# Difference between naive and Coriolis-force calculation

Consider the classical problem of dropping a coin from a tower at the equator of a planet without atmosphere and with spin $\Omega$: where in relation to a plumb-line will the coin land? When doing this in a naive way (using the difference in horizontal motion due to rotation) or using the coriolis force, one obtains answers that differ by a factor 2/3. What is fundamentally wrong with the naive approach? I'm looking for a simple answer that avoids intricate mathematical arguments as much as possible.

The total force (in the frame of the tower) is the sum of gravity, centrifugal, and coriolis forces. If the height $h$ of the tower is much smaller than the radius of the planet, the first two of these forces are constant (described by $g$) to very good approximation during the fall. The vertical motion is thus $$\ddot{y} = -g,\qquad\dot{y} = -gt,\qquad y=h-\tfrac{1}{2}gt^2.$$ with travel time $T=\sqrt{2h/g}$. The coriolis acceleration is $\boldsymbol{a}=-2\boldsymbol{\Omega}\wedge\boldsymbol{v},$ giving (with $\boldsymbol{\Omega}=\Omega\hat{\boldsymbol{z}}$) $$\ddot{x} = 2\Omega\dot{y}.$$ Inserting $\dot{y}$ from above, integrating twice and inserting $T$, one finds for the horizontal offset $$x = \frac{2}{3} \Omega \sqrt{\frac{2h^3}{g}}.$$

Alternatively, one may use a naive method by simply stating that $$x \approx T \Delta v_{x} = T \Omega h = \Omega \sqrt{\frac{2h^3}{g}}$$ with $\Delta v_{x}$ the difference of the azimuthal (horizontal in the frame of the person on the tower) velocity between the top and bottom of the tower. This method obtains the correct answer if the coin travels down at constant speed.

There is an even more discrepant situation where a coin is thrown vertically (in the direction of the plumb-line) upwards from the equator and falls back. Here, the naive argument obtains $x=0$ (since $\Delta v_x=0$), while the coriolis force gives $$x = \frac{8}{3} \Omega \sqrt{\frac{2h^3}{g}}.$$ This is obtained using $\dot{y}=v_0-gt$ with $v_0=\sqrt{2hg}$ and $T=2v_0/g=\sqrt{8h/g}$ with $h$ the maximum height of the trajectory, and then proceeding as above (actually $x(t)$ is maximal for $t=T$). Apparently, the symmetry assumed in the naive method is not present in the real system – can that be seen easily?

• Duplicate? physics.stackexchange.com/q/249423 Jun 2, 2016 at 12:53
• Just want to point out that you are using $\Omega$ to mean "spin", but really you are treating it like $\omega \sin\theta$ for the latitude $\theta$. Jun 2, 2016 at 17:15
• @Farcher well spotted. The problem is indeed itentical, but not the question. I want a simple answer that exposes the fundamental flaw of the simple method w/o diving into details. Jun 2, 2016 at 18:46
• @levitopher $\omega$ or $\Omega$ does it matter? I put this on the equator, so $\sin\theta=1$. Jun 2, 2016 at 18:47
• @Walter: I see, didn't catch that assumption at the beginning. Jun 2, 2016 at 20:32

The relation $\Delta v_x = \Omega h$ correctly expresses the difference in the inertial velocity between the top and bottom of the building/tower; this is not the cause of the problem.

The naive approach is based on the (implicit) assumption that as the coin falls it moves from one inertial frame to another, each moving at different horizontal velocity (due to the rotation), and that there is nothing more to this.

However, a rotating frame is also inevitably an accelerated frame and the assumption of shearing inertial frames too simple. The acceleration (which arises entirely from to the choice of reference frame) must correctly account for the change of $\boldsymbol{v}$ relative to the rotating frame for an object that in a non-rotating inertial frame (in which the planet rotates at rate $\boldsymbol{\Omega}$) moves unaccelerated (on a straight line).

Here's how I get your $\Delta v=\Omega h$:

Move to a stationary (inertial) frame relative to the surface of the Earth, where the surface is now moving at a linear speed $\Omega R$ ($R$ is the radius of the Earth). If an object is dropped from a height $h$, it has an initial speed $(R+h)\Omega$. When it hits the ground, it has a speed $R\Omega$. It's average acceleration is then $a_{ave}=-h\Omega/T$. If you assume that acceleration is constant the entire way, you can use kinematics to get determine how far the object moves during the time of flight $T$:

$$\Delta x(T)=-\frac{1}{2}\frac{h\Omega}{T}T^2+(R+h)\Omega,$$

and since the point right below where the object started only moved $R\Omega T$, you get a shift of $h\Omega\sqrt{2h/g}$.

So the problem here is that your two frames are not inertial with respect to each other (the top of the drop and the bottom). If you don't include the Coriolis force, the bottom frame needs an acceleration of $h\Omega/T$ in order to satisfy the conditions of a pair of concentric rotating rings. Since that's not the same coordinate transformation you need to do to get between an inertial frame and the Earth's frame, you're not reproducing the same physical result.

For your second question, we should not expect a symmetric answer because although the time is twice as large, the displacement is given by

$$x(t)=\frac{1}{3}g\Omega t^3-\Omega v_0 t^2$$

e.g. it's not a linear function of $t$.

• I fully understand what you're saying (and I knew that), but what I want to know is the deeper reason why the naive model is wrong (not merely that it's wrong). In other words: what is the flaw in the naive logic? Btw, the problem is not the relation $\Delta v_x=\Omega h$ – that's a mere fact and hence not wrong – but it's use. Jun 4, 2016 at 9:23
• I mean, if you work out the kinematics you get $\Delta v_x= 2\Omega h$, right? Jun 4, 2016 at 21:25
• In any case, the reason your second answer is not right is because of using $x=T\Delta v$. $\Delta v=\Delta x/\Delta t$ is an expression for the average velocity. But here, the acceleration in the x-direction is not constant so that expression is not true any more. You must use $x(t)$ given by the expression in my answer. Jun 4, 2016 at 21:26
• I think we misunderstand each other regarding the meaning of $\Delta v$. I meant this to be the difference in the azimuthal/horizontal velocity between different heights and not $\Delta x/\Delta t$. Jun 5, 2016 at 18:24
• what does that even mean? I mean, obviously symbols can mean anything you want them to, but do you mean "the horizontal velocity as a function of height"? When I work that out (by eliminating the time) I get $v_x(y)=2\Omega (h-y)(1-v_0\sqrt{2/(g(h-y))})$, so you simply get $\Delta v_x=v_x(0)-v_x(h)=2\Omega (h-v_0\sqrt{2h/g})$. Ok, I guess if $v_0\sqrt{2/(gh)}<<1$ then you're only off by a factor of 2, but in any case, it's not a good approximation. Jun 6, 2016 at 15:27