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What if inertial mass did NOT = grav. mass?

  1. Jul 10, 2006 #1
    What if inertial mass did NOT equal gravitational mass?

    How would our normal daily existence be different?

    Most interesting and creative answer wins !

    [Edit: This is intended to be a fun question to answer, but I'm hoping to learn something quite serious from it. The force of attraction between two charged particles is a function of their charge and distance apart. The resistance to that force is the inertial mass of the particles, which has little to do with the charge. The stronger the charges (assuming oppositely charged) and the shorter the distance, the greater the acceleration. The greater the inertial mass, the less the acceleration. Similarly with magnetic bodies. The acceleration due to magnetic attraction is resisted by the inertial mass, having little to do with magntism.

    Not so for accleration due to gravity, where the attraction and resistance are both due to mass (given the equivalence of gravitational and inertial mass).

    If they weren't equivalent, how would the simple act of walking, or throwing a baseball or driving a car, or flying a rocket to the moon, be different?]
    Last edited: Jul 10, 2006
  2. jcsd
  3. Jul 10, 2006 #2
    Eotvos would be coming to haunt you (being Transylvanian, he was related to Dracula)
  4. Jul 11, 2006 #3
    Let's put aside charge for a moment, and relativistic speeds. Mass is a measure of resistance to change in motion.

    When we talk about inertial mass, we're usually looking at a stationary mass and thinking about applying a force to get it moving. So we give it a push, and we measure the force applied and the time we applied it for, whereupon we see our object accelerating by a measured amount. We then calculate the inertial mass as a relation between the force, time and acceleration.

    When we talk about gravitational mass, the force acting on our object is gravity instead of pushing. It's a different type of force, but it's the same object. How can that object exhibit a different resistance to change in motion to one flavour of force compared with another? It can't. Force is force.

    OK, the magnitude of the force might vary with the distance from another mass, then catching a ball might come with a late rush and a click, like you were catching a magnet with a magnet. And then it would be real hard to throw the ball away. But that's a whole different kettle of fish from what you were asking. And I can't see how what you were asking makes sense. Even on the moon.
  5. Jul 11, 2006 #4
    Gravitational mass and inertial mass need not be equal, just proportional. If they are not proportional then the rate at which an object falls will depend on the inertial and gavitational mass. The equivalence principle woulf then no longer hold to be true.

  6. Jul 11, 2006 #5
    Heavier objects would fall faster than lighter objects.
  7. Jul 11, 2006 #6
    Now that's got me thinking. Does anybody know the answer to this:

    If we held two earth-sized planets ten miles apart and then let go, would their surfaces accelerate towards one another at 9.8ms-2 ?
  8. Jul 11, 2006 #7
    10 miles is barely any distance. But assuming an ideal situation, yes, they will have the same accelaration. Gm/r^2 will be same in both cases. It's symmetrical...you can't expect one to move faster than the other.
  9. Jul 11, 2006 #8
    I was wondering if they'd accelerate towards one another at a combined closing of rate of 19.6ms-2.
  10. Jul 11, 2006 #9


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    The gravitational forces cancel, so of course there would be no acceleration towards each other or in any other direction, at least for the surfaces that are 10 miles apart.

    By the way, holding the planets apart might be kind of tough on your wrists.

  11. Jul 11, 2006 #10
    Farsight - I think Neutrino is right. Unless they were in mutual orbit, the gravitational acceleration between two Earth-sized planets positioned only 10 miles apart should be approx. 9.8 m/s^2 in total. Both would move towards the center of mass of the two (halfway in between) at about 4.9 m/s^2.

    Reasoning: 9.8 m/s^2 is roughly the gravitation acceleration at the surface of the Earth for ANY OBJECT, regardless of its mass. 10 miles, compared to the 4,000+ miles radius of the Earth is not significantly different, so 9.8 is probably still correct to one decimal.

    An apple, or an Elephant, or the Moon, or another planet Earth, all if positioned just 10 miles above the Earth, would fall at the same speed (ignoring air resistance). And they all would fall toward their common center of gravity. But with the apple and the Elephant, that center is essentially the center of the Earth since it's so much more massive. And the apple and elephant do most of the accelerating and the Earth barely budges.

    But in your example of two Earths, they would both share the acceleration equally because of their identical masses.
    Last edited: Jul 11, 2006
  12. Jul 11, 2006 #11
    pym_phy and actionintegral - very insightful responses !!

    So, if A) gravitational mass was greater than inertial mass, then Aristotle would have been right all along. . . heavier objects would fall faster than lighter ones.

    But if B) inertial mass was greater than gravitational mass, then LIGHER objects would fall faster than HEAVIER objects !!! Wouldn't that be wierd !

    Objects falling from great heights might break apart on their fall as the heavier/denser parts seperated from the lighter/less dense parts because of the differential in gravitational forces and inertial forces. ??

    Would carrying a 15 lb. bowling ball in one hand and an 8lb. one in the other would feel quite strange, as perhaps the 8lb. ball would be easier to keep from dropping, but would offer more resistance to the change of direction the forward and back motion of your arms as you walked????

    Sounds ripe for an Isaac Asimov novel !
  13. Jul 11, 2006 #12

    Eotvos has demonstrated long ago the equivalence between the "two" masses. Reenactments of the Eotvos experiment have brought the error bars of the experiment towards 10^-20 (see the Eot-Wash experiment). So, if you continue to post, Eotvos' vampire will come hunting and haunting :surprised

    And he'll get you!
    Last edited: Jul 11, 2006
  14. Jul 11, 2006 #13
    Clj4. . . I think you may have misinterpreted the intent behind my question. :-) I have no quarrels with Eotvos, and certainly am not trying to deny that inertial and gravitational mass are equal. Of course they are equal.

