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Do we weight less during high tide?

  1. Nov 20, 2008 #1
    I had a high-school student ask me this question, and quite frankly, I don't know the answer. It seems obvious by drawing a free body diagram that the gravitational force vector on you, if you are standing on the side of the earth facing the moon, would be smaller than that if you are standing on the other side of the earth. But what is really confusing me is whether or not you could actually MEASURE this effect using a scale (even assuming perfect precision). It seems to me that you could measure it. But I'm not sure. Can someone please clarify?
  2. jcsd
  3. Nov 20, 2008 #2
    The force exerted by the Moon can be estimated easily relative to the weight . The distance is about 60*R. (R is teh radius of the Earth). So the force will be decrease by a factor of 60^2.
    The mass of the Moon is about 1/100 the mass of the Earth so this will give a factor of 1/100 for the force. Altogether will be 1/(60^2*100), about 10^(-6).
    One in a million (actually more like 3 in a million).
  4. Nov 20, 2008 #3

    D H

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    nasu, what you wrote is incorrect. Pacopag asked about what a scale will measure: apparent weight. You wrote about something that cannot be measured. Weight defined as mass times gravitational acceleration has some a useful applications (e.g. lift), but is inherently unmeasurable. Scales (spring scales, that is) measure something quite different. They directly measure the deflection of the scale platform. Knowing the scale's spring constant, this deflection translates directly to a force. When a person stands still on the scale, the normal force exerted by the scale on the person is that needed to keep the person stationary with respect to the scale. Since the scale platform is buoyed by springs, the platform deflection will be a direct measure of this normal force. This normal force is equal in magnitude to the person's weight less the centripetal acceleration resulting the Earth's rotation about it's axis. Your bathroom scale does not measure your "weight". It measures your apparent weight.

    So, how does the Moon (or the Sun) affect a person's apparent weight? The person accelerates toward the Moon (or Sun) due to gravity, but so does the Earth itself. The scale will effectively measure the difference between the person's acceleration and the Earth's acceleration. Note well: This is the exact same mechanism that causes the tides.

    If the Moon is directly overhead (or directly underfoot), a person's apparent weight will fractionally change from nominal by about

    &\frac {M_{\text{moon}}} {M_{\text{earth}}}
    \left(\frac 1 {({R_m}/{r_e})^2} - \frac 1 {({R_m}/{r_e}\mp1)^2}\right) \\
    &\quad \approx \frac 1 {81.3}\left(\frac 1 {(60.2)^2}- \frac 1 {(60.2\mp1)^2}\right) \\\quad &\quad \approx 1.1 \cdot 10^{-7}

    A person's apparent weight is a tiny bit smaller when the Moon is directly overhead (or directly underfoot). The math is a bit trickier when the center of the Moon, center of the Earth, and the person are not collinear. When the Moon is on the horizon a person's apparent weight will be a tiny, tiny bit larger than nominal.

    Finally, the answer to the question posed in the title of the thread ("do we weigh less during high tide?") is "it depends". High tide typically does not occur when the Moon is directly overhead. The phase delay between the Moon being overhead and the timing of high tide varies by location. In places where the phase delay is zero a person will weigh a tiny bit less at high tide, but in places where the phase delay is 6 hours a person will weigh a tiny, tiny bit more.
    Last edited: Nov 20, 2008
  5. Nov 21, 2008 #4
    I did not say anything about measurements. Just estimate the relative effects of Moon and Earth for a person. My idea was that it's too little to worry about it, from a practical point of view.

    Your analysis is of course more correct but also more complicated and I did not think he really wants to go into that. I for sure did not feel like it. The effect of the Moon (10^(-6) g) is even smaller that the effect of Earth's rotation (about 5*10^(-4) g).
    If you just want to talk "in principle", there is an effect from the other planets too (including Pluto :smile:).
  6. Nov 21, 2008 #5

    D H

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    Since the OP asked about tides and scales, the answer IMO should have been in terms of tidal forces. That little tiny difference in apparent weight manages to create 17 meter tides in the Bay of Fundy and 1/2 meter tides in the Earth itself.
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