Galileo's tower of Pisa experiment

[JD_PM]

Homework Statement

A ball is dropped from the top of the tower of Pisa (which of course is not vertical). Neglect air resistance and take into account Earth's rotation for a) and b).

a) Show that its path deviates from that of a plumbline and that the ball is drifted to the east a distance ($d_p$ stands for traveled distance from the tower of Pisa):


$$d_p = \sqrt{\frac{8 h^3}{9g}}\Big(\Omega \sin \beta \Big)$$


b) A ball is now dropped from a vertical tower. Show that the traveled distance ($d_v$ stands for traveled distance from the vertical tower) is

$$d_v = \sqrt{\frac{2 h^3}{g}}\Big(\Omega \sin \beta\Big)$$

Note: $\Omega$ is the angular velocity with respect to the inertial frame, the angle $\beta$ is the co-latitude of the origin $O$ and $h$ is the height from which the ball is released. The origin $O$ is taken at the point where the ball is released.

Reference: Gregory's Classical Mechanics; Rotating reference frames (chapter 17)

Relevant Equations

Projectile equation on a rotating Earth:

$$m\Big( \frac{d \mathbf v}{dt} + 2 \mathbf \Omega \times \mathbf v \Big) = -mg \mathbf k$$

Where $\mathbf k$ is the unit vector in the apparent vertical direction.

Earth's angular velocity (please see attached image if you want to know why):

$$\mathbf \Omega = -(\Omega \sin \beta \mathbf) \mathbf i + (\Omega \cos \beta) \mathbf k$$

a) Let's first analyse the plumbline.

When it is in equilibrium, it follows from the projectile equation on a rotating Earth that:

$$\mathbf 0 = -mg \mathbf k + \mathbf T$$

Thus, the tension in the plumbline is $mg$ and the string is parallel to the apparent vertical.

Let's now analyse the ball. Suppose that the apparent vertical through $O$ meets the ground at the point $(0, 0, -h)$. Let us have the following initial conditions: $\mathbf v = \mathbf 0$ and $\mathbf r = \mathbf 0$ when $t = 0$. As $\mathbf \Omega$ is constant, we can integrate projectile equation on a rotating Earth with respect to time to get:

$$\frac{d \mathbf r}{dt} +2 \mathbf \Omega \times \mathbf r = -gt \mathbf k + \mathbf C$$

Applying initial conditions we get:

$$\frac{d \mathbf r}{dt} +2 \mathbf \Omega \times \mathbf r = -gt \mathbf k$$

If we integrate again with respect to time we end up with:

$$\mathbf r (t) = - \frac 1 2 g t^2 \mathbf k - 2 \mathbf \Omega \times \int_{0}^{t} \mathbf r(t') dt' \ \ \ \ (1)$$

Let's get a solution for the above equation by using the mathematical method called iteration.

The zeroth order approximation corresponds to the non-rotating Earth case. Thus:

$$\mathbf r_0 (t) = - \frac 1 2 g t^2 \mathbf k$$

In order to get the first order approximation we just have to plug the zeroth one into $(1)$ and integrate to get:

$$\mathbf r_1 (t) = - \frac 1 2 g t^2 \mathbf k - 2 \mathbf \Omega \times \int_{0}^{t} \mathbf r_0 (t') dt'$$

$$\mathbf r_1 (t) = - \frac 1 2 g t^2 \mathbf k + \frac 1 3 g t^3 (\mathbf \Omega \times \mathbf k)$$

If we compute the cross product we get

$$\mathbf \Omega \times \mathbf k = (\Omega \sin \beta) \mathbf j$$

Thus the first approximation is:

$$\mathbf r_1 (t) = - \frac 1 2 g t^2 \mathbf k + \frac 1 3 g t^3 (\Omega \sin \beta) \mathbf j$$

Here comes the meat:

The interesting thing is that the first order approximation predicts that the ball will drift to the +ive $\mathbf j$ direction. Is this direction east? (based on the provided diagram). My book states so but I do not see it...

Now we get the $\mathbf j$ component of the distance. What we could do is solving the time for the zeroth approximation, $t= \sqrt{2h/g}$, and then plug it into the first approximation. This should be OK, as both approximations are small. We get:$$\frac 1 3 g \Big(\frac{2h}{g}\Big)^{3/2} (\Omega \sin \beta) = \sqrt{\frac{8 h^3}{9g}} (\Omega \sin \beta)$$

That's nice! The only thing I am missing is understanding why my book states it drifts to the east.

b) For this section I know I need to modify somehow the first order approximation to take into account that now the ball is released from a vertical tower, but I do not see the trick. Could you give me a hint in this one?

