Falling Eastward
Problem:
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.
Solution:
Let
= earth's angular velocity due to rotation. Then, (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
v
(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
and
Substituting
2.
into 1.
gives for the Coriolis acceleration
2
gt (cos
l) |
4.
|
Integrating
once give the eastward velocity as a function of time
vE
= gt2
(cos l) |
5.
|
Integrating
a second time gives the required deflection
xE
= (1/3) gt3
(cos l) |
6.
|
Using
3.,
t3 = (2h/g)3/2, so that for the entire fall from
height h,
xE
= (2h)3/2 g-1/2
(cos l) |
7.
|
Let
us now assume that the initial height h = 100 m, and l = 42 deg. With
= 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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