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Drop a ball

  1. Sep 4, 2004 #1
    Hi! I need som hints on what method to use when solving this problem:

    There is a point on the ground somwhere on the equator. 100 m right above the point a ball is dropped. Where will the ball land?

    I'm neglecting air resistance and assuming Earths' gravitational acceleration to be constant ~9.8 m/s^2. I've figured these two factors should be of importance:

    1. The ball has higher velocity than the point on the ground: it should, following simple rules, land in front of the spot (in the direction of Earths' rotation (eastward)).
    2. When the ball moves eastward the direction of Earths' gravitational pull upon it will turn westwards slightly as it is always pointing directly at the centre of the Earth. This will make the ball land further to the west (closer to the point) than predicted if only (1.) was to be considered.

    How should I proceed?

    Cheers.
     
  2. jcsd
  3. Sep 4, 2004 #2

    pervect

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    I'm a little skeptical about part 2 of your argument. But what you need at this point to proceed is, I think, a mathematical treatment of the Coriolis force.

    Try your textbook, and if it doesn't have such a treatment, try

    http://scienceworld.wolfram.com/physics/CoriolisForce.html

    If this isn't enough of a "hint", feel free to ask more questions - I'm trying to respect your request for a "hint" rather than a full answer.
     
  4. Sep 4, 2004 #3

    Tide

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    One only needs to invoke the Coriolis force if one is working in the rotating frame of the Earth. To me the real question is whether the ball has a "horizontal" velocity component to begin with.
     
  5. Sep 4, 2004 #4
    Pervect:

    But after what I understand the Coriolis force is a force acting on a body moving in a rotating reference frame. What I need to calculate is the balls' movment with respect to a rotating reference frame. The ball is rotating with the same angular velocity as Earth as long as it is held over the point. Once dropped it will try to maintain the same velocity and direction (a straight path) while gravity will try (and probably succeed) to pull the ball to the ground. Gravity is the only force acting on the ball. Or have I got the concept wrong?

    Why are you skeptic about 2? Gravity is commonly threated as a force parallell to the y-axis in euclidean space. That is not a problem when throwing pebbles across the schoolyard but this model breaks down if you want to calculate sattelite or planet orbits (or if you want to impress my physics teacher :smile: ). I'm just not sure what reference frame to choose and what maths to apply.

    Tide:

    Yes, I think it has initial horisontal velocity. Velocity = angular velocity * radius.
    As the ball is at a greater distance from the center of earth than the ground it's velocity is higher. Furthermore the point on the ground will curve away from the ball and I can't neglect earth's rotation during the fall.

    Thanks for the replies all of you!

    BTW it just hit me that we are neglecting Earths' movement around the sun and the sun's movement around the centre of our galaxy and our galaxys movement around... ehm, around... yes, around what? <--- One of the big questions of our time actually. Can somone answer this?

    Cheers.
     
    Last edited: Sep 4, 2004
  6. Sep 4, 2004 #5

    HallsofIvy

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    There is insufficient information. If you carry the ball upward on a tower, helicopter, or airplane, then the ball already has motion to the east associated with a point at that height:
    Calculate the distance from the center of earth to the initial height of the ball, find the circumference of a circle of that radius and divide by 24 hours (or 86400 seconds) to find the balls speed (to the east). Call that vball Now do the same with a point on the surface of the earth to calculate its speed to the east. Call that vearth. Find the time the ball takes to fall to the earth (that is independent of its horizontal motion). Call that T. The ball will fall a distance T(vball-veath) to the east of the point on earth directly below the initial position of the ball.

    On the other hand, if the ball suddenly "appears" above the earth with no circular motion to the center of the earth, the earth will rotate while the ball does not. The ball will land a distance Tvearth to the west of the point that was directly below the initial position of the ball.
     
  7. Sep 4, 2004 #6
    Well I guess I'd have to start from there. But that still leave me with the problem with gravity. The acceleration will still not be in the same direction during the entire fall. Would it help if I drew a picture to show what i mean?

    And let's suppose the ball was dropped from a tower.

    Cheers.
     
  8. Sep 4, 2004 #7

    pervect

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    There are two approaches you could take. The first approach is to do physics in a rotating coordinate system. This is the approach I was suggesting - but on second thought, it may not be the best approach. In this approach, you need to include "pseudo-forces", such as the centrifugal force, and the coriolis force. These are sometimes called general forces. If you choose to do the problem in a non-inertial coordiante system, you must use pseudo forces. If you choose to do the problem in a truly inertial coordinate system, you must _not_ use pseudoforces. Sometimes beginning students are cautioned strongly not to use non-inertial coordinate systems, if this is the case you should follow your instructors guidelines.

    If you do use the non-inertial rotating coordinate system, you can write down equations for the drop of the ball which say that the total acceleration on the ball is given by the force of gravity (downwards), the centrifugal force due to the rotation of the earth (upwards), and the coriolis force (sidewards, proportional to the velocity of the ball). You would have to integrate the coriolis force over time to get the total displacement.

    But it may be simpler to do the problem in an inertial coordinate system, as Tom Matheson suggested. This avoids the concept of pseudo-forces at all, which may be a bit confusing. When you do the problem in an inertial coordinate system, you are correct in saying that the only force on the ball is gravity. In this approach, you stick an x-y coordinate frame through the center of the earth. You then say that the earth's surface follows the equations of motion x = r*cos(w*t), y=r*sin(w*t). Note that the earth's surface is accelerating due to it's rotation. You then say that the ball moves with an initial velocity of vy=(r+h)*w. The total force downwards on the ball will be G*m*M/r^2. Note that this force will be greater than the weight of the ball at rest on the equator, due to the very small centripetal acceleration of the earth's surface. In theory, the dirction of the gravitatoinal force will always be towards the center of the earth, which is changing, which is what I think you were worried about. But the variation of the direction of the gravitational force in this problem is going to be very small, so you can safely approximate it as a constant acceleration in the 'x' direction, unless you are really ambitious. If you are really ambitious, you can solve the Kepler problem for the orbit of the body moving with the initial velocity specified to describe the motion of the body.
     
  9. Sep 4, 2004 #8
    So you think the easiest approach is to use a frame in which the earth is stationary in the centre but not rotating with the frame? Thanks, I'll go for that.
    I'm primary interested in solving this for a constant g. It may not be as accurate but I still find it challenging enough... The gravitational law can wait. Still what I'm really interested in is finding the effect of the altering direction of the gravitational acceleration.
     
  10. Sep 5, 2004 #9

    pervect

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    Here's something that can help you estimate the error.

    If you assume a central force law, you know that the angular momentum around the center of the earth is constant, because a central force never contributes to the angular momentum of a body. So if the earth's radius is R and the height above the ground is h, we expect that v_htop = (R+h)*w, and the angular momentum L = v_htop*(R+h), and that therefore horizontal velocity upon impact with the ground will be v_bot = L/R = v_htop(R+h)/R.

    By approximating gravity as not changing direction, we can say that the horizontal velocity is approximately constant, which is a good approxiimation as long as h << R, which is true in this case.

    You'll have to do a mildly messy intergal to find the exact average velocity in this case, because v_h is a function of height, and the height is changing quadratically with time, not linearly. This approach still isn't exact, because one has neglected to take into account the manner in which the force of gravity varies with height. For practical purposes, though, both the variation of gravity with height and the small change in horizontal velocity with height can be neglected.
     
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