Thanks.

 

[kuruman]

I don't see how you got this$$\mathbf r_1 (t) = - \frac 1 2 g t^2 \mathbf k + \frac 1 3 g t^3 (\mathbf \Omega \times \mathbf k)~~~~~~(1)$$from this$$m\Big( \frac{d \mathbf v}{dt} + 2 \mathbf \Omega \times \mathbf v \Big) = -mg \mathbf k~~~~~~(2)$$Equation (2) provides three coupled equations,$$\ddot{x}=2 \Omega \left(\dot{y} \cos (\beta )-\dot{z} \sin (\beta )\right)$$ $$\ddot{y}=-2 \Omega \dot{x} \cos (\beta )$$ $$\ddot{z}=2\Omega \dot{x} \sin (\beta )-g$$How did you get from these three to your equation (1)? Normally, one starts from the zeroth order equations ($\Omega \approx 0$) $x(t)=x_0+v_{x0}t$; $y(t)=x_0+v_{y0}t$; $z(t)=v_{z0}t-\frac{1}{2}gt^2$. Then one would substitute these in the three acceleration equations and integrate twice to get the first order equations for $\vec r(t)$. Did you do that?

 

[JD_PM]

I don't see how you got this$$\mathbf r_1 (t) = - \frac 1 2 g t^2 \mathbf k + \frac 1 3 g t^3 (\mathbf \Omega \times \mathbf k)~~~~~~(1)$$from this$$m\Big( \frac{d \mathbf v}{dt} + 2 \mathbf \Omega \times \mathbf v \Big) = -mg \mathbf k~~~~~~(2)$$

I took your $(2)$ and integrated twice with respect to time (with the given initial conditions $v=0$ and $r=0$ when $t=0$) to get:

$$\mathbf r (t) = - \frac 1 2 g t^2 \mathbf k - 2 \mathbf \Omega \times \int_{0}^{t} \mathbf r(t') dt'$$

And then I applied iteration up to first order.

 

[JD_PM]

Section a) is solved in Gregory's, which I have exactly typed.

However, for sake of clarity, I will also upload pictures of the solution.

 

[kuruman]

Yes, I see now how it is derived. It's a more compact way of doing what I suggested. To understand why the mass drifts to the east, consider two facts: (a) a point near the surface of the Earth moves from west to east (sunrise occurs in Washington, DC before Los Angeles, CA); (b) the mass at the point of release has a higher linear velocity than the base of the tower.

I don't understand why the answer in part (b) should be different from part (a). Here is my reasoning: The direction of the plumb line is the $\mathbf k$ direction; you calculated your answer relative to a frame where the z-axis is along the plumb line direction. A mass released from rest at some height will follow the same trajectory whether the tower is leaning or not. If the tower is "vertical", it was erected using a plumb line therefore the vertical tower itself is along the z-axis. Unless (a) by "vertical" the author means "in the direction of $\mathbf g$" and not "in the plumb-line direction" or (b) I am missing something which will not be the first time.

 

[JD_PM]

Sorry for the late response kuruman.

Yes, I see now how it is derived. It's a more compact way of doing what I suggested. To understand why the mass drifts to the east, consider two facts: (a) a point near the surface of the Earth moves from west to east (sunrise occurs in Washington, DC before Los Angeles, CA); (b) the mass at the point of release has a higher linear velocity than the base of the tower.

Oh so if the Earth were to rotate clockwise, the ball would drift to the west.

The direction of the plumb line is the $\mathbf k$ direction; you calculated your answer relative to a frame where the z-axis is along the plumb line direction. A mass released from rest at some height will follow the same trajectory whether the tower is leaning or not.

I totally agree with your reasoning.

The drift component $\sqrt{\frac{8 h^3}{9g}}\Big(\Omega \sin \beta \Big)$ makes sense to me, as the term $\sqrt{\frac{8 h^3}{9g}}$ is obtained out of $t = \sqrt{2h/g}$, which obeys the well known kinematic equation $z(t)=v_{z0}t-\frac{1}{2}gt^2$ (taking the initial velocity to be zero and taking absolute value we get $t = \sqrt{2h/g}$, where $h = z(t)$ in this case).

