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mjsd

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do u know why the shuttle launch site is situated in Florida? Well, it is closer to the equator!

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uhmmm, what does that mean. So do u think the golf ball thing might have some results?

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arildno

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Your effect will be most strongly present the closer you are to equator.

That's why Florida is used rather than, say, Alaska.

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Isn't the centrifugal and coriolis forces fictitious?

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Danger

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They may be introduce small corrections, but I think those are the the very effects the OP is asking for (ie, the effects of the rotation of the earth).

If you want to calculate the actual effect, try this: take the w-axis to be the z-axis. calculate Fcor = -2m(w x v) and Fcent = -m wx(wxr) using suitable axes. Then use F = ma and calculate the motion. You might want to refer some book on undergrad classical mech, which has a chapter on fictitious forces.

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Dale

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Isn't the centrifugal and coriolis forces fictitious?

The Coriolis force is a ficticious force. Ficticious forces are forces that are added to an accelerating reference frame in order to make Newton's first and second laws work correctly. Because of the earth's rotation you can consider a person on the earth to be in a rotating reference frame where the two ficticious forces are the centrifugal and Coriolis forces.The Coriolis force is real,

When working in a rotating reference frame you cannot ignore the ficticious forces any more than you can ignore gravity. In the rotating reference frame they can accelerate projectiles, do work, cause stress and strain, and basically everything else that you can think of a force doing. The only way that they do not behave as a real force is that they violate Newton's 3rd law, i.e. there is no equal and opposite reaction force to the Coriolis force.

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Danger

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Oops! I misunderstood what the term meant (I have no formal education). I thought he was suggesting that there is no such thing. Sorry.The Coriolis force is a ficticious force. Ficticious forces are forces that are added to an accelerating reference frame in order to make Newton's first and second laws work correctly.

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arildno

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Often, they are called auxiliary forces instead (at least in Norwegian)

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Then the centrifugal force (which points away from the rotational axis) is m*r*w^2 , where R is the radius of the earth, and w is the angular acceleration of the earth. At the equator, this points directly vertically at the surface, so all it does is make a tiny correction to gravity. we'll call that corrected value g (it'll actually be the gravitational g minus r*w^2).

The coriolis force, at the equator, is zero... so it simplifies the discussion somewhat. (The coriolis force depends as the cosine of the angle from the north pole. It is maximum at the poles, zero at the equator, and in opposite directions on the northern and southern hemispheres).

So, all we have is that gravity is slightly reduced. What will be the effect? Look at the formulae for projectile motion, both the maximum height and range are proportional to 1/g... so if g is reduced, both will be increased. R*w^2 is 3.39 x 10^-2, which is quite a bit smaller than g. The correction is therefore very small, about 0.3 % of g.

I dont think the rotation of the earth has any effect like the one you asked about, that is, if you throw a ball in the direction opposite to the earth's rotation, the distance traveled

There will also be some effect from the curvature of the earth, etc, but it'll be very small. All these effects will be negligible against air resistance.

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Ok, I was kind of wrong there. The rotating earth isn't exactly like a moving car, because, um, well, it rotates. There are some terms in the Coriolis force which depend on cos of the angle from the pole, and some which depend on sine. The sine terms do matter that the equator. (The cos terms are zero at the equator, and opposite on both hemispheres. )The Coriolis force, at the equator, is zero... so it simplifies the discussion somewhat. (The Coriolis force depends as the cosine of the angle from the north pole.

...

I dont think the rotation of the earth has any effect like the one you asked about, that is, if you throw a ball in the direction opposite to the earth's rotation, the distance traveledwon'tbe increased.

The sine terms give effects directed towards the west, and also vertically upwards. So, a body will be deflected towards the west, and there will be some further correction to gravity.

The west deflection depends on the vertical velocity, and the vertical Coriolis correction term depends on the eastward velocity.

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Dollars to donuts the Abrams M1A1 corrects for a rotating Earth in calculating a firing solution.

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Sorry to bump an old thread , but I think the simple answer to this one is yes. Throw something East it flies for longer simply because going east it has a higher orbit speed around the centre of gravity of Earth (you get some free from already present rotation of earth).

So it will fly

Talking about coriolis and centrifugal corrections just adds unnecessary complexity. The way to analyse it IMHO is to look from a point out in space. Not from one bound to the surface of the Earth.

..edit :.. I just realised the guy answered his own question in his original post. lol

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The coriolis force, at the equator, is zero...

The horizontal components of the coriolis force for objects that only move horizontally is zero. This is important for emptying bathtubs and for the weather (the reason you do not get hurricanes on the equator)

There is vertical movement, and a vertical components of the coriolis force.

In general the equation is [tex]F_c = -2m \omega \times v [/tex]

where \omega is the angular momentum vector of the earth, pointing in the direction of the north pole.

For movement on the equator

If you move north, the coriolis force is 0. [tex] \omega[/tex] is parallel with [tex]v[/tex]

if you move east, the coriolis force is directed upwards

if you move south, the coriolis force is 0

if you move west, the coriolis force is downwards

if you move up, the coriolis force is directed west.

if you move down, the coriolis force is directed east.

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Absolutely. Recall the big guns on WW-II battleships. To hit a target which is miles away they had to account for the coriolis force.

Pete

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There would be no difference in the distance the balls travelled on Earth.

Imagine you were on a moving conveyor belt.

Throw a ball in front of you in the direction of travell.

Throw another with the same force in the opposite direction.

Both balls will travell the same distance on the conveyor.

They will only appear to travell different distances to an observer not on the conveyor.

Or in the case of the golf balls from a reference point in space.

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Bet there are no variables when fireing a shell in direct east or west direction at the equator.

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