# Effect of Earth's rotation on moving vehicles

The earth rotates at a rate of 460 m/second. I wish to predict the position and approach for landing an UAV (quadcopter) on the vehicle. I have a free-moving landing platform on vehicle that aligns itself with UAV.

• Compare friction on a car with friction on a bullet fired from a gun. Mar 13 at 6:59
• Or are you think if a car wants to travel in the same direction of rotation it has to do 460m/s plus the speed it needs to travel and to go the other way it just waits? Sadly my car is always where I left it… Mar 13 at 7:01
• what drift are you talking about? Mar 13 at 7:08
• what about the orbit around the sun and around the galaxy? Mar 13 at 7:14
• You have described the problem vaguely. But you have not included any details of the symbols, variables, diagrams, conventions, and equations that you have already worked out. Also, make the question specific.
– AJN
Mar 13 at 12:00

The answer to the effects of Earth's rotation of moving vehicles:

First, Earth doesn't rotate at a rate of 460 m/s. In fact, m/s is not the unit for the angular speed. It's like using Liters for mass. Earth's angular speed is 1/24 rounds per hour or 0.0000727 rad/s. The surface of the Earth at the equator moves at a speed of 460 meters per second which is not the ultimate parameter that shows how fast the Earth rotates. It sounds fast because the radius of Earth is huge.

There are two fictitious forces that an observer feels on a rotating body.

1. Centrifugal force (affects stationary and moving observers)
2. Coriolis force (only affects if the observer is moving and getting closer or farther away from the rotation axis of the Earth)

Those two forces don't affect our daily lives because the Earth rotates very slowly, and their magnitudes are directly related to Earth's rotation speed. However, there are exceptional cases where those two must be considered carefully.

The two forces can be calculated with the equations below.

$$\overrightarrow{F_{\text{coriolis}}} = -2 \overrightarrow{\Omega} \times (\overrightarrow{v})_{\text{rel}} \\ \overrightarrow{F_{\text{centrifugal}}} = - \overrightarrow{\Omega} \times (\overrightarrow{\Omega}\times(\overrightarrow{r})_\text{rel})$$

where $$(\overrightarrow{r})_\text{rel}$$ and $$(\overrightarrow{v})_\text{rel}$$ are relative position and velocity with respect to any point on the rotation axis of the Earth and $$\overrightarrow{\Omega}$$ is the angular velocity vector of Earth. Note that velocity due to rotation is not included in $$(\overrightarrow{v})_\text{rel}$$ because the observer rotates with the earth as well.

A detailed explanation of the equations can be found here: http://mechanicsmap.psu.edu/websites/12_rigid_body_kinematics/12-8_rotating_frame_analysis/rotating_frame_analysis.html)

• Why is a centrifugal force “fictitious”? Even on a bicycle that force acts… F1 drivers, jet pilots… Mar 13 at 12:36
• It depends on how you define the force. If you define force as the fundamental interaction between particles which must be one of the four fundamental forces of nature, then it is fictitious. Again, if you define the force as an interaction that holds Newton's third law, it is fictitious. But if you define force as a cause of every acceleration you observe, then you can claim that it is not fictitious. But the definition of force is very clear in mainstream physics, and these two forces turn out to be fictitious. Fictitious force is not a force, but it is just a side effect of inertia. Mar 13 at 15:37
• @SinaAtalay In a relativistic world, both gravity and the force exerted by a magnetic field on a moving charge are fictitious by the same standard. Do you routinely call them fictitious? Mar 15 at 12:44
• I am not an expert on electromagnetics or relativity. Therefore, I will skip the question. However, I was trying to point out that it's a matter of definition, and without defining what a force is, it helps no one to call them fictitious or not. The mainstream definition is clear and centrifugal, and Coriolis forces are always considered fictitious. So the useful question is not "Why is it fictitious?" but "What do you mean by a force?". Mar 15 at 12:58
• The stated formulae produce vectors not scalars, so those are the forces, not merely the magnitude of the forces. Mar 15 at 13:02