Let's check b) as well. As we know the answer, let's see how $t$ looks.

$$\frac 1 3 g t^3 = \sqrt{\frac{2 h^3}{g}}$$

$$t = 3^{1/3} 2^{1/6} \sqrt{\frac{h}{g}}$$

The above time doesn't satisfy $z(t)=v_{z0}t-\frac{1}{2}gt^2$.

We would expect to get $\sqrt{2}$ instead of $3^{1/3} 2^{1/6}$. I will ask the original author of b) about this, as I think the term $\sqrt{\frac{2 h^3}{g}}$ is wrong.

 

[kuruman]

I will ask the original author of b) about this, as I think the term $\sqrt{\frac{2 h^3}{g}}$ is wrong.

I am curious to see what the author has to say about (b). Please post it here if you get a chance.

 

[256bits]

If the tower is "vertical", it was erected using a plumb line therefore the vertical tower itself is along the z-axis. Unless (a) by "vertical" the author means "in the direction of gg \mathbf g" and not "in the plumb-line direction"

The plumb line sets the apparent vertical, as the author mentions, the extension of which does not pass through the centre of the earth, for a rotating earth.
By vertical I assume the author means in the radial direction, or perpendicular to a tangent to the Earth's surface.
The plumb line would necessarily have a southward tilt to it if measured in the Northern hemisphere.

Would the plumb line also have an eastward tilt to it? The top of pole to which it is attached travels a greater distance than a point on the surface, the point and top of pole on a line in the radial direction passing through the centre of the earth. The forces acting on the plumb, as mentioned in the article, are mg ( gravitational force of the Earth ) and T ( the tension in the wire ) . If for a thought example, we position the pole and plumb at the equator, and rotate the Earth faster and faster, the plumb has to rise up from the surface, up to a maximum rotation so that the plumb would point in the radial direction away from the earth. I think this rise is in the easterly direction.

For the problem, the angle the plumb makes with vertical in the easterly direction is to be accounted for, knowing the latitude, and rotation rate of the earth.

Make sense.

Edt Or maybe the plumb lags the point on the surface as it rises... hmmmm.

 

[kuruman]

By vertical I assume the author means in the radial direction, or perpendicular to a tangent to the Earth's surface.

I disagree with this interpretation. If the author had "radial" in mind, the author should have said so. Also, "perpendicular to a tangent to the Earth's surface" doesn't work very well if you want to build a vertical wall on a hillside as people have been doing for several thousands of years. People use plumb bobs to build vertical walls. If they didn't do so, the walls would probably look tipped relative to a lake surface especially at a 45^o latitude where the discrepancy between the radial and the plumb bob directions is the greatest. It is much easier to determine the vertical relative to a plumb line than relative to the radial direction assuming that the Earth is a perfect sphere. I can't think of a single example where people need to pay attention to the radial direction when they go about their everyday business.

In any case, let's wait and see what @JD_PM has to say after contacting the author.

 

[JD_PM]

@kuruman My professor said he will be out the following months and if I have any question I better contact him 'somewhere later in spring' I think he means I do not bother him until April

I will wait until then.

 

[kuruman]

@kuruman My professor said he will be out the following months and if I have any question I better contact him 'somewhere later in spring' I think he means I do not bother him until April

I will wait until then.

Or maybe he expects that you will forget all about it by that time.

 

[CPW]

I have two textbooks from when I took this course (over a decade ago): Fowles & Cassiday and also Thornton & Marion.

In the first book, the drift to the East for free fall (example calculation for experiment on Earth's surface at latitude 45 deg), is explained in terms of the Coriolis force and the direction is given by a vector cross product. After the example problem, the authors ask the reader to think: "Because Earth turns to the east, common sense would seem to say that the body should drift westward. Can the reader think of an explanation?"
So I think it is good that detail caused you to pause. Do you feel you have this part figured out now?

