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Can rotational force of the earth have any effect on flying projectiles?

  1. Feb 26, 2008 #1
    Well we know guys that the Earth spins to the east. So i was thinking, if you launch projectiles (lets say golf balls) of equal speed and angle to the east and launch some to the west, will the ones thrown to the east land farther from the launching point compared to the ones thrown to the west. The idea here is that since the earth is spinning to the east, it will give an extra force exerted on the ball going to the east. Contrastingly, the balls launched to west will have to deal with the opposite force of earth acting to the east which will cause it to land closer to the origin. Opinions?
     
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  3. Feb 26, 2008 #2

    mjsd

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    do u know why the shuttle launch site is situated in Florida? Well, it is closer to the equator!
     
  4. Feb 26, 2008 #3
    uhmmm, what does that mean. So do u think the golf ball thing might have some results?
     
  5. Feb 26, 2008 #4

    arildno

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    Indeed it does.
    Your effect will be most strongly present the closer you are to equator.
    That's why Florida is used rather than, say, Alaska.
     
  6. Feb 26, 2008 #5
    Oh great! I am planning to do this experiment here in singapore, just 1 degree north of the equator! yay!
     
  7. Feb 26, 2008 #6
    I dont know if it gives a force as such (neglect air resistance), but rather that the time the ball is in the air, the earth has rotated, so it lands in a different spot than it would have if the earth were not rotating.
     
  8. Feb 26, 2008 #7
    In the earth's frame, assuming we're in the northern hemisphere, we will have two forces in addition to gravity... the centrifugal force, and the Coriolis force. The centrifugal force is directed away from the axis of rotation, and it's effect is that the direction of g is not exactly towards the centre, but slightly south of it. The coriolis force is directed to the right of the velocity vector. What effect you get depends of the direction you're throwing the ball.
     
  9. Feb 26, 2008 #8
    Isn't the centrifugal and coriolis forces fictitious?
     
  10. Feb 26, 2008 #9

    Danger

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    The Coriolis force is real, but the effect is negligible for any small-scale system (such as water going down a sink). It starts to make itself known in large storms or ocean currents.
     
  11. Feb 26, 2008 #10
    @ Danjroman: fictitious or not, if you're on the rotating earth, you cannot neglect them. They're fictitious in the sense that from a frame outside the earth, it'll appear as though you're correcting for unreal forces. On the earth, however, the forces will appear to be real. As Danger said, the Coriolis force is the reason why hurricanes rotate.

    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.
     
  12. Feb 27, 2008 #11
    thanks a lot rahuldandekar but im just a grade 11 student in high school. Do u think u can give me an easy step by step way in calculating for this with what the level of physics i should have. thanks
     
  13. Feb 27, 2008 #12

    Dale

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    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.

    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.
     
  14. Feb 27, 2008 #13

    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. :redface:
     
  15. Feb 27, 2008 #14

    arildno

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    Often, they are called auxiliary forces instead (at least in Norwegian)
     
  16. Feb 27, 2008 #15
    OK. To make the discussion easier (and as was mentioned earlier), I'll assume you're on the equator.

    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 won't be increased. Why? For the same reason that a ball throw vertically from a moving car lands in the car again. Think about it.

    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.
     
  17. Feb 28, 2008 #16
    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 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.
     
  18. Mar 4, 2008 #17
    Dollars to donuts the Abrams M1A1 corrects for a rotating Earth in calculating a firing solution.
     
  19. Mar 25, 2008 #18
    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 higher for the same energy input even though relative to earth's surface the initial launch speed looks the same.

    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
     
    Last edited: Mar 25, 2008
  20. Mar 25, 2008 #19

    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.
     
  21. Mar 25, 2008 #20
    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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