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Mass vs. Gravity and what our teachers have told us!

  1. Oct 12, 2004 #1
    It seems that I have uncovered a great contradiction in modern thoughts on gravity. First, 2 statements that teachers have often used:

    1) Gravity is dependent on the mass of an object. The higher the mass the stronger the gravitation field.

    2) Two objects will fall to the earth with the same acceleration no matter what their mass.

    These two statement seem to contradict each other. The only possible resolution that i have come up with is that statement 2 is not actually correct. Instead there are slight differences in acceleration but these are not noticeable due to the infinitesimal mass of say a bowling ball and a feather compared to the earth as a whole. Does this mean that if there were another "earth" that had approximately the same mass and was dropped at the same time as the feather and the bowling ball, shouldn't the other "earth" fall at a much greater rate due to the immensely stronger gravitational field as opposed to the bowling ball or feather?

    I may be totaly out of it here but if anyone can shed some light onto this subject, I believe that it will make a most interesting conversation.
  2. jcsd
  3. Oct 12, 2004 #2


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    The difference is how the various objects affect the earth, not the other way around.

    In other words, you can't notice the earth falling towards the feather while the feather falls to earth. However, you will notice the first earth falling towards the second earth while the second earth falls towards the first!
    Last edited: Oct 12, 2004
  4. Oct 12, 2004 #3
    I see what you're saying, but there's an equation that answers your problem dealing with two "earths".

    Force of gravity = gravity constant[(mass of object1 * mass of object2) / (distance between center of masses)^2]

    I've forgotten the gravity constant, but a quick serach should be able to locate it.
  5. Oct 12, 2004 #4
    Both statments are correct. To see how, first consider a mass we will call m1.

    Then we can write the equation for the gravitational attraction between m1 and the earth:

    F1 = Gm1Mearth/r^2

    By Newton's Second Law: F1 = m1a

    Substitute this in for F1 in the gravitational equation:

    m1a = Gm1Mearth/r^2

    Notice we can divide m1 out of both sides leaving:

    a = GMearth/r^2

    In other words the acceleration (or the rate at which a body falls) is independent of its mass.
  6. Oct 13, 2004 #5
    #1 That's circular logic. You're using the equations to prove the equations. I think he wants an intuitive explanation of how the equations would still hold under the common ideas he listed. (Which they won't.)

    #2 Your equation, F = ma = GmM/r^2, only holds if there are two masses in the system. When you introduce a third mass you can no longer claim that the Net Force on m is equal to the force of gravity. The equation you used presupposes that the ONLY force on the object is the force of gravity caused by the mass M

    The original poster is right, anyway.

    This statment:

    Would need to be applied to ALL objects.

    Certainly Mars doesn't fall to the earth with the same acceleration as a bowling ball? Of course not, other forces interfere.

    In the same vein, dropping a bowling ball and a feather truly AREN'T isolated with the Earth. They each have an effect on each other and the gravity of the sun and the moon effects them and so does every other star in the universe. But these interferences are so small that it is a good general rule that stuff falls towards the earth at the same acceleration.

    But it doesn't even always hold, in our daily earth-bound lives.

    Tides for example. They defy the Earth's gravity and don't accelerate towards the Earth at the same rate as a bowling ball. This is because the moon inteferes of course. Tides are proof that statement 2 is only correct under certain circumstances.
  7. Oct 13, 2004 #6
    The higher the masses the higher the gravitational force between them, according to:

    [tex]GravitationalForce = \frac{GravitationalConstant*MassA*MassB}{DistanceBetweenMasses^2}[/tex]

    but also, the higher a mass of an object the less a fixed force will acccelerate it according to:

    [tex]GravitationalForce = MassObject*Acceleration[/tex] or

    [tex]Acceleration= \frac{GravitationalForce }{MassObject}[/tex]

    so on earth,
    [tex]GravitationalForceBowlingBall = \frac{GravitationalConstant * MassBowlingBall*MassEarth}{RadiusEarth^2}[/tex]
    [tex]AccelerationBowlingBall = \frac{GravittionalForceBowlingBall}{MassBowlingBall}[/tex]

    will result in:
    [tex]AccelerationBowlingBall = \frac{GravitationalConstant * MassBowlingBall*MassEarth}{RadiusEarth^2*MassBowlingBall}[/tex]
    which means:
    [tex]AccelerationBowlingBall = \frac{GravitationalConstant*MassEarth}{RadiusEarth^2}[/tex]

    Thus, as you can see although the force does depend on the mass, the acceleration is independent of the mass, because the higher the mass the less it accelerates if a certain force acts on it.
  8. Oct 13, 2004 #7


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    What is happening here is that you are using the word "gravity" for two DIFFERENT things. There is a gravitational force, which is the WEIGHT, and the gravitational acceleration, which is usually designated as "g" (not "G", which is the universal gravitational constant).