In the second book, the same sample example calculation is provided (an object dropped from a height of 100 m at latitude 45 deg is deflected approximately 1.55 cm to the East; negelecting air resistance), but then more interesting details are provided as a footnote:
"According to M.S. Tiersten and H. Soodak, Am J Phys 68, 129 (200), the southerly deflection is on the order of a million times smaller than the easterly deflection for a drop of about 100 m, and there is no credible evidence that the sourtherly deflection has ever been correctly measured, despite many attempts. The easterly deflection was predicted by Newton (1679), and several experiments (notably those of Robert Hooke) appeared to confirm the results. The most careful measurements were probably those of F. Reich (1831; published 1833), who dropped pellets down a mine shaft 188 m deep and observed a mean deflection of 28 mm. The is smaller than the value calculated in Equation 10.35, the decrease being due to air resistance effects. In all the experiments, a small southerly component of the deflection was observed - and remained unaccounted for until Coriolis's theorem was appreciated (see problems 10-13 and 10-14)."

With regard to the plumb line vs the dropped object, the deviation of the static plumb line from a line directed to Earth's center is a static effect of the rotation of Earth. The dynamics of a projectile (dropped object in free fall in this discussion) require the use of a term for Coriolis force in addition to gravity. It helps me to keep those two cases (static vs dynamic) separate in my mind for this discussion, since the static plumb line's direction analysis does not require the term for Coriolis force, but the direction of the deflection of the dropped object does include the Coriolis term.

 

[JD_PM]

Or maybe he expects that you will forget all about it by that time.

Well, it turns out I did not eventually discuss the problem with him... Let's give it a last shot here though!

b) A ball is now dropped from a vertical tower. Show that the traveled distance ($d_v$ stands for traveled distance from the vertical tower) is

$$d_v = \sqrt{\frac{2 h^3}{g}}\Big(\Omega \sin \beta\Big)$$

Note: $\Omega$ is the angular velocity with respect to the inertial frame, the angle $\beta$ is the co-latitude of the origin $O$ and $h$ is the height from which the ball is released. The origin $O$ is taken at the point where the ball is released.

The statement looks ill-defined to me. That's why I'll present a study of the general fall of the ball, see how many potential cases we have and check if any of those match the provided answer.

Let's go step by step.

Instead of using what Gregory called 'projectile equation' right away as I did above I'll show how to actually get it.

We start off by writing Newton's second law in a non-inertial rotating frame (which is beautifully derived in Tong's notes; Galileo's tower experiment is explained at 6.4.2.)

$$m \Big(\frac{d^2 \mathbf r}{dt^2} \Big)_{S'} = \mathbf F - 2m \mathbf \Omega \times \Big(\frac{d \mathbf r}{dt} \Big)_{S'} - m \mathbf{\dot \Omega} \times r- m\mathbf \Omega \times (\mathbf \Omega \times \mathbf r)$$

Time to simplify. Let's assume that Earth's rotation angular speed is constant (which means we drop Euler's force term $- m \mathbf{\dot \Omega} \times r$). Besides, the centrifugal force has a tiny effect on the falling ball (we can see this fact in Tong's worked out problem 6.3.1. in apparent gravity, where he shows how even at latitude's value where the deflection angle due to the centrifugal force is maximum is ,in first order approximation, $\sim 10^{-4}$) so let's drop $-m\mathbf \Omega \times (\mathbf \Omega \times \mathbf r)$ as well.

Thus the simplified equation of motion for the falling ball ends up as follows (which is labelled by Gregory as 'projectile equation on a rotating Earth'; let's get rid of the masses though ;))

$$\mathbf{\ddot r} = \mathbf g - 2m \mathbf \Omega \times \mathbf{\dot r}$$

Next we basically do the same as above: integrate once, iterate and integrate once again to end up getting a first order solution for the equation of motion

$$\mathbf{r}(t) \approx \mathbf{r_0} + \frac 1 2 \mathbf g t^2 - \frac 1 3 t^3 \mathbf{\Omega} \times \mathbf g \tag{*}$$

Which is basically equivalent to the one I got at #1

$$\mathbf r_1 (t) = - \frac 1 2 g t^2 \mathbf k + \frac 1 3 g t^3 (\mathbf \Omega \times \mathbf k) \tag{**}$$

The differences are due to: 1) To get $(*)$ I did not apply initial conditions but dragged the initial initial position $\mathbf{r_0}$ 2) To get $(*)$ I worked with $\mathbf{g}$, enclosing the -g constant together with its corresponding unit vector, which I introduce below.

