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Same gravitational acceleration of unequal masses

  1. May 1, 2009 #1
    I understand that objects of different masses will accelerate equally in a vacuum. This is because the force of acceleration from gravity is the same regardless of mass.

    I also know I can manipulate the F=ma equation to a=f/m, which also shows the above to the true.

    Intuitively, though, I have trouble reconciling this with the notion of gravity "causing" the items to have different masses when sitting still.

    I was wondering if anyone had a plain-English explanation. Thank you for considering my question.
     
  2. jcsd
  3. May 1, 2009 #2
    Masses? It's weights, be careful which you use. Gravity doesn't give anything mass, it gives it weight, which is a force, think about that;

    F = ma, as you can clearly see the thing that gravity is GIVING the mass is force.

    Incidentally, the reason why acceleration due to gravity is constant for any body is due to newton's eqation of gravity;

    (Mass of Object 1 x Mass of Object 2 x Gravitational Constant) / Distance between them squared.

    Or to find the pull of 1 on the other only use the mass of the one pulling.

    As you can see, if you take you to be the first object, the centre of the earth to be the other, now equal this to F = ma, your mass simply cancels out with the mass in newton's equation.

    http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation


    Edit : If your question is really related to how things have mass, see Higgs Boson etc, much more abstract & I don't think anyone could give you a real answer right now
     
  4. May 1, 2009 #3
    Here's my way of thinking of it - in plain English, with no equations. Objects accelerate toward each other (toward the center of mass), not just one toward the other. In a vacuum, the Earth and a massive "ordinary" object WILL have a higher acceleration toward each other than would the Earth and a less massive "ordinary" object. But that difference in acceleration cannot be detected because the difference in the combined mass of the Earth and the two "ordinary" objects is too small to be detected. The rate of acceleration of any two objects toward each other depends on the TOTAL mass of both objects combined. So if you have a 1kg object combined with the mass of the Earth and then you double the objects mass to 2kg, how much have you increased the total mass (as a percentage)? It's a very very small number. So that's why all objects appear to accelerate at the same rate.

    Edit:
    Because of what I have learned in this thread, I now know that the above statement is incorrect. It is true that two free falling objects of different mass will have an extremely small and undetectable difference in acceleration. But it is not the great mass of the Earth that prevents us from detecting this difference. In fact, the Earths mass has nothing to do with it. Only the difference in mass of the two objects and their distance to the center of the Earth affect their difference in acceleration, expressed as G(M1-M2)/r2. I guess sometimes you just have to use a little math to see things as they really are. :)
     
    Last edited by a moderator: May 9, 2009
  5. May 1, 2009 #4
    I'm afraid you're wrong, we don't 'appear' to be attracted more or less, we are all attracted exactly the same, regardless of mass.

    You work out the gravitational field (using newton's law's) to be;

    Mass of Object * G / r²

    Yes, we also attract the sun as well, but that's entirely separate. As I understand it the force due to gravity is constant (at sea level) for all bodies regardless of mass, another way to look at it is as I said;

    [tex]\frac{M1 * M2 * G}{r^2}[/tex] = M1 * A (M1 bieng my mass for example)

    The M1's cancel, (our) mass does not affect it.
     
  6. May 1, 2009 #5
    Yes, my interpretation may indeed be wrong. But I don't think it is. And I welcome anyone who can explain to me why it isn't. The OP asked for a plain English explanation and that was what I gave, stating clearly that it was my way of thinking. I admit that it's not complete because I said that the acceleration of two bodies toward each other depends on the total mass of both bodies. That's true, but I left out that it also depends on the distance between them. Anyway, here's the problem I have with your explanation (which is the one I usually hear):

    You're saying that I am wrong because of this:

    A = M2 * G / r2 where M2 = mass of Earth
    (M1 * M2 * G) / r2 = M1 * A where M1 = my body mass
    The M1's cancel, my body mass does not affect it.

    If that's true then why isn't this true:

    A = M1 * G / r2 where M1 = my body mass
    (M1 * M2 * G) / r2 = M2 * A where M2 = mass of Earth
    The M2's cancel, Earths mass does not affect it.
     
  7. May 1, 2009 #6

    sylas

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    The "it" here is the acceleration of the Earth towards you, and yes, the Earth's mass doesn't affect that.
     
  8. May 1, 2009 #7
    I see your point. have to go right now but will think about this and reply later. Thanks for the input.
     
  9. May 1, 2009 #8

    D H

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    Your explanation is looking at things from the perspective of a non-inertial reference frame. If you look at things from the perspective of an inertial frame, all objects, regardless of mass, will undergo exactly the same acceleration.
     
