24.6. CORIOLIS FORCE ON THE ROTATING EARTH 459

Note thatΩ×(Ω×R) = (Ω ·R)Ω−|Ω|2R

and so g, the acceleration relative to the moving coordinate system on the earth is not di-rected exactly toward the center of the earth except at the poles and at the equator, althoughthe components of acceleration which are in other directions are very small when com-pared 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 theacceleration relative to a coordinate system moving with the earth’s surface.

aB = g−2(Ω×vB)

While the vector Ω is quite small, if the relative velocity, vB is large, the Coriolis acceler-ation 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 surfacetaken 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  ρ sin(φ)cos(θ)ρ sin(φ)sin(θ)

ρ cos(φ)

 .

It follows,i= cos(φ)cos(θ)i∗+ cos(φ)sin(θ)j∗− sin(φ)k∗

j =−sin(θ)i∗+ cos(θ)j∗+0k∗

andk= 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 shownearlier, ωk∗ is the angular velocity vector Ω. To simplify notation, I will suppress thedependence of these vectors on t. Thus the following equation needs to be solved for a,b,cto find k∗ = a i+bj+ ck

k∗︷ ︸︸ ︷ 001

= a

i︷ ︸︸ ︷ cos(φ)cos(θ)cos(φ)sin(θ)−sin(φ)

+b

j︷ ︸︸ ︷ −sin(θ)cos(θ)

0

+ c

k︷ ︸︸ ︷ sin(φ)cos(θ)sin(φ)sin(θ)

cos(φ)

 (24.23)

The solution is a =−sin(φ) ,b = 0, and c = cos(φ) .Now the Coriolis acceleration on the earth equals

2(Ω×vB) = 2ω

 k∗︷ ︸︸ ︷−sin(φ) i+0j+ cos(φ)k

× (x′ i+ y′ j+ z′k).

This equals2ω[(−y′ cosφ

)i+(x′ cosφ + z′ sinφ

)j−

(y′ sinφ

)k]. (24.24)

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′ sinφ

)j−

(y′ sinφ

)k]