Let's now have a closer look at the figure 17.4 I attached at #1 (which happens to show the above-equator case). Note that the unit vector $\mathbf{i}$ points South, the unit vector $\mathbf{j}$ points East and the unit vector $\mathbf{k}$ points radially outwards. OK. By trigonometry we see that the formula for $\mathbf{\Omega}$ in the $Oxy$ plane is

$$\mathbf{\Omega} = -\Omega \sin \beta \mathbf{i} + \Omega \cos \beta \mathbf{k}$$

We also note that

$$\mathbf{g} = -g \mathbf{k}$$

$$\mathbf{r_0} = (R+h) \mathbf{k}$$

OK. Now I'd like to check out the key cases and see what I get.

Case 1: $O$ is located at the equator.

Here we have $\beta = \pi /2$ so

$$\mathbf{\Omega} = -\Omega \mathbf{i}, \ \ \ \ \mathbf \Omega \times \mathbf g = -\Omega g\mathbf{j}$$

Thus the equation of motion becomes

$$\mathbf{r}(t) \approx \Big(R+h - \frac 1 2 g t^2\Big)\mathbf{k} + \frac 1 3 t^3 \Omega g \mathbf{j} $$

We get the expected result, as the ball drifts east. Let's write ball's east drift (i.e. the $\mathbf{j}$ term) in function of the height of the tower $h$. We know the time the ball takes to reach the floor, $t = \sqrt{\frac{2h}{g}}$, so we get

$$d=\frac{2 \Omega}{3} \sqrt{\frac{2 h^3}{g}}$$

Case 2: $O$ is located at The North (South) pole.

Here we have $\beta = 0 (\pi)$ so

$$\mathbf{\Omega} = \Omega (-\Omega) \mathbf{k}, \ \ \ \ \mathbf \Omega \times \mathbf g = 0$$

Thus there's no drift from the vertical (i.e. $d=0$ here).

Case 3: $O$ is located above the equator at an angle $0 < \beta < \pi/2$

$$\mathbf{\Omega} = -\Omega \sin \beta \mathbf{i} + \Omega \cos \beta \mathbf{k}, \ \ \ \ \mathbf \Omega \times \mathbf g = -\Omega \sin \beta$$

So the ball drifts east as given by

$$d=\frac{2 \Omega}{3} \sqrt{\frac{2 h^3}{g}} \sin \beta$$

Plugging some 'nice' angles one gets

$\beta=\pi/3$

$$d=\Omega \sqrt{\frac{2 h^3}{3g}}$$

$\beta=\pi/4$

$$d=\frac{2\Omega}{3} \sqrt{h^3}$$

$\beta=\pi/6$

$$d=\frac{\Omega}{3} \sqrt{\frac{2 h^3}{g}}$$

Conclusion: Note that I did not get the provided answer. What I think is that the exercise is simply flawed; I would expect b) to be formulated as follows:

At what angle one gets the east drift (for instance) $d=\frac{2\Omega}{3} \sqrt{h^3}$?

Mmm but this is too simple... So I may be the one wrong. What do you all think?

 

[haruspex]

a) Show that its path deviates from that of a plumbline
b) A ball is now dropped from a vertical tower

I don't believe there is any difference. Vertical towers are built using plumblines.

Fwiw, here's a nonvectorial way:
We can think of the ball as having been thrown eastward at $h\Omega \sin(\beta)$. For simplicity I'll leave out the trig from hereon.
After time t, it has traversed an arc $t\Omega$ relative to the Earth's centre, so it is as though the angle of gravity has changed by that much. That doesn't alter the component parallel to the original vertical much (second degree), but it adds a component opposing the original velocity of $-gt\Omega$. That velocity is thus reduced to $(h-\frac 12gt^2)\Omega$, and the distance covered is $(ht-\frac 16gt^3)\Omega$.
Substituting $t=\sqrt{\frac{2h}g}$ yields the $\Omega\sqrt{\frac{8h^3}{9g}}$ result.

 

[JD_PM]

Hi haruspex, thanks for your reply.

We can think of the ball as having been thrown eastward at $h\Omega \sin(\beta)$. For simplicity I'll leave out the trig from hereon.

