As shown above, on the rotating earth, Ω is a constant and so 17.17 reduces
to
a = a (t)+ 2(Ω(t)× v (t))+ Ω (t)× (Ω (t)× r(t)) (17.18)
B B
(17.18)
Since r_{B} + R = r,
a = a − Ω× (Ω× R )− 2Ω × v − Ω× (Ω × r ). (17.19)
B B B
(17.19)
In this formula, you can totally ignore the term Ω×
(Ω × rB)
because it is so small
whenever you are considering motion near some point on the earth’s surface. To see this,
note ω
seconds in a day
◜(24)◞(◟3600◝)
= 2π, and so ω = 7.2722 × 10^{−5} in radians per second. If you are
using seconds to measure time and feet to measure distance, this term is therefore, no
larger than
( )
7.2722× 10−5 2|rB|.
Clearly this is not worth considering in the presence of the acceleration due to
gravity which is approximately 32 feet per second squared near the surface of the
earth.
and so g, the acceleration relative to the moving coordinate system on the earth is not
directed exactly toward the center of the earth except at the poles and at the equator,
although the components of acceleration which are in other directions are very small
when compared with the acceleration due to the force of gravity and are often neglected.
Therefore, if the only force acting on an object is due to gravity, the following formula
describes the acceleration relative to a coordinate system moving with the earth’s
surface.
a = g− 2 (Ω × v )
B B
While the vector Ω is quite small, if the relative velocity, v_{B} is large, the Coriolis
acceleration could be significant. This is described in terms of the vectors i
(t)
,j
(t)
,k
(t)
next.
Letting
(ρ,θ,ϕ )
be the usual spherical coordinates of the point p
(t)
on the surface
taken with respect to i^{∗},j^{∗},k^{∗} the usual way with ϕ the polar angle, it follows the
i^{∗},j^{∗},k^{∗} coordinates of this point are
i = cos (ϕ) cos(θ)i∗ + cos(ϕ)sin(θ)j∗ − sin (ϕ )k∗
j = − sin (θ)i∗ + cos(θ)j∗ + 0k∗
and
k = sin(ϕ)cos(θ)i∗ + sin(ϕ)sin(θ)j∗ + cos (ϕ)k ∗.
It is necessary to obtain k^{∗} in terms of the vectors, i
(t)
,j
(t)
,k
(t)
because, as shown
earlier, ωk^{∗} is the angular velocity vector Ω. To simplify notation, I will suppress the
dependence of these vectors on t. Thus the following equation needs to be solved for a,b,c
to find k^{∗} = ai+bj+ck
′ ′ ′ ′
2ω[(− y cos ϕ)i+ (x cosϕ+ z sin ϕ)j− (ysinϕ)k]. (17.21)
(17.21)
Remember ϕ is fixed and pertains to the fixed point, p
(t)
on the earth’s surface.
Therefore, if the acceleration a is due to gravity,
aB = g− 2ω[(− y′cosϕ)i+ (x′cosϕ+ z′sin ϕ)j− (y′sinϕ)k ]
where g = −
GM-(R+r3B)
|R+rB|
− Ω×
(Ω × R)
as explained above. The term Ω×
(Ω ×R )
is
pretty small and so it will be neglected. However, the Coriolis force will not be
neglected.
Example 17.7.1Suppose a rock is dropped from a tall building. Where will itstrike?
Assume a = −gk and the j component of a_{B} is approximately
− 2ω(x′cosϕ+ z′sin ϕ).
The dominant term in this expression is clearly the second one because x^{′} will be small.
Also, the i and k contributions will be very small. Therefore, the following equation is
descriptive of the situation.
′
aB = − gk− 2zω sinϕj.
z^{′} = −gt approximately. Therefore, considering the j component, this is
2gtωsin ϕ.
Two integrations give
( 3 )
ωgt ∕3
sinϕ for the j component of the relative displacement at
time t.
This shows the rock does not fall directly towards the center of the earth as expected
but slightly to the east.