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Heavenly Gravity

  1. Dec 3, 2006 #1
    If we, say had something the mass of a rock, and then one of a small asteroid, which one would fall faster to the earth?
     
  2. jcsd
  3. Dec 3, 2006 #2
    If the asteroid was small (so that you can ignore the difference of the gravity across it's length) then by the equivalence principle, both will fall at the same rate. The mass of an object does not alter it's gravitational acceleration.

    It's essentially that in the following equations

    [tex]F = \frac{GMm}{r^{2}}[/tex] and [tex]F = \tilde{m}a[/tex]

    that [tex]m = \tilde{m}[/tex].
     
  4. Dec 3, 2006 #3

    Danger

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    Also, most asteroids are rocks.
     
  5. Dec 4, 2006 #4

    LURCH

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    Of course, this is all ignoring air resistance; if both are the same size and one has far geater mass, then the one with greater mass will fall much faster because it is more aerodynamically streamlined.

    It should also be emphasized that the answers given so far ignore the gravitational effect of the object itself. A rock (the size one could hold in the hand) will have a very small, but non-zero, gravitational field of its own. A much larger rock, like the size of some of the bigger asteroids, will have a significant gravitational field. So, while Earth is pulling on each rock, the rock is pulling the Earth as well, and one pulls much harder than the other.

    But I'm only talking about a very slight difference. Although the asteroid will have a gravitational pull many millions of times stronger than the "rock", even the largest asteroids like Vesta or Cerus have a gravitational field that is negligable compared to that of Earth.
     
    Last edited: Dec 4, 2006
  6. Dec 4, 2006 #5
    All objects regardless of their size or mass will accelerate towards the earth at the same speed, about 9.8 m/s2. meaning an object will fall 9.8 meters the first second and then double its distance every second there after. Air resistance will stop an objects acceleration at its terminal velocity speed, it will continue to fall at this speed but not accelerate. Terminal velocity is subject to the objects density like for instance a styrofoam ball will reach terminal velocity before a lead ball of the same size due to the resistance of the air.

    Dropping the same styrofoam and lead balls in a vacuum will show an equal acceleration of about 9.8 m/s2 and they will hit the ground at the same time. I find this very interesting to see that F(force) is equal to GM(gravitational constant, whatever that is) multiplied by t2(time squared) and the m(masses of the falling objects, small 'm') are irrelevant. Think of dropping a penny and a huge cargo ship, logicaly we might think that the ship will fall faster and hit the ground sooner then the penny but that isn't what really happens. Well, not like i've tried that experiment myself but if you have the chance try the one in a vacuum with objects of different density. And be carefull when you ask what the terminal velocity in a vacuum is, there seems to be a problem with finding a correct answer there.

    When I said GM or 'G'(gravitational constant, whatever that is) I meant that the speed of which objects fall is a number gained by actually watching and timing them. Is the actual mass of the earth used in this equation? Or do we just add in 9.8 m/s2? Of coarse this 'constant' will be different for other planets. Not to try and confuse anyone here but I would like to know why things fall at the same speed regardless of their mass.
     
  7. Dec 5, 2006 #6
    Even so, if an object with a larger mass is particularly spread out and thin, compared to a smaller mass which is compact, then because the surface area being resisted against by the air is larger the acceleration from air resistance could also be larger.

    Although, in the case of an asteroid it would require a spectacularly low density I would imagine...
     
  8. Dec 5, 2006 #7

    arildno

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    This is (theoretically) incorrect, devino!

    We wish to find an object's RELATIVE acceleration to the Earth, using only the law of gravitation between the object and the Earth.

    In an inertial frame, we have:
    [tex]a_{obj}=\frac{F}{m_{obj}}, a_{earth}=-\frac{F}{m_{earth}}[/tex]
    where the notations should be self-explanatory.

    But, the relative acceleration of the object, as seen from the Earth, is therefore:
    [tex]a_{rel}=a_{obj}-a_{earth}=-\frac{G(m_{earth}+m_{obj})}{r^{2}}[/tex], where negative direction means towards Earth.

    Thus, the object's acceleration is perceived, from the Earth, to depend on the object's mass, since the Earth will experience a stronger absolute acceleration interacting with an object with greater mass.
     
    Last edited: Dec 5, 2006
  9. Dec 5, 2006 #8
  10. Dec 6, 2006 #9
    What is theoretically inccorect? I think I understand you in that there are discrepancies in gravitational theories but empirical observations are not theories. The first theory was objects with more mass accelerate faster then ones with less. Galileo proved this wrong some time ago, then it was said that air resistance caused them to equal out but this was also proven wrong as shown in scott1's movie link (empirical evidence). Now I have found an explanation that it's the objects inertia that equals out their acceleration (more mass more inertia), it seems that we are replacing one bad explanation for another here. How do we test this last one?

    Chutzpah's post, as simple as it may seem, brings out some interesting questions that I find myself thinking about alot. I still haven't found a good explanation why objects fall at the same rate. It seems that the theory makes a case that the two masses are nessecary to find its acceleration rate (the earth and the dropped object). As we observe objects falling toward a large mass like the earth or the moon they fall with equal acceleration regardless of there mass. Just look up Gravitational Acceleration .

    Arildno, I have to confess I'm not good at math and I don't understand your equations, but I think I see that the earths gravity is pulling all objects toward itself with equal force depending on the distance. The weight of the objects pull on the earth, which is negligible, and doesn't add to their acceleration. and if that were the case then why is the Terminal Velocity of these objects different? Or maybe the ratio of the earths mass compaired to the objects are so great that the difference between a 1 gram and 20 ton object is negligible, in other words the 20 ton object falls faster but its so small we can't yet see it.

    The answer I have found to be true is that we don't know the cause of gravity and only assume it's mass. This is only one of many inconsistancies of our understanding of gravity which is what intrigues me.
     
  11. Dec 6, 2006 #10

    vanesch

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    What arildno pointed out, is that if you call "acceleration" the acceleration wrt an observer on the earth's surface (and not the acceleration wrt an inertial observer somewhere in outer space), then, that accelation will be function of the object's mass when the object's mass is not totally neglegible wrt the earth's mass, simply because the object then gives a non-negligible acceleration to the earth which you have to add to the acceleration of the object itself.

    All this is entirely within the scope of Newtonian physics.
    (that said, even a small asteroid is neglegible as compared to earth).
     
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