OK what I understood is that you assumed as the initial velocity of the ball from tower's top:

$$\mathbf{v} = h\Omega \sin(\beta) \mathbf{j}$$

After time t, it has traversed an arc $t\Omega$ relative to the Earth's centre, so it is as though the angle of gravity has changed by that much. That doesn't alter the component parallel to the original vertical much (second degree), but it adds a component opposing the original velocity of $-gt\Omega$. That velocity is thus reduced to $(h-\frac 12gt^2)\Omega$, and the distance covered is $(ht-\frac 16gt^3)\Omega$.
Substituting $t=\sqrt{\frac{2h}g}$ yields the $\Omega\sqrt{\frac{8h^3}{9g}}$ result.

I do not fully understand your reasoning. Is this the arc you meant?

 

[haruspex]

Hi haruspex, thanks for your reply.
OK what I understood is that you assumed as the initial velocity of the ball from tower's top:

$$\mathbf{v} = h\Omega \sin(\beta) \mathbf{j}$$
I do not fully understand your reasoning. Is this the arc you meant?

View attachment 267155

Yes, and yes.
I should clarify that the thrown velocity is relative to the surface of the earth, but that is tiny compared with the speed of the Earth's surface, so I ignore it in the traversed arc $\Omega t \sin(\beta)$.

 

[kuruman]

So it has been shown that one gets the required result if the ball is "thrown" with some initial velocity and not "dropped" (from rest relative to the point of release) as stated in the problem. I wish I knew we had this latitude (the pun is intentional) when we looked at the problem in January.

 

[etotheipi]

I thought that @haruspex's solution is essentially considering the problem in an inertial space frame in which the Earth is rotating (as opposed to a rotating frame, which the solution manual uses), but surely it is an equivalent analysis of the same scenario? A ball at rest on a tower rotating with the Earth has initially non-zero horizontal velocity w.r.t. this inertial space frame.

 

[kuruman]

I thought that @haruspex's solution is essentially considering the problem in an inertial space frame in which the Earth is rotating (as opposed to a rotating frame, which the solution manual uses), but surely it is an equivalent analysis of the same scenario? A ball at rest on a tower rotating with the Earth has initially non-zero horizontal velocity w.r.t. this inertial space frame. I think it's also necessary assume that the distance through which the base of the tower moves is negligible

Yes, of course. Thanks for pointing that out.

 

[etotheipi]

Actually I have realized there is one bit @haruspex's solution that I missed also!

In the space frame, the initial velocity of the ball (dropping trig factors) would be $(R+h)\Omega$, not $h\Omega$. Following the working, the horizontal acceleration is $-gt\Omega$ which integrates to$$v(t) = (R+h)\Omega - \frac{1}{2}gt^2$$and the distance covered is$$x(t) = (R+h)\Omega t - \frac{1}{6}gt^3$$then we subtract the distance traveled by base of the tower, $R\Omega t$, which yields the required answer. So we can sort of "ignore the $R$ term" through the calculation

 

[haruspex]

Actually I have realized there is one bit @haruspex's solution that I missed also!

In the space frame, the initial velocity of the ball (dropping trig factors) would be $(R+h)\Omega$, not $h\Omega$. Following the working, the horizontal acceleration is $-gt\Omega$ which integrates to$$v(t) = (R+h)\Omega - \frac{1}{2}gt^2$$and the distance covered is$$x(t) = (R+h)\Omega t - \frac{1}{6}gt^3$$then we subtract the distance traveled by base of the tower, $R\Omega t$, which yields the required answer. So we can sort of "ignore the $R$ term" through the calculation

Yes, that's why I said it was equivalent to being thrown at speed $h\Omega$; it is the speed relative to the Earth's surface.

 

[etotheipi]

Yes, that's why I said it was equivalent to being thrown at speed $h\Omega$; it is the speed relative to the Earth's surface.

You're always one step ahead...

 

[JD_PM]

Hi! Sorry for the late reply!

OK I get it now, thank you all!

Actually, I think it's worth considering the same problem but from a non-inertial (rotating) frame.

We expect to get a larger distance, as the gravity dragging the particle in the horizontal direction is now weaker due to the centrifugal acceleration.

