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Serious answer from a physicist required (please no random answers)

  1. Jan 6, 2007 #1
    Serious answer from a physicist required (please no random answers)

    Had an argument in the pub tonight whether 2 objects of equal size and shape one 5kg and one 100000 kg would fall from a height of 40k feet and hit the ground at the same time or not?

    This would happen in perfect lab conditions i.e. no weather but in air.

    I won’t say in which direction I argued please provide an answer with working/proof.

    Note that perfect lab conditions does not indicate a vacuum just no external factors other than still air or another arbitry gas i.e. with a mass that could provide a frictional force.

    Cheers D

    P.S. please provide worked proof!
  2. jcsd
  3. Jan 6, 2007 #2
    The heavier one will hit the ground first (sorry, Im not going to provide a working proof but if you want one you can open university physics by sears and Zemanski and turn to chapter, errrr 4 or 5?).
    Last edited: Jan 6, 2007
  4. Jan 6, 2007 #3


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    Try dropping a tissue box and a brick of the same size and shape, from a height of 4 meters, and see which one, if any, hits the ground first. Or better still, hold a coffee filter paper basket in one hand, and in the other hand, stack a few of them together, and them drop them both from the same height at the same time. Which hits first, the single filter or the multiple set??
  5. Jan 6, 2007 #4


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    If there is air resistance, the denser object falls faster.
  6. Jan 6, 2007 #5
    The heavy object will fall faster.

    Even though their is no wind, the falling objects feel the force of wind because they are moving past the air (no different then the air moving past them).

    If you are asking for a proof, then I hope you know about force. Then the total force on the object towards the ground is:

    [tex] F_{Total} = F_{Gravity} - F_{Air Resistance} [/tex]

    The force of gravity involves the mass [tex] m [/tex] (in kilograms) and the acceleration due to gravity [tex] g = 9.8 [/tex] (measured in meters and seconds)

    [tex] F_{Gravity} = m g [/tex]

    But the force of air resistance does not involve the mass, but only the area and form of the object, as well as its velocity:

    [tex] F_{Air Resistance} = \rho A v^2 [/tex]

    Where [tex] A [/tex] is the area, [tex] \rho [/tex] is a constant depending on shape, and [tex] v [/tex] is the velocity.

    As you can see, of two objects of the same area and shape it will be the one with the larger mass that feels a stronger force towards the ground, and hence falls to the ground faster then the lighter object.
  7. Jan 6, 2007 #6
    rho does not depend on shape.
  8. Jan 7, 2007 #7


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    I was thinking, if one really light object was dropped to the ground, and one really heavy, let's say an object with the mass comparable to the moon. The large object would fall faster than the lettle object even if we exclude air resistance. Because heavier objects creates their own gravitational force that pulls the earth towards them. Meaning it would go faster. (of course the heavy object must not be on earth as you drop the light, because then the heavy object will add it's gravitational force to the light object.)
  9. Jan 7, 2007 #8


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    Crosson has the correct equations that need to be understood to resolve this.

    Note the force due to gravity is mg which is proportional to the cube of some characteristic length of the object (ex: diameter of a ball).

    In contrast, the force due to air resistance is proportional to the area which is the square of some characteristic length of the object.

    So the force due to gravity rises as a cube as the size of the object increases, while the air resistance only rises as a square of the size of the object. This means the air resistance will be a smaller proportion of the overall force as the size increases assuming the object's density remains the same.

    For example, I calculate a ball with a 1" diameter and a density of 100 lbm/ft3 will have a terminal velocity of 106 ft/s. A ball with a 10 foot diameter and same density will have a terminal velocity of 1200 ft/s. I won't post the analysis because it's done on a spreadsheet I have, but you can do the same thing by following the equations provided by Crosson. The only additional information you need is the coefficient of drag as a function of Reynolds number which is also available on the net, for example - here: http://farside.ph.utexas.edu/teaching/329/lectures/node42.html
  10. Jan 7, 2007 #9
    That is simply not true, do you think that a dome falls at the same speed when it is face up (like a cereal bowl) as when it is face down (like a parachute)?

    You think [tex] \rho [/tex] depends on the viscosity of the medium, but in my statement of this equation I did not mention units (in order to simplify things), avoiding the need for this "fudge factor" (which does not effect the physics).

    In any case, the shape (and not just the area) of an object affects its drag force.
  11. Jan 7, 2007 #10


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    that's almost unbelievably presumptuous. your argument in the pub deserves or requires arbitration from only bona fide physicists with "worked proof" provided? you're funny. :rofl:

    denser objects (e.g. lead ball) falls faster than less dense objects (feathers, parachutes, snowflakes). i think heavier objects of the same shape and material than lighter objects (big lead ball vs. little lead ball) have less portion of their total graviational force lost to air resistance. this is because of smaller surface area to volume ratio and, for a given density, volume is directly proportional to mass and downward weight (which increases as [itex]r^3[/itex]), and air resistance will be proportional to surface area or cross-section area (which increases as [itex]r^2[/itex]).

    if air resistance is not a factor, then it sorta depends on if they were dropped together (simultaneously) or if the experiement was run twice with an (impractically) accuate timepiece recording the time of fall.

    the heavier object pulls the Earth toward it more than does the lighter object. if they were dropped together in a perfect vacuum, the heavier object is pulling the Earth up which also benefits the lighter object in the race, so they finish simultaneously (unless you try to include the even more negligible effect of the tilting of the Earth's "fall" slightly toward the heavier object, slightly away from the lighter obect). if the two experiments were done separately and timed with infinitelyprecise and perfectly accurate timers, the heavy object would finish in less time than the lighter object, but there will never be any practical measurement of the difference.
    Last edited: Jan 7, 2007
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