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Mass of falling objects vs. their acceleration

  1. Aug 25, 2014 #1
    As a non-scientist, this question has been bothering me, but probably has a laughably simple answer. In high school physics (which, for me, was long time ago) we were taught that 2 objects dropped from a height fall toward earth with the same acceleration regardless of mass (the so-called Tower of Pisa experiment).

    But, at the same time, we're taught that Newton's universal law of gravitation, says that the force between two masses (in this case, the earth and whatever object we are dropping) is proportional to the product of the masses of the two objects (and inversely proportional to the square of the distance between them).

    So why do a canon ball and a feather fall with the same acceleration (neglecting air resistance) and therefore take the same amount of time to reach the ground if they each have a different amount of force between themselves and the earth?
  2. jcsd
  3. Aug 25, 2014 #2


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    Because of newton's other law, namely, the acceleration that a force causes on a mass is inversely proportional to the mass (a=F/m)

    So the force is larger by a factor of the mass, but the acceleration from that force is smaller by a factor of the mass, so it balances out regardless of the mass.
  4. Aug 25, 2014 #3
    Thanks Nathanael. After posting my question, I found some relevant replies, but didn't fully understand them. Yours seems to explain it clearly (though I may only be deluding myself that I understand). So does this mean that, with the earth as one endpoint, a=F/m will always solve to 32 ft/sec2 (or whatever the metric equivalent is)?
  5. Aug 25, 2014 #4


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    Only on the surface of the earth (because gravity also depends on distance)

    But, at a fixed distance, the mass will be irrelevant to the acceleration.

    I'll show you mathematically why the mass is irrelevant. Maybe it will be helpful.
    I'll do this by combining two equations (each of which describes one of newton's laws that we've mentioned)

    Equation 1: (This equation is essentially the definition of Force, except I've rearranged it)
    (m is the mass of the object, "a" is the acceleration)

    Equation 2: (This equation is 'Newton's Law of Gravity')
    (M is Earth's mass, r is the distance to the center)

    So "plug equation 1 into equation 2" (or simply divide equation 2 by "m," same thing) and you will get:


    Which, you can see, only depends on the mass of Earth (and the distance)
  6. Aug 25, 2014 #5


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    Say you have two objects in space, free from any other gravitational effects except for each other. Both objects accelerate towards a common center of mass. The force on each object is G m1 m2 / r^2. The acceleration of object 1 towards the common center of mass is (G m1 m2 / r^2)/m1 = G m2 / r^2, and likewise the acceleration of object 2 towards the common center of mass is G m1 / r^2. So the rate of acceleration towards the common center of mass for each object is a function of the mass of the other object. The rate of closure acceleration (the second derivative of r) is the sum of the two accelerations towards each other:

    $$\ddot r = \frac{-G (m1 + m2)}{r^2} $$
    Last edited: Aug 25, 2014
  7. Aug 25, 2014 #6


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    Nothing wrong with the answers so far, but you might also want to take a look at the FAQ entry: https://www.physicsforums.com/showthread.php?t=511172 [Broken]
    Last edited by a moderator: May 6, 2017
  8. Aug 26, 2014 #7
    Consider this set up, i give the usual suspects different values for ease of calculation, but the theory is good.

    G = 1.0
    m1 = 100 kg
    m2 = 1 kg then 2 kg ( two seperate experiments )
    d = 10 m

    On release the two objects approach the stationary barycentre, accelerating at different rates tending to arrive at the barycentre at the same time, the total acceleration (ta) of the two objects is given by :
    ta = G *( m1+m2 ) / d²

    The acceleration of each body is given by :
    a (m1) = ( m2 / ( m1 + m2 ) ) * ta
    a (m2) = ( m1 / ( m1 + m2 ) ) * ta

    What you find is that the acceleration of m2 toward the barycentre remains unchanged in both experiments, but the total acceleration between m1 and m2 does.

    So, if you measure m2's acceleration from outside the system ( from the barycentre say ) its acceleration remains the same in both experiments, but if you measure it from the earth. it doesnt.

    So, strictly speaking there is a difference, but in an experiment on earth, because of the huge difference in mass between m1 and m2, the mass of m2 becomes negligable.
  9. Aug 26, 2014 #8
    Good post dean barry. You've got it right. I would just like to add that the reason the acceleration of m2 at 1kg remains the same as it does for 2kg is because when you change the mass of m2 you also change the location of the barycentre relative to the two bodies. If you increase the mass of just one body, say m2, then the relative acceleration between the two bodies increases proportionally. However, the location of the barrycentre also changes proportionally (moves closer to m2), keeping m2's acceleration constant (relative to the barrycentre).
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