56 Kinematics and Dynamics of a Point Mass
of both frames are at the Earth center. Suppose that a point mass m is in
the Northern hemisphere and moves East along a parallel ( a circle cut on
the surface of the Earth by a plane perpendicular to the line through the poles).
Denote by fi the velocity of the point relative to the frame 0\ so that rx
is perpendicular to the meridian plane; see Figure 2.1.6. By (2.1.24), the
Coriolis force acting on the point mass is Fcor = —2mu)h x f.
Fig. 2.1.6 Motion Along a Parallel(a)c Assume that both ||ri|| and 9 stay (approximately) constant during
the travel and that there is zero propulsion force Fp, as after a missile has
been fired. Ignore air resistance.
(i) Show that the point mass is deflected to the South, and the magnitude of
the deflection isu\ri\ sin9t^2 , wheret is the time of travel. Hint: verify that
Fcor • 9 = — 2mw||ri||sin0) and so 2mw||ri||sin6| is the force pushing the point
mass to the South, (ii) Suppose that 9 = 41°, ||ri || = 1000 meters per second,
and the point mass travels 1000 kilometers. Verify that the point mass will
be deflected by about 50 kilometers to the South: if the target is due East,
the missile will miss the target if aimed due East. Hint: 2o>sin0 « 10~^4.
(b)A Compute the deflection of the point mass taking into account the
change of \ri\ and 9 caused by the Coriolis force.
We now summarize the effects of the Coriolis force on the motion of a
point mass near the Earth.
- The force is equal to zero on the equator and is the strongest on the poles.
- For motion in the Northern Hemisphere: