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Universal Law of Gravitation

  1. May 12, 2009 #1
    Here is the succinct story of apple which fell on Sir Newton. Since both apple and earth attract each other with a force equal in magnitude from both sides but apple move faster towards earth because of its greater acceleration than negligible acceleration of earth. So Newton derived the equation of univesal law of gravitation (F= GMm/R2) by putting a mass of 1 kg of sphere on the surface of earth.

    ACTIVE PART OF EQUATION - IMAGINE 1 kg SPHERE ON EARTH

    F1=Weight of sphere=W=GMm/R2=mg1 = 9.8 Newton,
    Where M= mass of earth, m= mass of sphere, R = is c/c distance b/w sphere and earth, G= gravitational constant, g1=GM/R2 = gravitational acceleration say 9.8 m/s/s

    REACTIVE PART OF EQUATION - IMAGINE EARTH ON 1 kg SPHERE

    F2=Weight of Earth on sphere=W=GmM/R2=Mg2 Newton
    Where g2=Gm/R2 = sphere's gravitational pull at height R or accelaration of earth due to gravity of sphere = 1.6395685e -18 m/s/s.
    F2=W=g2M= 1.6395e-18 x 5.9742e+24 = 9795110.168782 Newton
    OR
    It would be difficult to find gravitatinal pull of sphere if we put another same sphere on aforementioned sphere. But anyway if g2 = Gm/r2=6.67x10-11 m/s/s
    where m= 1 kg and radius of sphere = 1 m then
    F2= 398479140000000 Newton

    So in both cases F2>F1, Since both forces F1 and F2 should equal in magnitude (g1m=g2M)
    why F2 > F1 ?
     
  2. jcsd
  3. May 12, 2009 #2

    Nabeshin

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    Umm, I calculated g2 to be 1.64*10^-24, not your value.

    Using G=6.67*10^-11, m=1kg, r=6.37*10^6m.

    I have no idea what you're doing to get your 2nd F2 value, but I don't think it relates to the first calculation.

    Edit: Looks like you used r=6.67*10^3m (perhaps confused units, because this is the radius of the earth in km, not m)
     
    Last edited: May 12, 2009
  4. May 12, 2009 #3

    D H

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    It's not. You used the wrong value for R. That should be the distance from the center of the Earth, not the height above the Earth's surface.
     
  5. May 12, 2009 #4
    yeh, your g2 value is correct. thanks
     
  6. May 13, 2009 #5
    Thank you Nabeshine for the corection
    Ok, now F1 = g1m =g2M = F2= 9.8 Newton, So both forces are equal in magnitude but opposit in direction and hence cancel out or face to face at the common tangent point of earth and sphere of 1 kg.
    Now here is my question that how come this “ F=GMm/R^2” which still exist between earth and the sphere of 1 kg when both aforementioned forces F1 and F2 which are equal in magnitude but opposite in direction cancel out (or meet) at the common point of earth and sphere.
     
  7. May 13, 2009 #6

    Doc Al

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    Huh? The forces are equal and opposite but they don't "cancel out" since they act on different bodies. The earth attracts the sphere and the sphere attracts the earth. So?
     
  8. May 13, 2009 #7
    Consider an apple (or sphere) at any height “h” and is falling with g1 on the surface of the earth. Both apple and earth attract each other with a force equal in magnitude (F1=F2) but opposite in direction.
    The g1 of apple (or sphere) is 9.8 m/sec/sec towards earth while
    The g2 of earth is 1.64 x 10^-24 m/sec/sec towards apple (or sphere)
    Since g1>g2 therefore apple (or sphere) move (fall) faster towards earth and both bodies comes to rest when they touch each other. At rest neither apple (or sphere) is accelarating (at any rate m/sec/sec) further towards earth nor earth towards apple (or sphere).
    The velocity of any object is zero at its rest position and since accelaration is the rate of change of velocity therefore, technically/ theoritically at rest, accelaration of any object is also equal to zero.
    Therefore, when both apple (or sphere) and earth rest on each other the value both g1 and g2 should also equal to zero because of no further rate of change of velocity towards each other.
    Therefore F1=g1m=g2m=0=F2
    Just like, we don’t feel the weight of earth nor earth feel our’s (at rest, F1=F2)
    Please just ignore in case someone dislike it. Thanks.
     
  9. May 13, 2009 #8

    D H

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    Deleted. My post was about a totally different thread.
     
    Last edited: May 13, 2009
  10. May 13, 2009 #9

    Doc Al

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    No, the reason why an object on the earth remains at rest (with zero acceleration) is because the net force on the object is zero. The surface of the earth pushes up (with a non-gravitational contact force) just enough to balance the earth downward gravitational pull. It has nothing to do with the fact--from Newton's 3rd law--that the gravitational pull of earth and object on each other are equal and opposite.
     
  11. May 13, 2009 #10
    No, the reason why an object on the earth remains at rest (with zero acceleration)….
    If accelaration=g=0 at rest then how can we calculate weight of an object on the surface of earth with equation w=mg.
     
  12. May 13, 2009 #11

    ZapperZ

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    You find the amount of force needed to counterbalance mg! That's what weighing scales do!

    Zz.
     
  13. May 13, 2009 #12

    Doc Al

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    "g" is the acceleration that an object would have if it were freely falling (ignoring air resistance) near the earth's surface. Think of it as a measure of the strength of the earth's gravitational field. When you place an object on a table, its acceleration goes to zero (because other forces besides gravity are at work--namely the contact force of the table), thus a = 0; this does not mean that g = 0! Near the earth's surface, g is always about 9.8 m/s^2--never zero. We calculate the weight using w = mg, not w = ma!
     
  14. May 13, 2009 #13

    D H

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    I'm being a bit picky here.

    The net force on an object at rest on the surface of the Earth is not zero. The Earth is rotating about its axis so there is a centripetal force on an object on the Earth's surface. This centripetal force is zero only at the poles. The Earth is also orbiting the Sun (and the Moon), so there are (rather small) tidal forces; these are zero only at the center of the Earth.

    What's the difference? A 200 lb person weighs (scale weight) about 1.06 pound-force more at the poles than at the equator, with about 0.69 lbf arising directly from the Earth's rotation in the form of centripetal force. (The remaining 0.37 lbf arises from the Earth's equatorial bulge, which also arises from the Earth's rotation).

    This is exactly why some scientists (mostly geologists) distinguish between gravitation and gravity.

    An ideal scale measures the magnitude of the total of all non-gravitational forces acting on some object. A weighing scale comes close to this ideal.
     
  15. May 13, 2009 #14

    Doc Al

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    Do you seriously think that this detail will be appreciated by the OP? :wink: (I think adding the complication of the earth's rotation will just confuse him more.)
     
  16. May 14, 2009 #15
    Ummmm, no leniency for hocus-pocus pelting.

    The net force on an object at rest on the surface of the Earth is not zero....
    This means that an object at rest doesn’t obey the first law of Newton and nor at its uniform motion because of Newton law of gravitation which states every two objects attract each other. As there is no single object in the universe therefore an object moving with uniform motion is always attracted by other object due to F=GMm/R^2

    If you want to consider all other aforementioned forces too then this means that this law needs modifying and you are also forgetting the weight/mass of vertical column of air on an object too.

    The acceleration g or a is acting along the line of action of force. Both cease when F = 0at the contact point. If I sit on the table still I don’t feel the weight of table or earth because both forces F1=F2 become equal at the contact point of table surface and me.

    The weighing scales tell us only that how much mass of an object is on the earth. It has nothing to do with the Newton’s law of gravity.

    Uhhhhh, another question because of your weight scale
    Should the weight scale shows some reading because of the theory, If we put earth on the weight scale?

    Furthermore,
    Equation was derived on the assumption that earth is a homogeneous sphere while it is not and composed of many different things with variant densities. So the actual center of gravity of earth is somewhere else but not at its center. So this means that thing should fall on the ground surface at an angle to the normal except at two location where the real center of gravity of the earth is close and far to its surface.

    So why an object is falling straight on the ground surface not an angle???
     
  17. May 14, 2009 #16

    Doc Al

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    No, it means an object "at rest" on the surface of the earth is actually accelerating due to the earth's rotation. I recommend that you first understand how to handle gravity for the simpler case of a non-rotating earth before adding complications.
    Nope. Again, g will not equal zero.
    When you sit on a table, you feel the table push up against you. That force is not a gravitational force. Ignoring the earth's rotation, that upward contact force will balance your weight. The fact that your body exerts an equal gravitational force on the earth is irrelevant.
     
    Last edited: May 14, 2009
  18. May 14, 2009 #17

    ZapperZ

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    This is getting to be utterly ridiculous.

    Turn off gravity. Do you think the weighing scale will STILL read the same value? After all, that object you're measuring still has mass. By your definition, if I stand on such a scale in an accelerating upwards, I've GAINED MASS just because the weighing scale reads a larger value. You don't find this to be absurd?

    This has become nothing more than your desire to learn or figure out what's wrong with your idea. Instead, you are trying to promote a very faulty understanding of basic mechanics. This is the wrong place to do it.

    Zz.
     
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