Irodov 1.117: Every Falling Object Lands East – Amit & Deepali Sinha Foundation

Problem Irodov 1.117

At a point on the Earth’s equator, a body is released from rest, relative to the Earth, at height $h$ above the ground, and falls freely with air drag neglected. Find the horizontal distance, and the compass direction, by which the body’s landing point is displaced from the point directly below its release point.

In the vertical-east plane: the body is released from rest at the marked point, a height h above the ground, with the dashed plumb line below it. At the equator ω points north, into the page, hence the crossed-circle symbol. — In the vertical-east plane: the body is released from rest at the marked point, a height $h$ above the ground, with the dashed plumb line below it. At the equator $\omega$ points north, into the page, hence the crossed-circle symbol.

Given

$h = 500$ m
body released from rest relative to the Earth, at the equator

Find

The landing point’s horizontal displacement, 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, so $\vect\omega$ is the Earth’s rotation, which at the equator lies along the local north, horizontal. The real force $\vect F$ is the body’s weight (drag is neglected), and $\vect\rho$ points from the polar axis straight up through the release point, so the centrifugal term is vertical here and merely joins gravity in the measured value of $g$ : it cannot deflect the body sideways at all. The deflection is entirely the Coriolis term, with $\vect v'$ the fall velocity relative to the Earth, written simply $\vect v$ from here on since no other velocity appears. It is vertically downward and therefore perpendicular to the northward $\vect\omega$ , so $\sin\theta=1$ and the term has magnitude $2mv\omega$ pointing due east.

Step 2The dominant fall is unaffected by the small deflection

To leading order, the body falls straight down with the ordinary constant acceleration $g$ , reaching speed $v(t)=gt$ downward at time $t$ after release.

Step 3The Coriolis acceleration, eastward and growing with time

Dividing the Coriolis term of (1) by $m$ gives the deflecting acceleration. At the equator $\vect\omega$ is horizontal, pointing along the local north; the body’s velocity $\vect v$ is vertical (downward). Their cross product is horizontal, along the local east.

\vect a_{Cor} = -2\vect\omega\times\vect v = 2\omega v\,\hat{\vect e} = 2\omega gt\,\hat{\vect e}

Step 4Integrating twice for the eastward drift

Integrating the eastward acceleration once gives the eastward velocity, and again gives the eastward displacement, both starting from zero.

v_e(t) = \int_0^t 2\omega g t'\,dt' = \omega g t^2

x(t) = \int_0^t \omega g t'^2\,dt' = \frac{\omega g t^3}{3} \tag{2}

Step 5Substituting the time of fall

The fall, exaggerated for clarity: the curve bows east of the plumb line as the body drops the height h under the eastward Coriolis force F_Cor (drawn at one representative point), landing the distance x from it. — The fall, exaggerated for clarity: the curve bows east of the plumb line as the body drops the height $h$ under the eastward Coriolis force $\vect F_{Cor}$ (drawn at one representative point), landing the distance $x$ from it.

The time to fall the height $h$ , from $h=gt^2/2$ , is $t=\sqrt{2h/g}$ . Substituting into (2),

x = \frac{\omega g}{3}\left(\frac{2h}{g}\right)^{3/2} = \frac{\omega g}{3}\cdot\frac{2h}{g}\sqrt{\frac{2h}{g}}

Numerically,

x = \dfrac23\times7.27\times10^{-5}\times500\sqrt{\dfrac{2\times500}{9.8}}\text{ m}

x = \dfrac{2}{3}\omega h\sqrt{\dfrac{2h}{g}} \approx 0.24\text{ m}

(3)

The body lands that far east of the plumb line.

Check

Dimensionally, $\omega h\sqrt{h/g}$ has units $(\text{s}^{-1})(\text{m})(\text{s})=\text{m}$ , a length, as required. As $\omega\to0$ (a non-rotating Earth), $x\to0$ and the body falls straight down, and as $h\to0$ there is no time to deflect either, both as they must.

Remark

The deflection grows as $x\propto t^3$ in the fall time $t$ , not merely $x\propto t$ as a sideways drift at some fixed rate would give: the Coriolis acceleration is not constant during the fall but grows linearly with the ever-increasing fall speed, so the eastward velocity it produces grows as $t^2$ and the eastward displacement as $t^3$ , a longer drop deflecting disproportionately more than a fixed-rate drift ever could.