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Method to determine surface gravity (g) at home?

  1. Nov 10, 2013 #1
    Though standard gravity is 9.80665 m/s2, it must vary from city to city, since the Earth is neither a perfect sphere nor of uniform composition.

    I'm looking for an experiment to accurately determine the value of g at my current location. The method needs to be precise enough to yield different values when performed in different places, for some value of "place." I think the minimum is to accurately obtain g to 3 decimal places.

    I've come up with the following unusable methods:
    • Measure a very precise weight with a very precise scale (expensive)
    • Measure the time it takes a standard weight to fall (imprecise)
    • Use Wolfram Alpha (not interesting)
    • Measure the friction between two surfaces with a known coefficient of friction (imprecise)

    Is there a way to use Newtonian physics to accurately find g without spending a lot of money?
     
  2. jcsd
  3. Nov 10, 2013 #2

    jtbell

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    How about measuring the period (over many cycles) of a pendulum of known length?
     
  4. Nov 10, 2013 #3

    hilbert2

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    When I was doing some early physics lab exercises, we measured ##g## with a pendulum (long string attached to the ceiling with a mass hanging from it). We measured the time needed for a large number of oscillations of the pendulum to get a relatively accurate value for the period of oscillation ##T_{0}##. When we know the period of oscillation, and the length ##l## of the string, we can solve ##g## from the equation ##T_{0}=2\pi\sqrt{\frac{l}{g}}##.

    This kind of measurement takes some time, though, as the period of oscillation has to be measured pretty accurately.
     
  5. Nov 10, 2013 #4
    I'd forgotten about pendulums! Do you think 4π2L / T2 will be accurate enough for my purposes, or do I need to work harder? If the latter, how can I deal with lessening swings?
     
  6. Nov 10, 2013 #5

    hilbert2

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    The period of oscillation does not depend on amplitude. The damping of motion due to air resistance does not change the results significantly. You just measure, like, the time needed for 40 full swings.
     
  7. Nov 10, 2013 #6

    AlephZero

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    Both those statements are wrong, but to be fair, for practical measurements the second one is much less wrong than the first.

    To keep the pendulum swinging longer, make the mass heavier and the string as long as possible. You should be able to get a long pendulum to swing for say 30 minutes, if you are patient enough to count for that long.

    Alternatively, time the period of one swing very accurately - e.g. using an electronic timer that is triggered by the pendulum itself, eliminating your human reaction time.

    A pendulum will certainly show differences at different locations and at different altitudes (even within a tall building).

    Understanding why pendulum clocks didn't keep the same time at different places was one of the big scientific questions in Newton's day, and he answered it with his theory of gravitation. One of the alternative theories in Newton's time was the effect of gravity reduced as temperature increased, which was consistent with the fact that pendulum clocks ran slower near the equator.

    Of course the length of the pendulum did change because of thermal expansion, but since a fixed scale of temperature and practical thermometers had not yet been invented, it was hard to do any quantitative experiments.
     
  8. Nov 10, 2013 #7
    Thanks all. I'll set up a very tall, heavy pendulum and see what I get. I might also do the experiment a few times at different angles to see how much effect the angle has on the measurement of g.

    In the interest of theoretical knowledge, is there a way to deal with the lessening swings? For example, will the angle decrease in some predictable fashion, allowing some calculus to take the time and initial and final angle, and return g?
     
  9. Nov 10, 2013 #8

    Nugatory

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    That solution comes from making the approximation ##sin(\theta) = \theta##. That approximation is valid only when ##\theta##, the angle the pendulum makes from the vertical, is small - it's just the first term of the series expansion of ##sin(\theta)##. The solution also assumes that the pendulum itself is massless and all the weight is concentrated in a point at the end. Whether it's good enough will depend on how much accuracy you need, how small of a swing you can get away with, and how much you can concentrate the weight at the end of the pendulum. All else being the same, the errors will be smaller if you use a longer pendulum with a heaver weight at the end.
     
  10. Nov 10, 2013 #9

    AlephZero

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    There is a formula to correct the period for the amplitude of the swings. $$T = T_0(1 + \frac{\theta^2}{16} + \frac{11\theta^4}{3072} + \cdots)$$ where ##T_0## is the period for small oscillations and ##\theta## is the angle (in radians). See http://en.wikipedia.org/wiki/Pendulum_(mathematics [Broken])

    There are several different ways the pendulum could be losing energy, e.g. air resistance, flexibility in the way it is supported, or stretching of the string as the tension varies with time during each oscillation. Each way will have a different "formula" for how it affects the rate of decay. Measuring the rate of decay of your particular pendulum will be much easier than trying to predict it.
     
    Last edited by a moderator: May 6, 2017
  11. Nov 10, 2013 #10

    Vanadium 50

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    You should be thinking about how accurate you need this to be. If you want a 10% measurement, this is easy with pendula. If you want a 1% measurement, it's possible but will take some work. If you want a 0.1% measurement, you'll need to visit a machine shop to make the pendulum and set up something automatic to time the oscillations.
     
  12. Nov 23, 2013 #11
    Like I said,

     
  13. Nov 24, 2013 #12
    Cool thread was thinking of this myself but I'd like to know why two light gates that record time and then dropping a object through them wouldn't be precise why is that? I mean I know you got air resistance I suppose but wouldn't there be some way to rule that out through working with equations?
     
  14. Nov 24, 2013 #13

    Vanadium 50

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    If you need 1 part per 1000 precision, then you need to visit a machine shop.
     
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