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Gravitational pull of manmade satellites

  1. Sep 16, 2014 #1
    Hi there,

    to answer some people believing in astrology, I wanted to know what was the order of magnitude of the difference of the gravitational pull between famous stars & the man-made satellites...
    I have no knowledge in math or so I have no idea...

    I just thought that I would find it funny if the lost screwdriver from an astronaut orbiting earth would be 100x more influent than any well know distant star... :p
    Or if there is a comparison between pluto pull & the ISS pull...

    Anyone able to do the math easily ??
  2. jcsd
  3. Sep 16, 2014 #2


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    Staff: Mentor

    You can do the math yourself. If an object has mass ##m## and is at a distance ##r## from earth, it's gravitational force will be proportional to ##m/r^2##.

    Here are two examples to get you started:
    - a moon-sized object twice as far away as the moon would have one-fourth the influence (same ##m## but ##r## is twice as great, ##r^2## is four times as great);
    - an object with one one-millionth of the mass of the moon would have to be one thousand times closer than the moon to produce the same pull.
  4. Sep 21, 2014 #3
    Better yet, do a calculation of the tidal acceleration that each sort of object creates. That's the difference in gravitational acceleration across some distance, like one's size. To lowest order, it's GMa/r^3 for size a. The next-order terms have relative size a/r and thus can be ignored.

    It should be easy to do this calculation.

    But although one can show that tidal forces of distant objects are teeny teeny teeny tiny, an astrologer can get around that problem by positing some otherwise-unknown force that makes astrology work. This issue can be addressed in these ways:
    1. Independent evidence of such a force
    2. Empirical tests of astrological predictions
    For the first one, ask an astrologer about how one could detect the astrological force, for lack of a better name. What effects independent of astrological effects might it have?

    In principle, an astrologer could win in the second one, by showing empirical evidence of astrological effects. I say that an astrologer could win there because there are numerous phenomena and effects that were observed long before we developed a good theory for them, and I don't see how astrological effects are fundamentally different.

    But astrologers are aware of counterevidence to various astrological predictions, and they have a slogan, "The stars incline, but they do not compel." But there is a big branch of mathematics dedicated to testing for the occurrence of effects that incline without compelling. Statistics. So one ought to do statistical tests of astrological predictions. So far, such tests have failed.
  5. Sep 21, 2014 #4
    A lot of the visible stars will have more gravitational influence than a satellite in orbit (do the math yourself); however, I think you will find that the gravitational force gradient of a distant star is less than the gradient created by your location on earth based on altitude and density variations on the Earth's surface, although I haven't done the math on that.
  6. Sep 21, 2014 #5
    The tidal factor GM/r^3 is equal to ω^2, where ω is the angular velocity of an object in a circular orbit around it with radius r.

    A Sunlike star at 1 parsec would produce about 10^(-16) times as much tide as the Sun does. The more distant stars' tides tend to cancel out because of their different directions. Our Galaxy overall produces something like 10^(-17) times as much tide as the Sun does.

    Turning to the planets, Jupiter produces the highest tide at about 10^(-5) times the Sun's tides.

    The Sun and the Moon make tides on an Earth-surface object that are about 10^(-7) the tide that the Earth produces on it, and for a size of about 1 m, that is about 10^(-7) times the Earth's surface acceleration of gravity.

    A 1-ton satellite about 200 km upward makes a tide that is about 10^(-16) what the Earth makes.
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