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Gravitational forces between 2 masses

  1. Sep 8, 2010 #1
    22f17757801c9ae9afd6dd9374f877b1.png

    Its what Newton's law states. That one mass attracts another mass with the force equal to the product of this equation. The same should apply to the second mass.

    And now this is the part that confuses me. When we jump into the air and fall down on the ground, shouldnt we hit the ground with double the force which we calculated from Newton's law, since Earth accelerates towards our center of mass aswell?

    Thanks in advance,
    fawk3s
     
  2. jcsd
  3. Sep 8, 2010 #2

    ZapperZ

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    Do you know where the center of mass for the "you-earth" system is? How much do you think the earth has to move to get there?

    Zz.
     
  4. Sep 8, 2010 #3

    Doc Al

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    (1) Don't confuse the contact forces generated between the ground and your feet with the force of gravity.
    (2) The acceleration of the Earth is minuscule compared to your acceleration.
     
  5. Sep 8, 2010 #4
    The acceleration of Earth is indeed very small, but yet its mass is big compared to us. So shouldnt the force be the same?

    Im not sure what you mean by (1) though. Could you elaborate?
     
  6. Sep 8, 2010 #5

    Doc Al

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    I don't know what you mean. When two things collide, like you and the ground after you jump into the air, they exert equal and opposite forces on each other per Newton's 3rd law. So?
    The force of the collision is not the force of gravity. Jump up one half inch; then jump off a 10 foot ladder. Would you agree that the force between your feet and the ground, upon collision, is different in those two cases? Did gravity change?
     
  7. Sep 8, 2010 #6
    Oh ok. I understand what you mean. But when jumping from higher, Earth also has more time to accelerate.

    Ok, lets say Im standing on the ground. Im exerting a force on the ground due to Earth pulling me towards its center of mass. The ground is exerting an equal but opposite force on me due to elasticity.
    But what about the force Earth is exerting on my feet due to it accelerating towards me? Its supposed to be equal with the one Im exerting on the ground. Or am I completely wrong here?
     
  8. Sep 8, 2010 #7

    Doc Al

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    Please take a moment and actually calculate the acceleration of the earth. It is vanishingly small and plays no role in the collision. (When you jump up, the earth does move away with the same momentum you have going up. But for all practical purposes, the speed of the earth is zero.)

    You and the earth exert equal and opposite gravitational forces on each other.
    The ground resists your attempt to get closer to the center of the earth. You and the ground push against each other with equal and opposite forces. As long as you are at rest, the force the upward force of the ground happens to equal the downward gravitational force, so the net force on you is zero.
    I don't know what you mean. Again, the earth isn't accelerating!
     
  9. Sep 8, 2010 #8
    But why isnt the earth accelerating towards me when I am towards it? Im accelerating because of the gravitational force the earth is exerting on me. I am exerting a gravitational force on earth too.
     
  10. Sep 8, 2010 #9

    Doc Al

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    OK, the earth does accelerate. But only a teeny, teeny amount!

    Your mass (say) is 100 kg; the earth's mass is 6 x 1024 kg. Near the earth, you accelerate at 9.8 m/s^2; the earth's acceleration toward you would be smaller by a factor of 6 x 1022! I think we can safely treat the earth's acceleration and speed as zero for the sake of this problem.
     
  11. Sep 8, 2010 #10
    Ok so my feet are exerting a force of about 9800 N on the ground and the ground is exerting the same force on my feet.
    But also the earth is applying a force of 9800 N on my feet. (Right? And my feet are exerting the same on the earth.)

    So why cant I feel 19600 N on the bottom of my feet?

    Sorry if Im just being stupid but I think I just fail to understand at the moment.
     
  12. Sep 8, 2010 #11
    Imagine a spring-and-dial scale that is in-line with a rope. One end of the rope is tied to a hook on the ceiling. The other end has a pan. Put 10kg in the pan. The dial reads 10.

    Unhook the end from the ceiling. What happens?

    Now, thread the rope over a pulley, and hang another 10kg from the loose end of the rope. What does the scale read? Is that situation different in any way from the first?
     
  13. Sep 8, 2010 #12

    Doc Al

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    Right. That's why you 'feel' a force of 9800 N.
    The earth exerts a gravitational force on your entire body that adds up to 9800 N. (And your body exerts an equal force on the earth.) That's the reason why the ground needs to exert an equal upward force to support you.