  10. May 1, 2009 #9
    I'm sorry but that's where you're going wrong. The force is greater for larger masses. [itex]F=GMm/r^2[/itex] So, if you increase m (the mass of the object), then the gravitational force on that object increases too.

    I think the phrase "force of acceleration" is too confusing and has led you astray. There is force. And there is acceleration. They are different things, measured in different units.
     
  11. May 2, 2009 #10
    What do you mean by inertial frame?

    Ok, I checked Wikipedia and I don't get how my view is non-inertial. But I'll admit that I am having doubts about my view. Is it completely wrong? Or is it a matter of perspective?

    Why is it ok to use these equations and say that one object is accelerating toward the other, but it's not ok to say that both objects are accelerating toward each other (toward the center of mass)? It would appear that by setting A = M1 * G / r2 or A = M2 * G / r2 that you are setting your frame of reference to one or the other object and linking A to only one of the masses. Obviously any changes in the mass not linked to A will have no affect. Isn't it just as true to say that both objects are accelerating toward each other according to (M1 * M2 * G) / r2 and that A depends on both M1 and M2?
     
    Last edited: May 2, 2009
  12. May 2, 2009 #11
    Think of it this way, i'm floating in space, about 3,000 kilometres from the sun, wearing an astronaut suit so I don't die.

    Now, me and the sun attract each other due to gravity thanks to the equation

    [tex]\frac{M1 * M2 * G}{r²}[/tex] = Vector Force

    I accept this, now I think 'how fast am I moving?' I can work out the force on me, so now I decide to use another law of newtons - F = ma, since the above equation gives me the force, i can equal it to MA

    [tex]\frac{M1 * M2 *G}{r²}[/tex] = M1 a (M1 bieng my mass)

    I can then divide through by M1, meaning my mass is redundant, independant of my acceleration.

    I think maybe some of you are getting caught up on the fact that it's an attraction from the two, and not just one, and you'd be right, but you can see them as independent from each other, the effect of our attraction on the sun, and the sun's to us, when working out the acceleration of one (towards the other), it's independant of (the one that's your trying to work out the acceleration for's) mass
     
    Last edited: May 2, 2009
  13. May 2, 2009 #12

    sylas

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    Ignore this post. Originally I queried a comment which has now been fixed up in edit, so I am also removing what I quoted. We're all on the same page again.
     
    Last edited: May 2, 2009
  14. May 2, 2009 #13
    Yeah ok, didn't really read it my bad, the gravitational pull is greater for objects with larger mass's, yeah. f = ma with a being constant.
     
  15. May 2, 2009 #14

    D H

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    Your view is non-inertial because your reference frame is accelerating. Strictly speaking, Newton's laws of motion are not valid in a non-inertial frame. To make them appear to be valid one must add fictitious forces to the mix.

    There is nothing wrong per se with working in a non-inertial frame. Suppose you want to explain some phenomenon. Whether you do your derivations from the perspective of an inertial frame or non-inertial frame in a way is irrelevant. Do the math right and both perspectives will yield the same end result. The choice is very relevant if you take into consideration the difficulty of arriving at the end result. Explaining the weather from the perspective of an accelerating and rotating reference frame (i.e., a frame fixed to the rotating Earth) is rather difficult but doable. Explaining the weather from the perspective of an inertial frame (ignoring very tiny effects, a non-rotating frame with origin at the solar system barycenter) is a lot worse than rather difficult. On the other hand, trying to explain the interactions in some remote star system from the perspective of an Earth-fixed frame, while possible, is downright stupid.

    What you have done wrong is to think that because the acceleration of an object toward the Earth does depend on the object's mass somehow falsifies the equivalence principle. It doesn't. The equivalence principle now stands as one of the most precisely verified axioms in all of physics. It has been verified to about 1 part in 1013 (see the Physics World article cited below). Three future (in development or proposed) satellite missions, MICROSCOPE (CNES), Galileo Galilei (ASI) and STEP (NASA), intend to increase this accuracy by two, four, and five orders of magnitude respectively.


    References:
    Relativity at the centenary, Physics World (2005), http://physicsworld.com/cws/article/print/21148

    Microscope: Exploring the limits of the equivalence principle, CNES, (2008), http://www.cnes.fr/web/CNES-en/2847-microscope.php [Broken]

    Galileo Galilei, GG project home page, http://eotvos.dm.unipi.it

    STEP, STEP project home page, http://einstein.stanford.edu/STEP
     
    Last edited by a moderator: May 4, 2017
  16. May 2, 2009 #15
    Thanks for all the info. It has provided me with a lot to think about.

