Yes, the moon's effect is about -0.000034 m/s^2, or 0.00034%. Sensitive instruments (such as the CERN laboratories) have to make a calibration adjustment for this.
Also, I notice that the value in the question is the value of "standard gravity". At the equator, gravity is about 9.780327 m/s^2, and at the poles it is 9.832186 m/s^2. That a difference of 0.53%. So gravity varies a bit with how far north or south you are. Standard gravity is at latitude 45.49732 degrees north (or south) of the equator. The moon's effect is the same value as moving 4.2 km further south.
The amount depends on your latitude, but it's of the order of the tidal acceleration, which is 2*(earth radius)*(gravitational constant) * (moon mass)/(distance to moon)^3 = 1.2*10^-6 m/s^2.
So... essentially undetectable, especially considering that your statement of g=9.80665 lends you an implied accuracy that that figure simply doesn't have. The acceleration due to gravity at the level of 4-5 significant figures depends on your location on Earth. The tidal acceleration of the moon is 10-100 times less significant than that.
David F's analysis is incorrect, because it does not consider the acceleration of the Earth due to the moon, only your own. But in reality both you and the Earth are freely falling around the moon, in which case the difference in normal force between the two of you (which is what one really means by "weight" or "acceleration", i.e. weight as measured by a scale on the surface of Earth, or acceleration relative to the surface of Earth) would be zilch, since all bodies fall at the same rate, except for the fact that the center of mass of the earth is in a higher (lower) orbit than you are, and thus there is a tidal differential. It's that tidal differential that makes the, um, differential, not the simple addition of the moon's force of gravity.
Yes, and even more so the further up you go: up a ladder, up an elevator, or up in a helicopter.
Force due to Earth, Fe = - G Me m / re^2.
Force due to Moon, Fm = + G Mm m / rm^2.
Total force = Fe + Fm.
Where G is a gravitational constant, Me is mass of Earth. m is your mass, Mm is mass of moon, re is distance of m to centre of Me, rm is distance to moon from m.
Answers & Comments
Verified answer
Yes, the moon's effect is about -0.000034 m/s^2, or 0.00034%. Sensitive instruments (such as the CERN laboratories) have to make a calibration adjustment for this.
Also, I notice that the value in the question is the value of "standard gravity". At the equator, gravity is about 9.780327 m/s^2, and at the poles it is 9.832186 m/s^2. That a difference of 0.53%. So gravity varies a bit with how far north or south you are. Standard gravity is at latitude 45.49732 degrees north (or south) of the equator. The moon's effect is the same value as moving 4.2 km further south.
Yes, and also when the moon is underfoot.
The amount depends on your latitude, but it's of the order of the tidal acceleration, which is 2*(earth radius)*(gravitational constant) * (moon mass)/(distance to moon)^3 = 1.2*10^-6 m/s^2.
So... essentially undetectable, especially considering that your statement of g=9.80665 lends you an implied accuracy that that figure simply doesn't have. The acceleration due to gravity at the level of 4-5 significant figures depends on your location on Earth. The tidal acceleration of the moon is 10-100 times less significant than that.
David F's analysis is incorrect, because it does not consider the acceleration of the Earth due to the moon, only your own. But in reality both you and the Earth are freely falling around the moon, in which case the difference in normal force between the two of you (which is what one really means by "weight" or "acceleration", i.e. weight as measured by a scale on the surface of Earth, or acceleration relative to the surface of Earth) would be zilch, since all bodies fall at the same rate, except for the fact that the center of mass of the earth is in a higher (lower) orbit than you are, and thus there is a tidal differential. It's that tidal differential that makes the, um, differential, not the simple addition of the moon's force of gravity.
How about this:
Is the gravity less than 9.80665 m/s2 when the sun is directly above our heads?
Yes, and even more so the further up you go: up a ladder, up an elevator, or up in a helicopter.
Force due to Earth, Fe = - G Me m / re^2.
Force due to Moon, Fm = + G Mm m / rm^2.
Total force = Fe + Fm.
Where G is a gravitational constant, Me is mass of Earth. m is your mass, Mm is mass of moon, re is distance of m to centre of Me, rm is distance to moon from m.
near = 3.78[8] meters
far = 3.91[8] meters
M = 7.35[22] kg
G = 6.67[-1] N (m/kg)²
GM = 4.90[12] N m²/kg
Îγ = GM((f² - n²)/(fn)²) =
Îγ = 4.9[12]((3.91[8]² - 3.78[8]²)/(3.91[8] • 3.78[8])² =
Îγ = 4.9[12][16]/1.48[17]² =
Îγ = 4.9[28]/2.18[34] =
Îγ = 2.24[-6] m/s²
A mass of 100 kg has a weight* of 980.665 ± 1.22[-4] N depending on which side, near or far, of the Earth it is on.
Edit:
* An ideal weight if we must point this out.
.
In short - Yes
By how much - No idea.