The problem

An object is dropped from a helicopter, which is at rest relative to the Earth rotating at $\Omega$ at height $h=500\text{ m}$ above the ground at the equator.

Without using the Coriolis force (i.e. working in an inertial frame outside of the Earth), calculate the displacement from the point directly below the helicopter and the object when it hits the ground.


We can conserve angular momentum, $L$, of the object: $$L=m(R+h)^2\Omega=m(R+h-\frac{1}{2}gt^2)^2\omega$$ where $m$ is its mass, $R$ is the radius of the Earth, $g$ is the vertical acceleration due to gravity, $t$ is the time ellapsed since the object is dropped and $\omega$ is the angular velocity of the object at time $t$.

Rearranging gives $$\omega=\Omega\frac{(R+h)^2}{(R+h-\frac{1}{2}gt^2)^2}$$

You can do one of two things from this point...

(a) You can identify $\text{d}\theta/\text{d}t=\omega$, and integrate to obtain $\theta$ and hence get the displacement $\Delta x = R\Delta\theta$ (after remembering of course to factor in the Earth itself rotating by $\Omega R t$). This gives the correct answer of $\sim24\text{ cm}$.

(b) You can identify $\omega=v(R+h-\frac{1}{2}gt^2)^{-1}$, and $v=\text{d}x/\text{d}t$. Then you can integrate to get $x$. This gives the incorrect answer of $\sim47\text{ cm}$, which is twice the correct answer.

My question

I do not understand why it is wrong to do method (b). Could someone give some intuition as to why you cannot do this?

  • $\begingroup$ If you do not consider the Coriolis force, the stone will not cover any horizontal distance. The stone is in the atmosphere and will also be at rest with respect to Earth's rotating frame of reference. $\endgroup$ – Sam Feb 3 at 14:45
  • $\begingroup$ I am ignoring air resistance effects here - due to the conservation of angular momentum, the object must increase its angular velocity as it approaches the ground. The Coriolis force is relevant when looking at the problem in the frame of the rotating Earth, however I do not want to do that here. $\endgroup$ – Garf Feb 3 at 15:33
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    $\begingroup$ The second method does not make sense to me because dx is calculated at different heights, and then the equation $\Delta x=R \Delta \theta$ loses its meaning $\endgroup$ – Wolphram jonny Feb 3 at 15:33
  • $\begingroup$ @Sam Are you saying that you need to take into account Coriolis forces when working in inertial frames? $\endgroup$ – user253751 Feb 3 at 15:44
  • $\begingroup$ The answer to the post deviation of free falling objects may help you? $\endgroup$ – Farcher Feb 3 at 15:59

I agree with you that working in the inertial frame is straightforward and appropriate here.

We agree on why there will be a displacement, but for the sake of completeness let me write it explicitly:
When the object is released to free motion the object is from that point on in a Kepler orbit around the Earth's center. The altitude of release is the apogee of that orbit. After about 10 seconds the object impacts the Earth. From release to impact the motion of the object is orbital motion.

During the descent of the object the Earth's gravity is doing work, increasing the angular velocity of the object. The faster the object descends, the larger the increase of angular velocity. So we need an expression that relates sideways acceleration with radial velocity.

As we know, this is a case where angular momentum is conserved. So: the time derivative of angular momentum is zero. I will write down an expression for the angular momentum, then I take the time derivative of that expression, and I apply the constraint that that time derivative is zero.

Some definitions:
Radial velocity $v_r$: velocity in the direction of the Earth's center (hence velocity perpendicular to the local surface.

Sideways velocity $v_p$: velocity perpendicular to the radial velocity.

Sideways acceleration $a_p$: acceleration perpendicular to the radial velocity.

The general convention is to use the greek uppercase Omega, '$\Omega$' for an overall angular velocity that is constant. Here I will use the lowercase omega '$\omega$' to denote angular velocity because it does change a little.

We can readily work out what the radial velocity will be as the object descends, what we need is an expression that will get us from $v_r$ to $a_p$

$$ \frac{d(\omega r^2)}{dt} = 0 $$


$$ r^2 \frac{d\omega}{dt} + \omega \frac{d(r^2)}{dt} = 0 $$

With the chain rule we obtain a term $\frac{dr}{dt}$ which is the $v_r$ part of what we need.

$$ r^2 \frac{d\omega}{dt} + 2 r \omega \frac{dr}{dt} = 0 $$

Dividing by 'r', and rearranging:

$$ r \frac{d\omega}{dt} = - 2 \omega \frac{dr}{dt} $$

On the left we have a term $\frac{d\omega}{dt}$; that is angular acceleration. If the height over which the object falls is very small compared to the total distance to the center then we can treat the $r$ as a constant and move it inside the differentation

$$ r \frac{d\omega}{dt} = \frac{dv_p}{dt} $$

In all we have derived this expression from the conservation of angular momentum:

$$ \frac{dv_p}{dt} = - 2 \omega \frac{dr}{dt} $$

The angular velocity of the object changes over time, but compared to the total angular velocity the change is small.

In all: to a good approximation we can treat the object as subject to a sideways acceleration described by the following expression:

  • $a_p$ acceleration component perpendicular to the radial direction
  • $v_r$ velocity component in radial direction

$$ a_p = -2\omega v_r $$

For emphasis: the above is a dynamical expression.
This expression describes a property of the motion with respect to the inertial coordinate system.

Interestingly, the above expression has exactly the same form as the expression for the Coriolis term in the equation of motion for motion with respect to a rotating coordinate system.

The two expressions have the same form, but different origins.

With the above expression in place the calculation can be performed.

This is done in the answer already linked to by Farcher, in a comment. Deviation of free falling objects

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  • $\begingroup$ This is a nice derivation of the Coriolis acceleration, but does this actually answer my question above, which is why is what I'm doing in (b) wrong? $\endgroup$ – Garf Feb 9 at 19:11
  • $\begingroup$ @Garf To capitalize on the constraint that the angular momentum of the object remains the same I specified that the time derivative of the angular momentum must equal zero. The derivation follows from that. In your question you mentioned conservation of angular momentum, but as far as I can tell you didn't actually use that as constraint. At the equator I do expect a value of 24 cm, but I cannot explain why one of your attempts arrived at that number. My best guess is that it is a fluke. To try and figure out why it happened to arrive at 24 cm is not worth the time and effort, it seems. $\endgroup$ – Cleonis Feb 9 at 19:31
  • $\begingroup$ I use conservation of angular momentum in the line $$L=m(R+h)^2\Omega=m(R+h-\frac{1}{2}gt^2)^2\omega$$ This does lead to the answer of $x=24$cm if I integrate $\omega$ as a function of $t$ before converting angles to distances - I don't think this is a fluke. My problem is that integrating $x=r\omega$ gives me the wrong answer by (it seems exactly) a factor of two, and I cannot figure out why. $\endgroup$ – Garf Feb 9 at 19:37

Your v is the horizontal velocity of the object, (R+h)Ω, which without friction does not change. Integration is not required. The horizontal distance x = vt, where t is the time of fall. Compare this with the horizontal distance moved by a point on the surface, R Ω t.

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  • $\begingroup$ v does change because of conservation of angular momentum $\endgroup$ – Wolphram jonny Feb 4 at 17:59

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