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Falling Eastward

This page is under review.

Let a test object be at an initial height h, small enough that the variation in the value of g (= 9.807 m/sec2) may be ignored. Calculate the eastward deflection due to the Coriolis effect if the object is allowed to free-fall from an initial state of rest to the earth's surface.

Let omega = earth's angular velocity due to rotation. Then, omega(cos l) is the component of the angular velocity tangent to the earth's surface at latitude l, and is the component of concern to this problem since it is perpendicular to the plane of the test object's motion6. The Coriolis acceleration due to this component of angular velocity is

2 vomega (cos l)

and is directed eastward. The term v is the velocity of the falling test object as a function of time. If the eastward deflection is assumed sufficiently small compared to h, we may approximate v and h as

v = gt
h = (1/2) gt2

Substituting 2. into 1. gives for the Coriolis acceleration

2 gt omega(cos l)

Integrating once give the eastward velocity as a function of time

vE = gt2 omega (cos l)

Integrating a second time gives the required deflection

xE = (1/3) gt3 omega (cos l)

Using 3., t3 = (2h/g)3/2, so that for the entire fall from height h,

xE = (2h)3/2 g-1/2 omega (cos l)

Let us now assume that the initial height h = 100 m, and l = 42 deg. With omega = 7.27 x 10-7 /sec, we find that at the moment of impact,

t = 4.52 sec
v = 44.3 m/sec
vE = 1.08 x 10-4 m/sec
xE = 1.63 x 10-4 m = .16 mm

Comparing vE and v, we also see that the deviation of the velocity from the vertical is

arctan (1.08 x 10-4/44.3) = 1.4 x 10-4 deg.

With h = 1000 m, l = 42 deg, we find

t = 14.3 sec
v = 140 m/sec
vE = 1.08 x 10-3 m/sec
xE = 5.16 x 10-3 m = 5.16 mm
with arctan (1.08 x 10-3/140) = 4.4 x 10-4 deg.


6. The deviation of the test object's velocity vector from the vertical is assumed small enough to be ignored; hence, its cross product with the vertical component of the earth's angular velocity is also small enough to be ignored.

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