The Earth Presses Every Train Sideways

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

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 along the polar axis; $\vect\rho$ is the perpendicular from that axis to the train, of magnitude $R\cos\varphi$ and not along the local vertical; and $\vect v'$, written simply $\vect v$ from here on since no other velocity appears, is the train’s velocity relative to the ground, horizontal. Both fictitious terms are wanted, and each is needed as a vector rather than a magnitude, so the work below resolves them in a local east-north-up frame. Part (a) asks only for the sideways (east-west) part of the Coriolis term, for which the surviving factor is the vertical component $\omega\sin\varphi$ crossed with a meridional $\vect v'$; part (b) asks for the velocity that makes the two fictitious terms of (1) cancel as vectors.

At latitude $\varphi$, set up local unit vectors east $\vect e$, north $\vect n$ and up $\vect u$ (radially outward), a right-handed set with $\vect e\times\vect n=\vect u$. The angle $\varphi$ in the figure is exactly the tilt between this local frame and the equatorial one, so the Earth’s own spin vector, purely along the polar axis, splits into a horizontal (northward) and a vertical (upward) part.

The Coriolis term of (1), written with the cross product the other way round, is $\vect F_{Cor}=-2m\vect\omega\times\vect v$. The right-handed frame’s three cyclic products give every cross product the expansion below needs.

For a train on the meridian, $v_e=0$ and $v_n=\pm v$. Equation (2) then leaves only the east component, evaluated with the Earth’s mean solar angular velocity $\omega=2\pi/86400\text{ s}\approx7.27\times10^{-5}$ rad/s (the sidereal value changes this by under half a percent, immaterial at this precision). Numerically,

Whether $v_n$ is positive (heading north) or negative (heading south), the deflection is to the right of the direction of travel: east of a train heading north is to its right, and west of a train heading south is to its right too, so the Earth’s reaction always presses the same rail. By Newton’s third law the train presses on that rail with the force above.

Writing $\vect X$ for the fixed global direction perpendicular to the spin axis (through the train’s own meridian), $\vect\rho=R\cos\varphi\,\vect X$ and the centrifugal term of (1) is $\vect F_{cf}=m\omega^2R\cos\varphi\,\vect X$. The same rotation that resolved $\vect\omega$ in Step 2 resolves $\vect X$ into the local frame, $\vect X=\cos\varphi\,\vect u-\sin\varphi\,\vect n$.

Adding (2) and (4) and setting each of the three components to zero,

The east equation forces $v_n=0$ (a due east-west train), and the up equation then gives $v_e$ directly.

Substituting that value back into the north equation confirms it is satisfied automatically, so this single velocity kills all three components at once and not merely the vertical one that fixed it.

The negative sign means westward; evaluating the constants fixes the numerical value.

The train must head due west, at half the ground’s own rotation speed at that latitude.