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Mass doesn not effect speed of dropped objects

  1. Nov 6, 2009 #1
    Okay I am no physics pro, but Galileo claimed/proved that two objects of differing mass will fall at the same rate/speed.

    As I understand the force of gravity is a function of mass and distance. There is gravitational pull between all objects and this force is determined by the mass of the objects and their distance from each other.

    If I understand this then gravitational pull would be greater on an object of greater mass. Now I can see why Galileo's tests showed no difference as the the difference in mass of any two objects he tested would be insignificant compared to the mass of the earth.

    What am I missing here?

    Thanks for your help and please don't get too technical.
  2. jcsd
  3. Nov 6, 2009 #2


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    You need to be careful about which mass you're talking about. Also, it's the acceleration that is the same, not the speed. Although if both objects are dropped with the same speed, then they will continue to have the same speeds (although both speeds are constantly increasing) through their falls. The force a mass [tex]m_1[/tex], for example the Earth, exerts on a second mass, [tex]m_2[/tex], say for example a person, due to gravity is given by [tex]F = \frac{Gm_1m_2}{r^2}[/tex]. However, the force acting on an object is [tex]F = ma[/tex] so for the person, [tex]m_1[/tex], you can equate the forces, [tex]F = m_1a = \frac{Gm_1m_2}{r^2}[/tex] and you'll see that the acceleration depends only on the mass of the other object and the distance between them.

    As you can see, however, this works no matter what the scale of the objects. You can say instead of looking at the person, you look at the force the person exerts on the planet! Using the same equation, you see that it also gives an acceleration independent of it's own mass.
  4. Nov 6, 2009 #3
    It is true that the acceleration of an object in earth free-fall is determined solely by the mass of the earth and not by the mass of the object. It's a fact. However, you should keep in mind that the earth is also accelerated by the object. Their relative acceleration is the product of their masses as shown in the universal law of gravitation.

    If you increase the mass of the falling object then the acceleration of the earth will increase, but not the acceleration of the object itself. Obviously, as you've already pointed out, this increase would be undetectable (unless we're talking about an astronomical size object).
    Last edited: Nov 6, 2009
  5. Nov 6, 2009 #4
    Turtle, I do understand and agree with what you said. Pengwuino, can you tell me what all the letters represent? What are G r and a? I got F and m1 and m2. I will probably have further questions.
  6. Nov 6, 2009 #5
    G is a constant, r is a distance between two masses and a is acceleration.
  7. Nov 6, 2009 #6


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    Turtle, it is best not to confuse the OP with something that is a secondary issue.

    JeepinBob: If you look at the gravitational force equation that Penguino posted, you can see that the force of gravity scales linearly with the mass of your small object (m1). In other words, if you double the mass, you double the force. But then you plug them into f=ma and you find that if you double the force and double the mass, the acceleration stays the same.
  8. Nov 6, 2009 #7
    You did not state the relationship correctly: Objects of different masses accelerate at the same rate in a particular gravitational field. Any object will accelerate in a gravitational field at the same rate, regardless of the mass of the object. This has nothing to do with the relative mass of earth and relatively small masses. All masses accelerate at about g = 9.8 m/s/s at the earth's surface. If one mass is twice another, it is subject to twice the gravitational force resultring in the same acceleration g.
  9. Nov 6, 2009 #8
    Sorry, I wanted to be "not too technical" as the op requested. But I guess I did not understand the question.

    JeepinBob, this page has helped me.
  10. Nov 6, 2009 #9


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    It might also be helpful to add that, while the acceleration due to gravity is equal on both masses, galileo would probably not have seen both masses hit the ground at exactly the same time - no matter how careful he was to drop both masses from the same height and time.

