Irodov 1.111: The Rifle That Cannot Aim True North – Amit & Deepali Sinha Foundation

Problem Irodov 1.111

A rifle at latitude $\varphi=60^\circ$ fires a bullet horizontally due north, aimed at a vertical line on a target a horizontal distance $s=1.0\text{ km}$ away, with muzzle speed $v=900\text{ m/s}$ . Air drag is negligible, and the only effect considered besides gravity is the Earth’s rotation. Find the sideways distance $h$ by which the bullet misses the vertical aiming line, and the compass direction of the miss.

The Earth's angular velocity ω, along the polar axis, resolved at the rifle's latitude φ into a component ωφ along the local vertical Z and a component ωφ along the local horizontal, northward direction N. — The Earth’s angular velocity $\vect\omega$ , along the polar axis, resolved at the rifle’s latitude $\varphi$ into a component $\omega\sin\varphi$ along the local vertical $Z$ and a component $\omega\cos\varphi$ along the local horizontal, northward direction $N$ .

Given

$\varphi=60^\circ$
$v=900\text{ m/s}$
$s=1.0\text{ km}$

Find

h

and its compass direction

Solution

Step 1Reference: the equation of motion in a rotating frame

In a reference frame rotating with constant angular velocity $\vect\omega$ relative to an inertial frame, a particle of mass $m$ obeys

m\vect w' = \vect F + m\omega^{2}\vect\rho + 2m\,(\vect v'\times\vect\omega)

(1)

where $\vect w'$ and $\vect v'$ are the particle’s acceleration and velocity measured in the rotating frame, $\vect F$ is the resultant of the real (interaction) forces, and $\vect\rho$ is the position vector drawn perpendicularly from the rotation axis to the particle. The term $m\omega^{2}\vect\rho$ is the centrifugal force: it depends only on where the particle is and points away from the axis. The term $2m(\vect v'\times\vect\omega)$ is the Coriolis force: it depends only on how the particle moves, is perpendicular to both $\vect v'$ and $\vect\omega$ , and has magnitude $2mv'\omega\sin\theta$ , where $\theta$ is the angle between $\vect v'$ and $\vect\omega$ . Both vanish in an inertial frame ( $\omega=0$ ), and the Coriolis force does no work, since it is always perpendicular to $\vect v'$ .

In this problem the rotating frame is the Earth itself, so $\vect\omega$ is the Earth’s rotation about the polar axis and $\vect v'$ is the bullet’s velocity relative to the ground, written simply $\vect v$ where its direction matters: horizontal, northward, of magnitude $v$ . The real forces $\vect F$ are gravity alone (drag is neglected), and the centrifugal term is vertical to a very good approximation, so it is already absorbed into the local plumb vertical and the measured value of $g$ and deflects nothing sideways. The whole sideways miss therefore comes from the Coriolis term, and of the two components into which $\vect\omega$ splits at latitude $\varphi$ it is the vertical one, $\omega\sin\varphi$ , that crosses a horizontal $\vect v'$ into a horizontal deflection.

Step 2The vertical component of the Earth’s rotation

From the first figure, at latitude $\varphi$ the component of $\vect\omega$ along the local vertical is

\omega_z = \omega\sin\varphi.

Step 3The horizontal Coriolis acceleration

Dividing the Coriolis term of (1) by $m$ gives the bullet’s sideways acceleration. With $\vect v'$ horizontal (northward, magnitude $v$ ) and only the vertical component $\omega_z$ of $\vect\omega$ contributing a horizontal deflection, the angle between the two is $90^\circ$ there and the magnitude of the sideways acceleration is

a_c = 2\omega_z v = 2\omega v\sin\varphi

(2)

directed to the right of the direction of travel in the northern hemisphere, i.e. eastward for a bullet flying north.

Step 4The sideways deflection at the target

What this step works out, in plan view from above: the bullet, aimed along the dashed north line ON, follows the solid path instead and reaches the target's range a distance h east of that aiming line. — What this step works out, in plan view from above: the bullet, aimed along the dashed north line $ON$ , follows the solid path instead and reaches the target’s range a distance $h$ east of that aiming line.

The flight time to cover the range $s$ at the (essentially unchanged) forward speed $v$ is $t=s/v$ . The sideways deflection built up from rest under the constant acceleration (2) is

\begin{align*}h &= \frac{1}{2}a_c t^2 \\ &= \frac{1}{2}(2\omega v\sin\varphi)\left(\frac{s}{v}\right)^2\end{align*}

h = \dfrac{\omega s^2\sin\varphi}{v}

(3)

Step 5Numerical value

The Earth’s sidereal angular velocity is $\omega=2\pi/(86164\text{ s})$ , which evaluates to

\omega = 7.29\times10^{-5}\text{ s}^{-1}.

With $\varphi=60^\circ$ , $s=1000\text{ m}$ , $v=900\text{ m/s}$ , substituting into (3),

h = \frac{(7.29\times10^{-5}\text{ s}^{-1})(1000\text{ m})^2\sin60^\circ}{900\text{ m/s}} = 0.0702\text{ m}

h = \dfrac{\omega s^2\sin\varphi}{v} \approx 7\text{ cm}

(4)

The deflection is to the east of the aiming line.

Check

Dimensionally, $\omega s^2/v$ is $(\text{s}^{-1})(\text{m}^2)/(\text{m/s})=\text{m}$ , a length, as required. At the equator ( $\varphi=0$ ) the formula gives $h=0$ : a bullet fired due north along the equator experiences no eastward Coriolis deflection, since there the local vertical is entirely perpendicular to $\vect\omega$ and $\omega_z=0$ , consistent with Step 2.

Remark

The deflection is eastward specifically because the bullet travels north in the northern hemisphere; the same rule (deflection to the right of the direction of travel) is why a north-flowing river there erodes its eastern bank more than its western one (Baer’s law). A bullet fired due south from the same rifle would deflect west of its aim line by the identical magnitude, since reversing $\vect v$ in $-2\vect\omega\times\vect v$ reverses the deflection’s sense along with it.