Factor of 2 error in angular momentum / Coriolis force calculation

Question

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.

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Attempt

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.

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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?

Answer 1

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 $a_p$

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

Differentiating:

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

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

Answer 2

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.