems that you are claiming that the “parabola shape” path is shorter than the circular path. This doesn’t even make intuitive sense.
If there are two points on a sphere (launch point of the cannon, and the point where it is when the cannonball lands), then the only paths that are shorter than the circular arc that connects them are under the surface of the sphere (assuming the cannon is on the equator). So, since the projectile is clearly following a longer path, at a slower angular velocity, how do you figure they meet up?
Objects on the surface of the earth are subject to centripal acceleration and the earth can never be considered an inertial frame unless you are standing at one of the poles.
In regards to the original OP, the cannon ball will not land in same spot as so thoroughly explained by all the people using the angular velocity explainations but as stated early on by Q.E.D., the effect is very real and called the Coriolis Effect.
Here is a site and I hope you take the time to read it. You could learn some really interesting things.
Uh…OK… That will teach me to cite things before actually reading them. The effect is not the Coriolis Effect as that is specifically dealing with North South differences in flight, which the OP is not asking about.
The angular momentum arguements are correct though, that once the cannonball has been displaced on the way up, that it cannot make up the displacement on the way down, as that would require a tangential acceration of the ball.
Skeptico, here’s problem 2.3 from “Theoretical Mechanics of Particles and Continua”, by Alexander L. Fetter and John Dirk Walecka, Professors of physics, Stanford University:
Also, here is a PDF file with the problem solved (problem 4). They get the same answer as Fetter and Walecka.
Here’s a problem for everyone else: Fetter and Walecka have the same equation for a particle dropped from a height h as Chronos posted. Chronos said the horizontal distances for the two cases would differ by a factor of 2, but these answers differ by a factor of 4. Now, Chronos can’t be expected to be always correct, but I can’t come up with any reason why the difference should be a factor of 4, rather than a factor of 2. So who’s right, and if it’s a factor of 4 difference, why?
For the record, I have done this problem from scratch, but that was years ago, and I didn’t feel like going through it again, so I just looked up the answer in Fetter and Walecka. However, F&W only has the result for a particle dropped from a height, not for one launched from the surface. I remembered that there was a factor of -2 between the two answers, but I may have misremembered: It may have been -4. The fact remains, though, that the effect is real, and it’s miniscule.
That answer, by the way, does not violate or ignore conservation of angular momentum, since the calculation is based on the conservation of angular momentum. That’s why [symbol]w[/symbol], the rotational speed of the Earth, enters into the formula.
Skeptico, since you’re discussing the different paths taken by the ball and Earth: Your mention of the Earth moving in an (approximate) circle tells me that you’re considering the Earth to be moving around the Sun. Suppose we had a planet which was not orbiting a star, but just spinning all by itself in the middle of the Great Void. Would your answer be the same? Why or why not?
Obviously it was a misstatement. I meant angular velocity, of course.
As Dr. Quest, er, I mean Race Bannon said, the only path connecting two points on a sphere that are shorter than the sphere’s actual surface are inside the sphere itself. How can you claim that a parabola arcing 1000 km above the surface is shorter than the circular arc traveled by the cannon due to the earth’s rotation? That’s ridiculous.
OK, I figured out where the extra factor of 2 comes from. When the particle is dropped, it moves faster in angle than the Earth by an amount proportional to its distance below h. When it is shot upward, it moves slower in angle by an amount proportional to its distance above the Earth. Since it’s spending more time near h than near the Earth’s surface, there’s an additional factor of 2 for the case where it’s shot upward compared with the case where it’s dropped.
I think there are clear answers to both the OP, and the question of what does happen.
The different paths the earth and cannonball take are not an issue, because angular displacement is single-dimensional. There is not more than one path from A to B in a one-dimensional system. If you don’t believe me, see for yourself: draw a plot of angular displacement vs time for both the earth and the cannonball. You’ll see that in order for the ball to achieve the same angular displacement as the earth, it has to average the same angular velocity, which means that it’s either got to a) maintain the same angular velocity as the earth the whole time, or b) exceed the earth’s angular velocity part of the time. Neither of these happens.
If you disagree, I can only ask that you draw the plots I describe above. If you’re right, the plot will prove me wrong once and for all. If you’re not, it will prove you wrong.
The cannonball has a significantly reduced angular velocity compared to the earth. As it falls, its angular velocity increases until it again matches that of the earth.
However, because the cannonball has spent time moving at a lower angular velocity, while the earth has not, the cannonball cannot catch up.
The cannonball spends 903.5 seconds in the air.
During that time, the earth rotates 6.5884x10-[sup]2[/sup] radians, or about 420 kilometres at the equator.
The cannonball has a lower average angular velocity, 6.6142x10[sup]-5[/sup] radians per second.
The cannonball rotates through 5.976x10[sup]-2[/sup] radians, or about 381 kilometres at the equator.
The result is that the cannonball hits the earth 39 kilometres from where it started.