    Yes, I can see now that it is a matter of perspective and that my view would indeed have to be accelerating (unless both objects were the same mass).
    No. I do not think that. Equivalence experiments is one of the things I have read quite a lot about. I am especially interested in equivalence experiments involving active and passive gravitational mass. I've only found one (Lloyd B. Kreuzer 1966). See my first PF post: https://www.physicsforums.com/showthread.php?t=305189
     
    Last edited: May 2, 2009
  17. May 2, 2009 #16
    Never mind, I found my error. :)
     
  18. May 2, 2009 #17

    D H

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    Choose a different reference frame and you can get a vastly different view of things. What's so unintuitive about that?

    You're comparing apples to oranges.

    The equivalence principle is about the gravitational acceleration as observed by an inertial observer. You are talking about relative acceleration. It's well known that the masses of both objects contribute to relative acceleration. For example, this is one reason why Kepler's laws are only approximately correct. (Kepler's laws implicitly assume planets have negligible mass compared to that of the Sun, and that is not a valid assumption given with the precision with which we can measure orbits.)
     
  19. May 2, 2009 #18
    D_H, since you quoted something that I deleted I will re-post it here so that other readers can see it in it's entirety.

    Yes, I can see now that it is a matter of perspective and that my view would indeed have to be accelerating (unless both objects were the same mass). It just seems unintuitive to have to set my frame of reference to one or the other objects. Lets take the following example: An object in Earth free-fall will accelerate at the same rate regardless of it's mass. That is true. The key word being accelerate. But it is unintuitive, and in my opinion even deceiving, to say that any two objects will fall at the same rate regardless of their mass. And here is the reason I think so: Suppose we have an object (lets say 1kg) which is positioned at a certain distance from the Earths surface. We let the object free-fall and measure the time it takes for it to hit the surface (assuming there is no atmosphere). Now suppose we make the mass of the object 1/4 the mass of the Earth, set it at the same distance and let it free-fall. We once again measure the time it takes it to hit the surface. And keep in mind that regardless of what frame of reference we use to make the measurement (disregarding any minute differences due to relativity), the time lapse will not be affected. Now, will both the 1kg object and the 1/4 mass of the Earth object take the same amount of time to hit the surface? I don't think so. But of course you can get around this by saying that by increasing the mass of the object we increased the Earths acceleration toward the object. But I think you can see by this example how it is deceiving, and unintuitive to say the least, to say that any object, regardless of it's mass will fall at the same rate. It may be technically true to say that increasing an objects mass will not change it's acceleration, but it will indeed change the time it takes for it to reach the surface. The only reason this cannot be proven experimentally is that any object we create will be minute compared to the Earths mass. The difference in elapsed time would be undetectable.

    The unit of measure used in my thought experiment is time, not acceleration. And it works the same regardless of whether it's in an inertial or non-inertial frame of reference. Why isn't that more intuitive?

    D_H, I am not trying to dispute the equivalence principle. And I do not believe that my thought experiment shown above contradicts it. I have realized my error as far as the inertial frame is concerned. However, I still do not think my thought experiment is incorrect. I am willing to change my mind but I haven't been convinced yet. I have learned something from this thread and feel that I understand gravity better than I did before.
     
  20. May 2, 2009 #19

    D H

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    As perceived by whom? Let's ignore the Earth's rotation. Suppose we have two observers with infinitely accurate means of assessing acceleration. One is fixed with respect to this non-rotating Earth and the other is fixed with respect to some inertial frame. Both measure the acceleration of an object falling toward their Earth at the same time with their infinitely accurate sensors. The two will measure different accelerations.

    Now suppose we use a different test object whose mass is orders of magnitude greater than that of the first test object. Both observers measure the acceleration of this new test object after placing it at exactly the same position with respect to the Earth as the first test object. While the Earth-based observer will measure a different acceleration than measured for the first test object, the inertial observer will see this new object as undergoing exactly the same acceleration as the first test object.

    Another way to look at this: Imagine two cars, a blue one and a red one, in a drag race. From the perspective of someone in the grandstands, the blue car's acceleration is amazingly high. From the perspective of the driver of the red car, the blue car's acceleration is fairly small. In fact, the driver of the red car most fervent wish is that the blue car appears to be accelerating backwards.
     
  21. May 2, 2009 #20
    A = M2 * G / r2 where M2 = mass of Earth
    The objects mass has no affect.

    I'm not sure how to comment on this because I don't know what "some inertial frame" is. If this observer who is at "some inertial frame" measures the acceleration relative to himself then what he measures will depend on what his velocity and acceleration is relative to the objects being measured. However, I can use this to illustrate my point where I say that using A = M2 * G / r2 is unintuitive. If your observer at "some inertial frame" and the Earth observer measured the time lapse of the objects free-falls, wouldn't their results be the same - regardless of the frame of reference of the observer at "some inertial frame"? Your post seems bolster my argument!
     
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