Calculus of One and Several Variables

17.7 Coriolis Force on the Rotating Earth

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×

because it is so small whenever you are considering motion near some point on the earth’s surface. To see this, note ω = 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.

If the acceleration a is due to gravity, then

aB = a− Ω × (Ω × R )− 2Ω× vB =

◜-----------≡◞g◟-----------◝ GM (R + rB) − --------3---− Ω× (Ω× R )− 2Ω × vB ≡ g− 2Ω × vB. |R + rB|

Note that

Ω× (Ω× R ) = (Ω ⋅R)Ω − |Ω|2R

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

,j,k next.

Letting

be the usual spherical coordinates of the point p 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

( ) ρsin (ϕ )cos(θ) ( ρsin (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∗

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

,j,k 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

◜-k◞∗◟-◝ ◜------◞i◟------◝ ◜----◞j◟----◝ ◜-------k◞◟------◝ ( 0) ( cos(ϕ)cos(θ)) ( − sin(θ) ) ( sin (ϕ )cos (θ) ) ( 0) = a( cos(ϕ)sin(θ)) + b( cos(θ) ) + c( sin (ϕ)sin(θ) ) (17.20) 1 − sin (ϕ) 0 cos(ϕ )

(17.20)

The solution is a = −sin

,b = 0, and c = cos.

Now the Coriolis acceleration on the earth equals

( ) ◜--------k◞∗◟--------◝ 2 (Ω × vB ) = 2ω|( − sin (ϕ)i+0j+ cos(ϕ)k|) × (x ′i+y′j+z ′k ).

This equals

′ ′ ′ ′ 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

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 = −

− Ω× as explained above. The term Ω× is pretty small and so it will be neglected. However, the Coriolis force will not be neglected.

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

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.