    What you 'feel' is the contact force that the ground exerts on you.

    Try this example. Imagine you are holding a 100 lb barbell overhead. You are exerting an upward force of 100 lbs on the barbell, but the earth is exerting a downward force on it of 100 lbs. So, do you feel a force of 100 lbs on your hands or 200 lbs?
     
  14. Sep 9, 2010 #13
    Ok so say there's this scale-thingie, and lets say it has no gravitation.

    [PLAIN]http://img405.imageshack.us/img405/3331/scalething.png [Broken]

    There are two masses, doesnt matter if different or same masses. So they both exert equal but opposite gravitational forces on each other, lets say 1 N each. (The arrows are supposed to represent the forces.) How does the scale act? Does it read 1 N or 2 N ?

    If it reads 1 N, both ends of the scale moving by 0,5 N on the scale, then I think I understand.
     
    Last edited by a moderator: May 4, 2017
  15. Sep 9, 2010 #14
    The system is at rest, which means that each plate is applying on the respective mass a 1N force. At the same time each mass is applying to its plate an equal and opposite force.
     
    Last edited by a moderator: May 4, 2017
  16. Sep 9, 2010 #15

    Doc Al

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    The scale will read 1 N.

    In order for the scale to read 1 N, each end must be pushed with a force of 1 N. Just like when you stand on a bathroom scale. You push down on the scale with a force equal to your weight while the ground pushes up on the scale with an equal force.

    A similar example is tension in a (massless) rope. To create a tension of 1 N, each end must be pulled with a force of 1 N.
     
    Last edited by a moderator: May 4, 2017
  17. Sep 9, 2010 #16

    Doc Al

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    The system is at rest, which means that the net force on each mass is zero. On Mass 1 you have the 1 N force from Mass 2 pulling it one way and the 1 N force from the scale pushing it the other way.

    The scale reads 1 N.
     
  18. Sep 9, 2010 #17
    And the same happens on the bottom plate, where Mass 2 is pulled upwards by mass 1 pushing the bottom plate.

    Let's say that the scale is regulated so that it provides a reading equal to the intensity of the force applied on one side of the spring.
     
    Last edited: Sep 9, 2010
  19. Sep 9, 2010 #18
    My process of thought was something like this:

    If there was only one of those forces being applied on the scale, the scale would start moving with an acceleration and the scale wouldnt show anything if we left inertia out of the game. But if there was an equal and opposite force being applied, the scale would stand its ground and the springs would be compressed.

    But would it be correct to assume that both of the masses compress the scale by 0,5 N?
     
  20. Sep 9, 2010 #19

    Doc Al

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    Right. The same thing happens to both masses.

    The scale reads the force with which it is being compressed. In this case, 1 N.
     
  21. Sep 9, 2010 #20
    huge10, think of one of the masses as being Earth. The other one is 1/9.8kg. The scale reads 1N, not 2.

    Think of one mass as doing all the pushing and the other just making the whole contraption stand still, as opposed to being pushed away.

    fawk3s, no, each doesn't compress by half. If only one side is pushing, nothing can compress because the whole scale gets pushed out of the way instead! You must have a resisting force, and it is the same full amount. Trying to fathom what one side is doing by itself is like contemplating the sound of one hand clapping: it's not doing half the total work, but is part of a necessary complete system.
     
  22. Sep 10, 2010 #21
    I can agree that if the masses are different, the one with the smaller mass would do most of the work, because it gets a more acceleration, therefore more speed and according to the work formula A=Fs it will do more work. This was my initial question actually and Im happy I got an answer for it. Thanks guys. :shy:
    But lets assume for a minute that the masses are same. Equal forces, therefore equal acceleration on both. Wouldnt each of them compress the springs by 0,5 N? Because its not like one of them would randomly want to stand its ground, right? Of the 1 N exerted on one mass, wouldnt 0,5 N go for holding the scale in place since the other mass is pushing with 0,5 N, and vice versa?
     
  23. Sep 10, 2010 #22
    Yes, 2N is clearly wrong.
    As a matter of fact the force on the spring is zero, each mass applying a 1N force in opposite directions.

    The whole point is the sentence in bold.
     
  24. Sep 13, 2010 #23
    I agree that the spring will be compressed equally from each side, so that it would look the same if flipped in a mirror.

    But it can't be said that either side is compressing by itself. Without the other side to at least hold it, it would not compress at all.
     
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