    I just thought it was in interesting question to contemplate what it would be like if we had been born into a universe where they were somehow not equal.

    Plus, sometimes it helps me understand things better if I can contemplate what it would be like if the laws of nature were reversed somehow.

    But I'll offer my offical appologies to Eotvos if in case I've offended him (or you).

    Last edited: Jul 11, 2006
  15. Jul 11, 2006 #14
    Yes, I understood. I thought it was a contest for the funniest answers.

  16. Jul 12, 2006 #15
    Oh yeah. . . . in that case, you clearly win !

  17. Jul 14, 2006 #16
    No. The 9.8ms-2 refers to test particles falling in the gravitational field of earth assuming that the motion of the earth can be neglected. Thus a coffee cup would fall at 9.8ms-2 but Jupiter wouldn't. In fact Jupiter's mass is so much larger than earth's that earth would fall in Jupiter's gravitational field as if Jupiter was at rest. Then the earth would fall at a different acceleration. The rate a body falls depends on where the body is. A body such as the Earth would not fall as if the entire mass was located at its center unless the field was uniform enough so that spacetime curvature could be ignored.

  18. Jul 14, 2006 #17
    pmb, or anybody: Supposing I dropped a test particle into a black hole and timed its initial fall, then calculated its acceleration at 9.8ms-2. Now if I repeat this test replacing the test particle with a black hole identical to the other one, would the two black holes accelerate towards one another at 9.8ms-2?
  19. Jul 14, 2006 #18

  20. Jul 14, 2006 #19


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    There is a project under way at CERN to produce and study neutral anti-hydrogen. A key experiment in this project is the comparison of the gravitational infall rates of neutral antihydrogen and hydrogen. This is a critical test of the weak equivalence principal. Not a funny answer, but the most relevant I could come up with.

    http://alpha.web.cern.ch/alpha/overview.html [Broken]
    Last edited by a moderator: May 2, 2017
  21. Jul 20, 2006 #20
    If inertial mass is not equal gravitational mass, general relativity changes in general objectivity. Therefore uniformly accelerated elevator is not equal to homogeneous gravitational field.
  22. Jul 20, 2006 #21


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    Indeed. In fact, we would make them equal, by redefining Newton's gravitational constant, G.

    What needs to be, is that the ratio of inertial mass to gravitational mass is the same, for all bodies.
  23. Jul 20, 2006 #22


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    Well, assuming that we can replace a spherical mass by a point mass (there's a theorem on that, but I'm not sure it applies in exactly this circumstance), we have two point masses of the mass of the earth at a distance of 2 R (plus 10 miles).
    Now, we know that the gravitational attraction of earth1 on earth2 is G M^2 / (2 R)^2, so filling this in into Newton's second law in the inertial frame which is the center-of-mass frame:

    M.g1 = G M^2 / (2R)^2 or g1 = G M / 4 R^2

    g1 is the acceleration with which earth2 is accelerating towards earth1.

    Now, for a test particle at the surface of the earth, we have:

    m g = G m M/R^2 or g = G M/R^2 = 9.81 m/s^2, so we find that g1 = g/4

    By symmetry, g2, the acceleration by which earth1 is accelerating towards earth2, is also equal to g/4.

    So we have that both earths accelerate towards the center of gravity with an acceleration of g/4 ; so from the PoV of the surface of one, the other is accelerating 2 g/4 = g/2.

    So, as seen from the surface of earth1, earth2 is "falling" with an acceleration of g/2 = 4.9 m/s^2.

    Under the assumption that we can use the theorem that a spherical mass can be replaced by its mass concentrated in its center, of which I'm not 100% sure that I can apply it here...
  24. Jul 20, 2006 #23

    If you start with R>> than the radius of earth1 , earth2 then your calculation would work perfectly because the volume integrals can be shown to be independent (almost) of the earth radius.

    There is a problem, though: we all know that the test probe will experience a given acceleration (say 9.81m/s^2) only in the vicinity of the earth surface. So , now you need R to be very close to the earth radius. Thus, your proof contains a contradiction.

    So , we cannot solve this problem simplistically. What to do?
    Here is the calculations leading to the correct solution (within Newtonian physics)
    Last edited: Jul 20, 2006
  25. Jul 20, 2006 #24


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    Yes, well, what does it say ? For a full sphere, outside of the sphere ?
    (so R > b, and a = 0)

    We have that the potential is - 4 pi/3 b^3 rho G / R

    Now, 4 pi / 3 b^3 rho is nothing else but the total mass of the sphere, so we have that outside of the sphere, at a distance R from its center, the potential is - M G / R which is exactly the same as if we lumped the entire mass at its center. That's the theorem I talked about.

    What I'm slightly less sure of, is: does a sphere with density rho, bathing in such a potential, also act as if it were a point particle with its mass in its center ?
    That is, if we sum all the forces acting on all the parts of a sphere, exposed to the potential of a point particle (but of course off-center), does this sum equal the force that would act on a point particle in the center of that sphere ? It is somehow the reciproke of the above theorem and my gut feeling would tell me that it is ok, but I should work it out.
    This was what I'm not sure about.
  26. Jul 20, 2006 #25

    There was a more severe problem, I am not sure you caught what i was saying.:

    You are using R in two contradictory ways:

    1. To calculate the attraction between the two "earths" (R is big, much bigger than the "earth" radius)

    2. to calculate the acceleration of the test probe (R needs to be very close to the "earth" radius in order to get your predicted value of 9.81m/s62).

    You cannot have both 1 and 2 to be true.
    Last edited: Jul 21, 2006
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