    This is because of air resistance. You might think that air resistance is equal on both masses, but this isn't exactly so. Galileo used larger objects for the heavy mass, and smaller objects for the small mass. Their sizes are different. This means that more air is bombarding the larger object (resistance is a force that counteracts the force of the object due to gravity). But contrastingly, the larger object also has a higher mass and thus higher force due to [itex]F=ma[/itex]. Eventually, as the object keeps increasing speed, the air resistance force will keep increasing until the force due to gravity (which is constant as the mass and acceleration of gravity are constant) will be equal to each other. This is when the net force (total sum of forces - gravity and air) is 0 and thus there is no more acceleration. Then the object experiences a constant speed and this is what is meant by terminal velocity.

    Now, you'll probably find that even though the larger object experiences more air resistance, the increased mass is still more profound and thus the large object will have a high terminal velocity. So, if you drop the two objects from a very high location at the same time, the larger one will hit the ground first (the higher the tower, the bigger the difference).
  11. Nov 6, 2009 #10
    So if I understand correctly, the force of pulling on the heavier object would be greater but the added mass counters the force. that is the same force on a heavier object would cause less acceleration.
  12. Nov 6, 2009 #11


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    Exactly, an object that is twice as heavy feels twice as much gravitational force, but it also takes twice as much force to accelerate it.
  13. Nov 6, 2009 #12


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    I'm not sure exactly what you're trying to say simply because of the lack of terminology.
    Do you mean the air resistance force, or gravity force here?

    Anyway, I'll try explain it with example. Imagine you have two spheres (I'll refrain from using the term balls :smile:) and each sphere is the same size. However, the only difference is that one sphere is twice as heavy as the other (one sphere could be hollow).
    Now, looking at the equation F=ma, the acceleration is gravity and this is the same for all objects, but the only difference here is the mass. The heavier ball will have twice the constant force compared to the other.

    Since the two spheres are the same size, the air resistance that acts on them is also equal. As we drop both spheres from a large height, the accelerate (keep speeding up) and as their speed increases, the air resistance force which tries to counter-act the force of the spheres gets larger too. At some point, the air resistance force will become equal to the force of the lighter sphere. This means there is no more force being applied so no more acceleration. The speed of the lighter sphere is now constant and the air resistance force doesn't change anymore.

    There is still the force of gravity acting, but the air resistance is also acting so no acceleration.
    The twice as heavy sphere still has a larger gravity force so it continues to accelerate until the air resistance also becomes twice as much.
    Now, as you can probably visualize, the heavier sphere would hit the ground first because it manages to get to a greater speed and keep that higher speed.

    But, the heavier objects aren't always the first to hit the ground. Think of a man in a parachute versus an apple.
  14. Nov 6, 2009 #13


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    Oh I understand what you meant now after seeing mgb_phys' response.
  15. Nov 6, 2009 #14

    That's right. All objects are accelerated toward the earth at the same rate, regardless of diameter.

    There would not be any difference even if the masses he used were greater than those of the earth. Regardless of the mass of the object, the acceleration imparted to it by the gravity of the earth is still the same. The mass is the factor that determines the amount of momentum-changing force that the object experiences. Shape is also a factor when calculating for air resistance. But in freefall all objects can be treated as having equivalent masses. This means that if there were no air to provide resistance, a feather would fall at the same rate as a bowling ball.
  16. Nov 6, 2009 #15

    Why would an object that is twice as heavy feel twice as much gravitational force? I think what you mean is that a heavy object exerts more gravitational force than a lighter one.
  17. Nov 6, 2009 #16


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    That's how gravity works, it's been explained at least a couple times in this thread. On Earth, an object, A, that is twice as heavy as another object, B, will feel twice the gravitational force from the Earth as object B.
  18. Nov 6, 2009 #17
    There would still be acceleration, wouldn't there? Just no net force?
    Last edited: Nov 6, 2009
  19. Nov 6, 2009 #18


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    There's still two forces involved, but they exactly cancel at terminal velocity, giving a 0 net force. That means there is 0 acceleration: F = ma.
  20. Nov 6, 2009 #19

    Only when it's on the surface though. There's no heavy or light in freefall conditions.
    Last edited: Nov 6, 2009
  21. Nov 6, 2009 #20

    So air resistance changes gravitational acceleration?
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