Coriolis Acceleration and Inertial Frame

[Red_CCF]

Hi

In my textbook they were doing a derivation of the acceleration of a projectile flying on Earth. Although they used the center of the Earth as the frame of reference, they ended up with the linear combination of 3 quantities for the acceleration of the projectile, one of which they said was coriolis acceleration. However, I thought that coriolis force only exists in non-inertial frames and the center of the Earth should be an inertial frame shouldn't it?

Another question I have is that my physics text states that Earth is considered approximately an inertial frame because it's centripetal acceleration around the Sun is insignificant compared to Earth's gravity (almost direct quote). I don't really get this statement, can someone explain this?

Thanks.

 

[Petr Mugver]

The acceleration due to rotation around the Sun is

$$\omega^2R=\frac{4\pi^2R}{T^2}$$

where T = 1 year and R = distance Sun-Earth = 150 million Km. You get approx 0.0002 g.

 

[diazona]

To specify a reference frame, you need to specify not only where its origin is, but also which directions its coordinate axes are pointing. So for example, just saying that a reference frame is attached to the center of the Earth doesn't tell you which reference frame it is. You also have to specify whether the frame rotates with the Earth, or whether one axis always points toward the Sun, or to the center of the galaxy, etc. There are an infinite number of possibilities.

I believe that your textbook is using a reference frame which rotates with the Earth. This is an approximately inertial reference frame, but not exactly. It is possible to treat it as an inertial frame, and in doing so you'd get results that are pretty close to being correct, but not exact. In your textbook they have chosen to be more precise than that.

 

[Red_CCF]

To specify a reference frame, you need to specify not only where its origin is, but also which directions its coordinate axes are pointing. So for example, just saying that a reference frame is attached to the center of the Earth doesn't tell you which reference frame it is. You also have to specify whether the frame rotates with the Earth, or whether one axis always points toward the Sun, or to the center of the galaxy, etc. There are an infinite number of possibilities.

I believe that your textbook is using a reference frame which rotates with the Earth. This is an approximately inertial reference frame, but not exactly. It is possible to treat it as an inertial frame, and in doing so you'd get results that are pretty close to being correct, but not exact. In your textbook they have chosen to be more precise than that.

Yes I think that was it. They defined the axis x,y,z which are non-rotational but in their analysis they used a unit vector that rotate on the x-y plane, which would mean the actual axis the analysis was done on was rotating and thus non-inertial.

Here is a summary of the solution they gave: (From Advanced Mathematics and Engineering 9th Ed. no copyright intended)

Let x, y, z be a fixed Cartesian coordinate system in space, with unit vectors i, j k, in the direction of the axis. Let the earth, together with a unit vector b, be rotating about the z-axis with angular speed w > 0:
b(t) = cos wt i + sin wt j
Let the projectile be moving on the meridian whose plane is spanned by b and k with constant angular speed µ > 0. Then its position vector is:

r(t) = R cos µt b(t) + R sin µt k Where R is the radius of the Earth
After some calculations and simplifications, the acceleration of the projectile is:

a (t) = R cos µt b’’ - 2 µRsin µt b’ - µ2r

The middle term was labelled as Coriolis acceleration and is described as the interaction of the other two rotations. On the Northern Hemisphere acor has the direction of –b’ opposite the rotation of the Earth for sin µt greater than 0 (t and µ greater than 0). The magnitude of the Coriolis acceleration is maximum at the North Pole and 0 at the equator. The projectile experiences a force –macor where m is the mass of the projectile which tends to let B deviation from the meridian to the right on the Northern Hemisphere.

There's a couple of other things I was confused about though. For instance, they mention that the projectile would move right at the North Hemisphere due to a force that was -macor, why is this?

Also, the coriolis force is supposedly 0 at the equator and greatest at the Northern Hemisphere ,but this would only be true if the projectile was launched at the equator and travels through the Northerm Hemisphere first isn't it?

 

[Cleonis]

There's a couple of other things I was confused about though. For instance, they mention that the projectile would move right at the North Hemisphere due to a force that was -macor, why is this?

For visualizing try the following Java simulation: http://www.cleonis.nl/physics/ejs/ballistics_and_orbits_simulation.php" .
You can launch from any latitude, in any direction, with any elevation, with any velocity. The trajectory is computed with numerical analysis and displayed in 3D view. The simulation presents side-by-side two views, one as seen from an inertial point of view, one as seen from a co-rotating point of view.

 

[Red_CCF]

For visualizing try the following Java simulation: http://localhost/physics/ejs/ballistics_and_orbits_simulation.php" .
You can launch from any latitude, in any direction, with any elevation, with any velocity. The trajectory is computed with numerical analysis and displayed in 3D view. The simulation presents side-by-side two views, one as seen from an inertial point of view, one as seen from a co-rotating point of view.

The link doesn't work

 

[Cleonis]

The link doesn't work

My apologies, I replaced the link in the original message, it's correct now, checked it.