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G is calculated with g?

  1. Apr 17, 2012 #1
    i see G is calculated with g,or assumed to be the same,just other value,for all planets,stars,etc,variable being just mass and radius(they are also calculated with g)...
    so,on the moon gravity is x time smaller then on the earth,because the mass of the moon is x time smaller.on the sun is bigger,etc,etc,etc...
    i doubt very much...
    isn`t it tautology?
  2. jcsd
  3. Apr 17, 2012 #2
    It is the other way round. G is the universal constant and g is the derived one.

    Thus from the mass, radius and G you can calculate the local g

    G can be measured, look up 'cavendish experiment'

    So the 'g' for any planet can be different – and if it is spinning , as the earth does, is dependent on latitude because the 'centrifugal force' near the equator reduces 'g'.

    Hope this helps



    PS when I say G is universal I am neglecting unproven dark energy theories
  4. Apr 17, 2012 #3
    thank you...
    it`s the other way around because it`s reversed trough extrapolation...
    g was experimental measured on earth firstly...
    i agree with spinning and centrifugal force,not interested in dark or bright theories...
    in the cavendish experiment i see only what is meant for,ratio ground/water,the water being chose to have 1 unit traditional density...
    this experiment,like that with pendulum,are made on earth...
    i doubt that it`s as simple as that,`you sit at your desk,calculate `g` for where you want to go,manufacture a rocket,on with your costume...and off you go...`
    so,one experiment has to be made out of space to convince me...
  5. Apr 17, 2012 #4


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    The cavendish experiment does not use the gravity of the earth in any way. Only the gravitational force between two masses is tested, therefore you can calculate G with it.

    Using the known radius of the earth and the known acceleration g on the surface of the earth, you can calculate the mass of the earth.
    In a similar way, if you know the mass and the radius of an object, you can calculate the local gravitational acceleration g with your knowledge of the fundamental constant G. g=GM/r^2, neglecting rotation here.

    Look up details to Gravity Probe B, if you want to see precision tests of gravity in space.
  6. Apr 17, 2012 #5
    Something just occurred to me about the Cavendish Experiment, and I felt like posting it somewhere. So here it is.

    Won't it be affected by the Earth's rotation, sort of like the Focault Pendulum?
  7. Apr 17, 2012 #6


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    As the rotation is limited to one dimension, the two probe masses have a velocity in opposite directions, and the timescale of the experiment is much smaller than 24 hours, I don't see any significant effect.
  8. Apr 17, 2012 #7
    cavendish mechanical experiment is made on earth,but doesn`t use g in any way,after you...
    and you neglect earth rotation,which is the cause of gravity...then why you need RADIUS for?radius is for rotating objects...
    you simplify what you suit you to simplify for an unproven theory in the end which everybody knows and i i don`t agree with...
  9. Apr 17, 2012 #8

    Unproven theory? Technically, it has been disproven and a more general form of it, general relativity, has come into play, but this is a good enough approximation for anything. Newton's Universal Law of Gravity describes the motion of the planets very well. Almost perfectly, the largest deviation is in Mercury's orbit, which is something like a deviation of one part in [itex]10^7[/itex]. We don't need to know G to notice how well this works.

    Radius. I'd imagine that means the radius of the bar that the two balls are on. That's what radius is for, and, again, the Earth's rotation is negligible.
  10. Apr 17, 2012 #9


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    Do you need g to evaluate how long you have to accelerate with 1m/s^2 in horizontal direction to reach 10m/s? No.
    The cavendish experiment uses only forces perpendicular to the earth's gravity.


    Wrong. Take a ball from a random ball game. Do you have to rotate it to determine its radius?
    The radius of a spherical object is the distance between surface and center, and that is important to calculate g.

    No, I can do the math. And you?
  11. Apr 17, 2012 #10


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    So you have never actually taken a physics course, right?
  12. Apr 17, 2012 #11
    http://www.infoocean.info/avatar1.jpg [Broken]The cavendish experiment does not use the gravity of the earth in any way.
    Last edited by a moderator: May 5, 2017
  13. Apr 17, 2012 #12
    The earth's rotation is the cause of gravity? Lololol. Unproven theory? LOLOLOLOLOL. General relativity is one of the most proven theories of all time.
  14. Apr 17, 2012 #13
    i don`t see any counter argument( lol is not an argument),only quick childish not satisfy reactions...like a herd of dogs attacking a new one to show who`s the boss here...
    thank you very much for conversation ...
  15. Apr 18, 2012 #14

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    Looks like a good time to close this thread, and a good time to remind people of the PF Rules on Overly Speculative Posts.
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