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Galileo's Challenge Revised.

  1. Mar 10, 2012 #1
    Hi, new here just got registered to this forum since i have a physics question and Google had this forum in its first preferences. :)

    Well i got from my library an interesting book called: "Mad about Physics: Braintwisters, Paradoxes, and Curiosities", a very good book with all sort of interesting stuff until i met the problem 112 i will talk about here.

    I have to say that until problem 112, all the book, which was on the form of interesting physics questions of all kind and that the book had the solutions on the back, was of very good quality and the answers were correct from a scientifically point of view. Also the level of the book and its solutions was not for novice but had some relatively good mathematical analysis. It was all in all of high quality.

    But there was the damn problem 112. It asked:
    "Galileo's Challenge Revised."
    A person simultaneously drops a bowling ball and a much lighter plastic ball of the same diameter(and shape-sphere obviously) from the same height in the air. What do you predict?

    OK here i have to say that the book has many such problems having a double catch(trick). The first catch is for the beginner where the intuitive answer doesn't work and it needs a more advanced one, but sometimes the more advanced thought isn't correct also and you have to go a step further.

    Anyway in this question the intuitive answer of a person that doesn't know physics is of course that the heavier one(the bowling ball) will fall faster. Of course a more knowledgeable in physics person, knows that because of the identical shape the 2 balls have, the air resistance and the buoyant forces will be the same so the 2 balls will fall exactly on the same time.

    But here is the solution on the back of the book:

    The buoyant forces are equal, but the bowling ball weighs much more.
    Applying Newton’s second law, the bowling ball has the greater initial downward acceleration, and this condition is maintained all the way down.
    The bowling ball experiences a slightly greater air resistance force on the way down because it is moving faster, but the plastic ball never catches up.

    Nelson J. "About Terminal Velocity." Physics Teacher 22, 256–257.
    Toepker T.P. "Galileo Revisited." Physics Teacher 5, 76, 88.
    Weinstock R. "The Heavier They Are, the Faster They Fall: An Elementary Rigorous Proof." Physics Teacher 31, 56–57.

    What the hell is that?:confused:
    I'm specifically speaking about:
    ".....but the bowling ball weighs much more. Applying Newton’s second law, the bowling ball has the greater initial downward acceleration,"

    Oh my god!!!!
    We all know that both balls have the same acceleration and that is g(about 9.80665 m/s^2 but its value is irrelevant anyway).
    They both will fall with the same acceleration. And this is g. Or the equivalent [itex]\frac{G·M}{(R+h)^{2}}[/itex], where h the height they fall and M,R the mass and radius of Earth.

    Am i saying anything wrong here?
    No. I assume. Right?

    OK about the book, he made a "small" mistake then.:tongue2: BUT what about the 3 references he gave??????????????
    Are 3 independent scientists wrong also????
    Unfortunately i can't read the articles because this "Physics Teacher" magazine that offers them online asks for $30 per article and i can't afford that(actually every article is 2 pages and i don't want to spend $90 for 6 pages!!:yuck:). But for god's sake how 3 independent scientists and in a relatively renowned magazine, support the solution of the book??

    So it should be me that knows things incorrectly. But why?
    I would appreciate an answer....
  2. jcsd
  3. Mar 10, 2012 #2
    BTW i would really love to see this elementary rigorous proof the third reference claims:

    Weinstock R. "The Heavier They Are, the Faster They Fall: An Elementary Rigorous Proof."

    But as i've said. 2 pages, $30. :uhh:
  4. Mar 10, 2012 #3


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    I think the point is that the buoyant forces are equal, but since the masses are different, the accelerations due to these equal forces are different. Assume the buoyant forces are equal to Fb, the gravitational force on the bowling ball is mb*g, and the gravitational force on the plastic ball is mp*g. Then, since F= ma,

    ab = (mb*g - Fb)/mb = g - Fb/mb

    ap = (mp*g -Fb)/mp = g - Fb/mp

    Since mb > mp, the initial acceleration experienced by the bowling ball is slightly larger than that experienced by the plastic ball.
  5. Mar 10, 2012 #4


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    Forget about what what "we all know", and do the experiment.

    The last time I saw a kid let go of a helium balloon, it didn't fall downwards at all. It went up. But according to you, "we all know" it should have fallen down with the same acceleration as a bowling ball...
  6. Mar 10, 2012 #5
    Hehe, you are both right, i now get it.
    Thanks both of you!
  7. Mar 12, 2012 #6
    The difference produced by the buoyant forces will be negligible.The book is right! The real difference is caused by the drag force which causes greater decrease in acceleration in the plastic ball due to lower mass. The drag force keeps changing but we can assume it to be constant for better understanding. Try the maths by making all the forces.
  8. Mar 12, 2012 #7


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    2017 Award

    By the time a ping pong ball has fallen by 10m, the effect of drag is very measurable (great A level project) so a 20mm diameter lead ball would definitely hit the ground before a ping pong ball. They could have saved on the torture instruments if they'd only had a ping pong ball. No one would have bothered to listen to Sr G about the Universe.

    The posts above would have got to an answer quicker if, instead of 'acceleration' they had used the word 'force'.
  9. Mar 12, 2012 #8


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    Of course once the balls are moving, the impact of drag will be more significant on the lighter ball. But notice the book says, "Applying Newton’s second law, the bowling ball has the greater initial downward acceleration...". Drag cannot impact the initial acceleration, because the velocity of both balls is zero, so there is no drag. Only the buoyant force can cause this difference in initial acceleration. This difference is likely very small, but the question does not ask for the magnitude of the initial acceleration difference.
  10. Mar 13, 2012 #9
    Yes, the initial acceleration will be different. However difference in the times of falling down is mostly caused by drag.
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