Coriolis acceleration of a projectile launched at the equator

In summary, the conversation discusses the Coriolis acceleration experienced by a projectile launched from the surface of a rotating body, specifically at the equator. The question is whether the Coriolis acceleration is zero at the equator or if it is still present when the projectile is launched at an angle relative to the surface. The standard formula for calculating the Coriolis acceleration suggests that there is still a deflection present. The conversation also mentions the interest in long-range ballistic travel and the difficulty in finding information specifically addressing this question.
  • #1
jk185
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I have a question about the Coriolis acceleration experienced by a projectile launched from the surface of a rotating body.

Say a ball is launched at 45 degree angle relative to the surface at some initial velocity v0. Let's further specify that the ball is launched due north from the equator (i.e. latitude = 0). I want to calculate the Coriolis acceleration (and its x, y, z components) that is experienced by the ball. Since the launch occurs at the equator, most would say that the Coriolis acceleration is zero, but is this true? Isn't the Coriolis acceleration only 0 at the equator since the launch angle is assumed to be 0 (i.e. the motion is parallel to the surface), meaning the velocity vector is parallel to the rotational vector? If a ball is launched at some angle relative the surface, the velocity vector is no longer parallel to the surface, meaning that there should actually be a Coriolis acceleration that is experienced by the ball.

Also, I am particularly interested in long-range ballistic travel.

Is there something that I am missing here? I have looked everywhere for an answer to this question but haven't found anything that specifically addresses this question.

Any help is appreciated here. Thanks!
 
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  • #4
jk185 said:
Isn't the Coriolis acceleration only 0 at the equator since the launch angle is assumed to be 0 (i.e. the motion is parallel to the surface), meaning the velocity vector is parallel to the rotational vector? If a ball is launched at some angle relative the surface, the velocity vector is no longer parallel to the surface, meaning that there should actually be a Coriolis acceleration that is experienced by the ball.
That seems to be what the standard formula for calculating the Coriolis acceleration ##a_C=2\vec{\Omega}\times\vec{v}## says, yes.
 
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  • #5
https://www.eugeneleeslover.com/USNAVY/FC-APPENDIX-B-8IN-55.html said:
1652974773254.png
Look at the Latitude 0 section and the columns for a target at 0 degrees (North) or 180 degrees (South). Look at the row for 30,000 yard range. A shot at those angles is deflected 42 yards left (west) when firing north and 42 yards right (west) when firing south. This is presumably due to Coriolis acting on the arching shot.
 
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1. What is the Coriolis acceleration of a projectile launched at the equator?

The Coriolis acceleration of a projectile launched at the equator is zero. This is because the equator is the only location on Earth where the Coriolis force is equal to zero.

2. How does the Coriolis acceleration affect the trajectory of a projectile launched at the equator?

The Coriolis acceleration does not affect the trajectory of a projectile launched at the equator. This is because the Coriolis force is perpendicular to the direction of motion and therefore has no component in the direction of the projectile's motion.

3. Is the Coriolis acceleration of a projectile at the equator the same as at other latitudes?

No, the Coriolis acceleration at the equator is zero, while at other latitudes it is non-zero. The magnitude of the Coriolis acceleration increases as the latitude increases, reaching its maximum at the poles.

4. Can the Coriolis acceleration of a projectile at the equator be measured?

Yes, the Coriolis acceleration can be measured using specialized equipment such as a Foucault pendulum or a rotating platform. However, the effect of the Coriolis acceleration at the equator is very small and may be difficult to detect.

5. How does the Coriolis acceleration at the equator relate to the rotation of the Earth?

The Coriolis acceleration at the equator is a result of the Earth's rotation. As the Earth rotates, objects on its surface are subject to a Coriolis force, which is due to the rotation of the Earth and the motion of the object. At the equator, the Coriolis force is zero because the object is moving parallel to the Earth's axis of rotation.

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