    The weight W is defined as W = mg.

    Thus, already you can see it is dependent on the mass in question.

    However, the gravitational acceleration of that mass is g. It is independent of how large this mass is (at least, within this simple scenario where most terrestrial mechanics are applied).

    So next time, when you use the word "gravity", try to think what exactly it is that you are using it as. You will learn that in physics, "words" and phrases are ambiguous and vague. It is only when we realize the underlying mathematical description of it does it now become clear and unambiguous.

    So there are no contradictions.

  9. Oct 13, 2004 #8


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    They don't contradict each other. The first is a statement about the strength of the gravitational force, or the gravitational field. The second is a statement about acceleration.

    The first statement is that F=-GmM/r^2. Combine this with Newton's second law, F=ma, and you get the equation ma=-GmM/r^2. If you just divide both sides by m, this simplifies to a=-GM/r^2. So, the acceleration is clearly independent of the mass m.

    As you can see, the second statement does not only not contradict the first. It is a consequence of the first. (Assuming the validity of Newton's second law, and the equivalence between inertial and gravitational mass).
    Last edited: Oct 13, 2004
  10. Oct 13, 2004 #9

    Chi Meson

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    There is one thing that has (I believe) been overlooked in all the prior posts (it's a long thread, with a lot of good stuff in it, so I apologize if it was already covered):

    We know that the earth also accelerates (theoreticaly) toward the other object, it is just so small an acceleration that it cannot possibly be measured.

    Gravitational acceleration always turns out to be equivalent to the gravitational field strength that the object is placed in.

    Therefore the earth will have a greater acceleration toward the more massive object than it will toward the less massive object.

    There would then be an a greater net acceleration of the more massive object from the earth's frame of reference. Again, this extra amount is purely theoretical and could not possibly be detected.

    My pet peeve: I wish teachers and textbooks would stop calling g the "acceleration due to gravity." It makes more sense to call it from the start "gravitational field strength" using the unit N/kg. FOrce of gravity = mg whether it is accelerating or not. N's 2nd law says a = F/m . When F = weight, then a = mg/m = g.
  11. Oct 13, 2004 #10
    Wouldn't "dropping" a huge mass on the earth result in faster acceleration though? I thought the two masses would accelerate toward each other. (In this case the moon accelerating toward the earth at the earth's acceleration of gravity rate add to that the earth accelerating toward the moon at the moon's acceleration of gravity.)
  12. Oct 13, 2004 #11


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    Er.... you will notice that in my reply to this thread, I did mention:

    "It is independent of how large this mass is (at least, within this simple scenario where most terrestrial mechanics are applied)."

    Considering that this person is already confused between the simple concept of "gravitational acceleration" and "gravitational force", I didn't want to invoke a more complicated discussion of such things. So no, I didn't "overlook" such possibility. I just didn't want to complicate things further.

  13. Oct 13, 2004 #12

    Chi Meson

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    Oh, so you did.

    I was rushing and looking for key words. This is a discussion that comes up every year in my AP class, and I thought it was worth mentioning specifically here.
  14. Oct 13, 2004 #13

    Chi Meson

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    Both of these accelrations are (mostly) centripetal and do not result in either body getting any closer tho the other (assuming circular orbit).
  15. Oct 15, 2004 #14
    Chi Meson: Please explain this some more

    You've perfectly described the dilemma I just can't rectify. How can the feather and bowling ball accelerate towards the earth at the same rate, yet the earth accelerate towards the objects at different rates? Does one "really" touch the earth sooner than the other? This is such a fundamental issue, but I can't logically describe why this is OK or consistent. I'm tutoring an English as a Second Language student and am struggling to put these ideas into simple vocabulary and clear examples. The math and the manipulation of equations is not an issue at all. We get that, but I would appreciate a more in depth explanation of this situation or thoughts on how to describe it better.
  16. Oct 15, 2004 #15
    Maybe the best way to explain this is to look back at my post where I did the math. In that post, we looked at the situation from M1's (call M1 a bowling ball just for fun) point of view and we discovered that all objects will fall towards the earth at a constant rate proportional to the earth's mass divided by the distance between the object and the center of the earth squared.

    But, we can turn the equation around and look at it from the Earth's point of view. We will find that if we substitute Mearth in to the F = ma part of the equation we will be able to divide out the mass of the earth and find that all objects will fall toward the bowling ball at an acceleration independent of their mass, but much much smaller than the bowling ball will fall towards the earth because the bowling is so much less massive than the earth.