Let's check what the acceleration of the particle is (using polar coordinates; argh let's not drop trigonometry factors for now)

$$\mathbf{g_{\text{eff}}} := -g \mathbf{\hat r} - \mathbf{\Omega}\times(\mathbf{\Omega} \times \mathbf{r})=(-g+(R+h)\Omega^2 \sin^2(\theta))\mathbf{\hat r}-(R+h)\Omega^2\cos(\theta)\sin(\theta)\mathbf{\hat \theta} $$

We now move to the horizontal (by recalling what's the connection Polar-Cartesian: $\hat r=(\sin (\theta), \cos(\theta)), \hat \theta=(\cos (\theta),-\sin (\theta))$. Thus we have

$$g_{\text{eff}}=-g\sin(\theta)+(R+h)\Omega^2\sin^3(\theta)-(R+h)\Omega^2\cos^2(\theta)\sin(\theta)=-g\sin(\theta)-(R+h)\Omega^2 \sin(\theta)\cos(2\theta)$$

Where I've used $\cos^2 \theta - \sin^2 \theta =\cos(2\theta)$

Thus the velocity of the particle in the horizontal direction is

$$v(t)=(R+h)\Omega\sin(\theta)-\frac 1 2 g t^2 \Omega \sin(\theta)-(R+h)\Omega \sin(\theta)\cos(2\theta)$$ $$=2(R+h)\Omega\sin^3(\theta)-\frac 1 2 gt^2\Omega\sin(\theta)$$

Where I've used $1- \cos(2\theta)=2 \sin^2 (\theta)$

Thus the distance is

$$x(t)=2(R+h)\sin^3(\theta)\Omega t -\frac 1 6 gt^3\Omega\sin(\theta)$$

Dropping trig factors we of course get

$$x(t)=2(R+h)\Omega t -\frac 1 6 gt^3\Omega$$

Which is indeed greater than the distance obtained in the inertial frame.

For the sake of completeness, let's get $d$. Subtracting the distance covered by the base of the tower $2R\Omega t$ we get $5\Omega\sqrt{\frac{2h^3}{9g}}$

PS:

I wish I knew we had this latitude (the pun is intentional) when we looked at the problem in January.

Mmm what do you mean here? I had no idea what 'intentional pun' meant so I looked it up; were you trying to make a joke?

 

[hutchphd]

trying to make a joke

Folks afflicted with the curse of punning must be handled very carefully. A slight groan is usually sufficient to satisfy their reprobate tendencies, but in my experience some form of acknowledgment is always required lest the the puns increase in frequency while degenerating in quality. My father was seriously afflicted ...it may be hereditary

 

[kuruman]

Mmm what do you mean here? I had no idea what 'intentional pun' meant so I looked it up; were you trying to make a joke?

Disclaimer:
I will answer this being fully aware that explaining a joke is a fool's errand if one expects people to laugh after the explanation. (Fool's errand: A task or activity that has no hope of success.)

Yes, I was trying to make a joke by using the word "latitude" to exploit both its definitions simultaneously. It's a pun (Pun: a joke exploiting the different possible meanings of a word or the fact that there are words which sound alike but have different meanings.) My joke is a pun of the first kind.

The first meaning of "latitude" is "the angular distance of a place north or south of the Earth's equator, or of a celestial object north or south of the celestial equator, usually expressed in degrees and minutes." Example: the latitude of the tower of Pisa is 43.7230° N.

The second meaning of "latitude" is "scope for freedom of action or thought." Example: "I wish I knew we had this latitude when we looked at the problem in January."

So the intended joke (that is no longer a joke) was what did I mean by my use of the word "latitude"? Did I mean 43.7230° N or the freedom to think of the problem in a different way?

LOL anyone?

 

[haruspex]

explaining a joke is a fool's errand if one expects people to laugh after the explanation

Strangely, I found the explanation more amusing than the pun.
Puns occur in algebra. In $y(x)=\int_{x=0}^x\cos(x).dx$, x represents two different variables. To remove the pun, write $y(x)=\int_{t=0}^x\cos(t).dt$.
The y(x) is also problematic. Is y a variable or a function here?

 

[kuruman]

Strangely, I found the explanation more amusing than the pun.

I'm glad I succeeded in being amusing the second time around. The algebra puns that you mentioned are of the first kind and we see these all the time as attempts at a solution". See example below.
Problem
A rock is thrown straight up with initial speed $v_0$. Find the time of flight,
Solution$$y(t)=v_0t-\frac{1}{2}gt^2$$ $$0=t\left(v_0-\frac{1}{2}gt\right)~~\Rightarrow~t=\frac{2v_0}{g}.$$

Spoiler: Don't get it?

The independent variable $t$ is also used to denote the specific time $t_f$ at which the projectile returns to the launching height.