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Grav. acc. of object trade-off as vol. increases?

  1. Jul 5, 2013 #1
    Dear all,

    I am thinking of a simple situation:

    A cube is falling under the force of gravity. It is not in a vacuum, and so air resistance has an influence. However, to simplify, the air is still and the cube does not rotate, the cube has no surface texture, and other fluid dynamics are ignored. In other words, the cube is moving through the air so that only one of its faces contacts the air, and the resistance imparted is simply by the column of air molecules impacting the downward facing face.

    I know this situation is a simplification, but I made it up so that I can more easily picture the square to cube relationship (area to volume) and how there might be a trade off so to speak in terms of acceleration. In other words, what volume is the ideal in terms of fastest acceleration in the above scenario?

    I set up a very simple spreadsheet to get at this.

    I have a column that is simply length of side of the cube in cm (1 to 150 cm), and columns based off this one that are area of a single side, volume, and mass (volume * density of water in kg/cm3, I simply assume that density out of convenience).

    In the next column, I calculate the force of gravity using Newton's Univ. Law of Grav. and inserting values for the mass and radius of the Earth.

    I know that drag equations are complicated, but at any given velocity and air density, the drag is proportional to some area. I set up a force of air resistance at an arbitrary velocity by multiplying area of a side by a scalar constant that I could adjust.

    This force of air resistance was then subtracted from the force of gravity and the acceleration of the object was gotten by dividing this by mass.

    I then plotted acceleration vs. volume on a graph so that I could view the relationship and fiddle with the scalar constant that transforms side area into a force. So, obviously, I am looking at the curve at a snapshot of a single velocity across the board, whatever that may be and see what the acceleration would be at that velocity.

    What I expected to find was a unimodal curve, where there is some sweet spot where volume is high enough to supply sufficient mass to "knock" air molecules out of the way yet not so high as to provide a large side area to catch a great number of air molecules.

    What I got, no matter how I fiddled with either density of the falling object or the scalar relating side area to force of air resistance, is a graph that has its lowest values at small volumes and then very quickly increases to an asymptote (roughly at the expected ~9.8 m/s/s). It is basically an L shape rotated 90 deg. clockwise.

    I suppose this is due to the fact that adding a bit of side area increases mass quite a bit (it is added in all 3-Ds), and this is especially true as you add this bit to a larger and larger volume. So, my result does conform to common sense given hindsight and how I set this up.

    Has anyone else calculated or read of any potential trade-off when an object of set shape and density is increased in volume?

    Some shapes may give a different result.

    -W
     
  2. jcsd
  3. Jul 5, 2013 #2

    Bandersnatch

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    No, they won't show a different result, as long as the basic setup remains the same.

    You have created an equation of motion in which the net force on the object equals force of gravity minus the drag force. Force of gravity is mass dependent, while drag is here constant.

    So,

    [tex]F=ma[/tex]
    [tex]ma=F_g-F_d[/tex]
    [tex]a=\frac{F_g}{m}-\frac{F_d}{m}[/tex]
    knowing that [itex]\frac{F_g}{m}=g[/itex] we get:
    [tex]a=g-\frac{F_d}{m}[/tex]
    It should be obvious that for high m the last term goes asymptotically to zero, so the acceleration goes to g.

    You're essentially dropping longer and longer rods of constant cross-section. It doesn't matter what the shape of that cross section is, as long as you increase only the volume(i.e., mass) and keep the rest constant, you get an ever diminishing influence of drag.
     
  4. Jul 5, 2013 #3
    Thanks. In my case though, since the object increases its volume (and so mass) in all dimensions equally and simultaneously (not like a simple lengthening of a rod), then the force of drag does increase as mass goes up (larger surface facing air). However, the increase in mass seems to very quickly outstrip this increase in drag, as you pointed out. Interesting thought experiment, and my initial intuition (before any math) was wrong.

    P.S. Did anyone catch the error in last month's Discover magazine on the back page? The editor gave a list of things you don't know about gravity, and stated that objects of different masses would strike the ground at the same time if dropped from a building. No "in a vacuum" qualifier. Yep, the editor of Discover thinks that a bowling ball and balloon of the same size and shape will exactly match each other in motion when dropped somewhere outside in the atmosphere.
     
  5. Jul 5, 2013 #4

    Bandersnatch

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    Oh, right. I somehow managed to misunderstand that.
    In this case, as long as the force of drag can be simplified to a function of area of one face of the cube(or whatever):
    [itex]F_d=f(x^2)[/itex]
    (x is the length of a side) i.e., the drag will increase with the second power of the side length, while the mass is a function of volume,
    [itex]m=f(x^3)[/itex]
    so it increases with the third power of side length.
    All in all, you end up with the same final relationship as in my previous post. I.e., just as increasing only length of a rod made [itex]\frac{F_d}{m}[/itex] fall as [itex]\frac{1}{x}[/itex], so it remains now.
     
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