    So, the reason for the different accelerations is that the masses are different. The earth attracts the bowling ball with a force proportional to its mass and the bowling ball attracts the earth with a force prorportional to its mass which is many orders of magnitude less than the earth's mass.
  17. Oct 15, 2004 #16

    Doc Al

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    Because the feather and bowling ball are being pulled by the same gravitational field--that of the earth. But the earth is pulled by two different fields: a weak one due to the feather, and a stronger one due to the bowling ball.
    Of course, here I am talking about the acceleration of each object (feather, bowling ball, earth) with respect to an inertial frame (the center of mass frame). If you literally wanted the acceleration with respect to the earth (including the earth's acceleration) then you'd have to add the two accelerations. In that case, the bowling ball would hit the earth sooner than the feather (ignoring the vastly bigger effect due to air resistance) since the bowling ball accelerates the earth more. But realize this is a ludicrously small correction to the "acceleration due to gravity". g is typically given as 9.8 m/s^2; the additional acceleration due to the earth's response to a 5 kg bowling ball would be about 8.2E-24 m/s^2; and to a 5 g feather, 1/1000 of that. Just an amusing thought experiment.
  18. Oct 15, 2004 #17

    Chi Meson

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    I can't say it any better than that!
  19. Oct 15, 2004 #18


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    When two bodies attract each other and move towards each other their centre of mass stays steady. They fall toward the centre of mass and they will reach each other at the centre of mass.

    The distance between the centres is D. Then the distance of the bigger body from the CM is smaller than that of the smaller body. If the masses are M and m and the distances R and r, respectively, r=D*M/(M+m) and R=D*m/(m+M). As the distance between the bodies decreases, the bodies accelerate at different rates as they fall towards each other.

    [tex]\frac{d^2r}{dt^2}=\frac{M}{M+m} \frac{d^2D}{dt^2}\mbox{ and } \frac{d^2R}{dt^2}=\frac{m}{M+m}\frac{d^2D}{dt^2}[/tex].

    With a little math, we can write the second time derivative of D.

    For the smaller body:

    [tex] m\frac{d^2r}{dt^2}=-\gamma \frac{mM}{D^2}[/tex]

    Replacing r with D*M/(m+M),

    [tex] \frac{mM}{M+m} \frac{d^2D}{dt^2}=-\gamma \frac{mM}{D^2}[/tex]

    Rearranging a little:


    [tex] \mu=\frac{mM}{M+m} [/tex] is called "the reduced mass" of m and M.

    We get an equation of motion which looks like that for a single body of mass [tex]\mu[/tex] in the field of a central mass of (m+M). Its acceleration is proportional to the sum of the two masses. In this way, the acceleration of both a feather and the Earth towards their common CM is lower than those of a big stone and Earth - in principle.

  20. Oct 15, 2004 #19
    This is the language and the reasoning that makes explanations to ESL students so difficult. Which is it? Is acceleration independent of mass or do different accelerations depend on different masses?

    I understand the the explanations given in this thread, and I must complement everyone on their informed contributions. My issue is not with using equations to show that mass cancels from a ideal two body system. My problem is the assumption that equation manipulation communicates meaning. My work with this ESL student underscores our need, as teachers, to put problams and answers into terms students can understand to help shape their world view. I'm interested in exploring other language people have used in talking about this problem.
  21. Oct 15, 2004 #20


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    There are TWO separate issues here: How to answer your question as one would tackling a physics question, and how to EXPLAIN the question using [insert favorite language here]. In your case, you want this to be explained, not only logically, but clearly in English.

    I will put it to you that the latter isn't always possible. This is because ALL of physics use mathematics as its "natural language". The other human languages take a back seat. Your students may not be able to explain why Mass A and Mass B fall at the same acceleration, but I bet some of them can certainly prove it using mathematics as their language. It is the only unifying language of physics that do not require any translation between different societies.

    In your previous post, you asked :

    ehild has made an elegant explanation of the fact that when you switch perpective from "feather falling onto the earth" to "earth falling onto feather", you need to keep in mind that both "earth" and "feather" are actually unattached to anything else but each other. They are both suspended in nothingness, and thus, the issue of where the center of mass of the system is now comes into play. This is now no longer a trivial explanation, since you now have to invoke ideas of what a center of mass is, why would it be significant, what are the conservation laws, etc, etc., all to satisfy your requirement of a "satisfying answer". Your expectation of a reasonable and simple answer is then unrealistic, not just as a pure physics explanation, but especially as a physics explanation but in clear ESL-type english!

    I suggest you find a different, less complex problem for them to practice on shaping their world view....

    Last edited: Oct 15, 2004
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