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- Thread starter Pacopag
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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).

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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

[tex]\aligned

&\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}

\endaligned[/tex]

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.

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

[tex]\aligned

&\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}

\endaligned[/tex]

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.

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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 ).

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