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Physics LibreTexts

12.11: Free Motion on the Earth

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The calculation of trajectories for objects as they move near the surface of the earth is frequently required for many applications. Such calculations require inclusion of the noninertial Coriolis force.

10.11.1.PNG
Figure 12.11.1: Rotating frame fixed on the surface of the Earth.

In the frame of reference fixed to the earth’s surface, assuming that air resistance and other forces can be neglected, then the acceleration equals

a=geff2ω×v

Neglect the centrifugal correction term since it is very small, that is, let geff=g. Using the coordinate axis shown in Figure 12.11.1, the surface-frame vectors have components

ω=0^i+ωcosλ^j+ωsinλ^k

and

geff=g^k

Thus the Coriolis term is

2ω×v=2|^i^j^k0ωcosλωsinλ˙x˙y˙z|=2[(ω˙zcosλω˙ysinλ)^i+(ω˙xsinλ)^j(ω˙xcosλ)^k]

Therefore the equations of motion are

m¨r=mg^k2m[^i(˙zωcosλ˙yωsinλ)+^j˙xωsinλ^k˙xωcosλ]

That is, the components of this equation of motion are

¨x=2ω(˙zcosλ˙ysinλ)¨y=2ω˙xsinλ¨z=g+2ω˙xcosλ

Integrating these differential equations gives

˙x=2ω(zcosλysinλ)+˙x0˙y=2ωxsinλ+˙y0˙z=gt+2ωxcosλ+˙z0

where ˙x0,˙y0,˙z0 are the initial velocities. Substituting the above velocity relations into the equation of motion for ¨x gives

¨x=2ωgtcosλ2ω(˙z0cosλ˙y0sinλ)4ω2x

The last term 4ω2x is small and can be neglected leading to a simple uncoupled second-order differential equation in x. Integrating this twice assuming that x0=y0=z0=0, plus the fact that 2ωgtcosλ and 2ω(˙z0cosλ˙y0sinλ) are constant, gives

x=13ωgt3cosλωt2(˙z0cosλ˙y0sinλ)+˙x0t

Similarly,

y=(˙y0tω˙x0t2sinλ)z=12gt2+˙z0t+ω˙x0t2cosλ

Consider the following special cases;

Example 12.11.1: Free fall from rest

Assume that an object falls a height h starting from rest at t=0, x=0, y=0, z=h. Then

x=13ωgt3cosλ

y=0

z=h12gt2

Substituting for t gives

x=13ωcosλ8h3g

Thus the object drifts eastward as a consequence of the earth’s rotation. Note that relative to the fixed frame it is obvious that the angular velocity of the body must increase as it falls to compensate for the reduced distance from the axis of rotation in order to ensure that the angular momentum is conserved.

Example 12.11.2: Projectile fired vertically upwards

An upward fired projectile with initial velocities ˙x0=˙y0=0 and ˙z0=v0 leads to the relations

x=13ωgt3cosλωt2v0cosλ

y=0

z=12gt2+v0t

Solving for t when z=0 gives t=0, and t=2v0g. Also since the maximum height h that the projectile reaches is related by

v0=2gh

then the final deflection is

x=43ωcosλ8h3g

Thus the body drifts westwards.

Example 12.11.3: Motion parallel to Earth's surface

For motion in the horizontal xy plane the deflection is always to the right in the northern hemisphere of the Earth since the vertical component of ω is upwards and thus 2ω×v points to the right. In the southern hemisphere the vertical component of ω is downward and thus 2ω×v points to the left. This is also shown using the above relations for the case of a projectile fired upwards in an easterly direction with components ˙x0,0,˙z0. The resultant displacements are

x=13ωgt3cosλωt2˙z0cosλ+˙x0t

Similarly,

y=ω˙x0t2sinλ

z=12gt2+˙z0t+ω˙x0t2cosλ

The trajectory is non-planar and, in the northern hemisphere, the projectile drifts to the right, that is southerly.

In the battle of the River de la Plata, during World War 2, the gunners on the British light cruisers Exeter, Ajax and Achilles found that their accurately aimed salvos against the German pocket battleship Graf Spee were falling 100 yards to the left. The designers of the gun sighting mechanisms had corrected for the Coriolis effect assuming the ships would fight at latitudes near 50 north, not 50 south.


This page titled 12.11: Free Motion on the Earth is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform.

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