Effects of Earth's rotataion on a simple experiment

[Orodruin]

And you didn't put that information on the table. That's what I was getting at.

I beg to differ:

The relation is v = ωr. What does this tell you about the angular velocity if you keep v fixed and increase r?

The angular momentum of the ball is conserved and given by mvr, where v is the horizontal component of the velocity. What does this tell you about the horizontal velocity as r increases? What does it tell you about the angular velocity?

I gave all of the information necessary for the OP to work this out for himself/herself rather than simply providing the answer.

Qualitatively, yes, your explanation does explain why there's a westward drift. Quantitatively, good luck. That's voko's approach. Invoking the concept of the Coriolis effect and assuming gravity doesn't vary with height and it's just a matter of high school level calculus to arrive at the quantitative result that the ball thrown straight up at the equator will land a distance $d=\frac 4 3 \frac {\Omega v^3}{g}$ to the west.

If you reread the OP, you will see that it is not necessarily asking for a quantitative solution. I find it counter productive to meet a qualitative question of what happens and why with "there is this fictitious force in a rotating system", rather than trying to explain what is actually going on.

In addition, within your assumptions that the ball is not thrown to heights that require consideration of a varying g, it is also very easy to integrate the equations I supplied and arrive at the same result (assuming that your $g$ really should be $g^2$, which is necessary to make the dimensions consistent):

Conservation of angular momentum gives:
$$
\omega(r) = \omega_0 \frac{r_0^2}{r^2} \simeq \omega_0 - 2 \omega_0 \frac {h(t)}{r_0}
\quad \Rightarrow \quad
\frac{d\phi}{dt} = \omega(r) - \omega_0 \simeq - 2 \omega_0 \frac {h(t)}{r_0}.
$$
Given that $g$ is constant
$$
h(t) = v t - \frac{gt^2}2.
$$
Integrating from 0 to $2v/g$ yields
$$
\phi = - \frac{4\omega_0 v^3}{3r_0 g^2}.
$$
Multiply with $r_0$ to get the distance rather than angle:
$$
d = - \frac{4\omega_0 v^3}{3 g^2}.
$$

 

[Orodruin]

And just for curiosity reasons as to the size of the effect. Modern rifles can have a muzzle velocity of about 1200 m/s (source http://en.wikipedia.org/wiki/Muzzle_velocity). The resulting effect is of the order of a few hundred meters.

However, in order to have the effect as the dominant one, you must be able to aim the rifle to the vertical direction within some arcminutes. If not, your aim will be determining whether the bullet lands to the east or west.

 

[voko]

And just for curiosity reasons as to the size of the effect. Modern rifles can have a muzzle velocity of about 1200 m/s (source http://en.wikipedia.org/wiki/Muzzle_velocity). The resulting effect is of the order of a few hundred meters.

However, in order to have the effect as the dominant one, you must be able to aim the rifle to the vertical direction within some arcminutes. If not, your aim will be determining whether the bullet lands to the east or west.

Does that take the atmosphere into account?

 

[Orodruin]

Of course not, neither does anything else we have said. :)

 

[D H]

And just for curiosity reasons as to the size of the effect. Modern rifles can have a muzzle velocity of about 1200 m/s (source http://en.wikipedia.org/wiki/Muzzle_velocity). The resulting effect is of the order of a few hundred meters.

It's almost eighteen hundred meters if the bullet is fired from the equator, not just a few hundred. (I get (1787 meters using an orbital calculation, 1762 meters using $\frac 4 3 \frac {\Omega v^3}{g^2}$ as an estimate.)

Does that take the atmosphere into account?

Of course not. A drag-free bullet fired straight up at 1200 m/s will climb to almost 75 kilometers before falling back to Earth. A real 50 caliber bullet? I doubt it will even climb to 5 kilometers. A bullet shot straight up will most likely be tumbling on the way down, making for a lowish terminal velocity (a few hundred meters/second). The randomness that results from this tumbling will overwhelm the westward drift.

 

[voko]

Of course not, neither does anything else we have said. :)

There is a difference. We did not discuss the effects of the atmosphere, but we did not discuss the effects of imprecise aiming either. Saying that the latter will overwhelm the westward drift, without considering the former, as you did, seems rather strange to me.

 

[Orodruin]

Of course, but it is the easier one to take into account. All I wanted to convey was that the rotational effect was minor. You could also just as well say that considering the rotational effect without considering atmospheric or aiming effects would seem pretty strange.

 

[Orodruin]

It's almost eighteen hundred meters if the bullet is fired from the equator, not just a few hundred. (I get (1787 meters using an orbital calculation, 1762 meters using $\frac 4 3 \frac {\Omega v^3}{g^2}$ as an estimate.)

Yes, I probably forgot to divide by 2pi when computing the angular velocity ... But the point remains, I would even say the derivation using the conservation of angular momentum is easier as it allows me to get away with one integration less in the tangential direction than if just starting from the forces.