Can you think of an experiment without teleporting? Or a vacuum?
Something you can do at home...
Depends on how much you want to spend.
There are a couple ways to something homemade - your tide idea is pretty good.
If you have an accurate scale, you could set something on it and look up the "tide charts" for your area (not sure if there are such things for like Kansas, but there might be) and see when the moon is overhead and directly away.
The tradeoff is that a very accurate scale, think 0.001grams, can't measure a lot of mass, and only a little mass probably won't change by too much.
In any case, put something on the scale, I would suggest whatever the maximum measurable mass is. If it is a 50.000g scale, put 49g on it, etc.
Then take 8 readings, one for each high and low tide time (4) and then inbetween those.
the 4 inbetween times should give you something of a "neutral" tide reading, and then the highs and lows should give you measurable changes from the inbetween value.
To be most accurate, you should set up your mass at an 'inbetween' time, and then measure to make sure that number reappears at the other 3 'inbetween' times.
The problem with a low reading scale, is that the mass of the scale is probably the same or higher than the mass you are trying to measure, meaning the tide would pull the scale upwards stronger than the mass itself, giving weird readings.
But any difference in readings is a step in the right direction.
So what if something heavier, like a person, stood on the scale? Would they weigh less when the moon is above them?
If my math is correct LOL
A rough calculation shows that the moon would have about a 0.0003% effect on the weight of an object.
F=(GMoon massSubject mass)/D2
G=6.67x10-11 Moon mass=7.3x1022kg Subject mass=100kg (good for theory) D=distance between moon and subject. The mean distance between the earth and moon is taken to be 3.84x108 meters, although the distance varies a lot, change as appropriate. This distance is between the centers, so to get the effect on someone standing on the earth's surface, subtract 6.4x106meters from that, leaving 3.78x108 meters between the center of the moon and a person on earth.
(6.67x10-11 x 7.3x1022 x 100)/ ( (3.78x108)2 ) = 0.0034Newtons, which would be the equivalent of 0.35 grams
This means a 100kg person would see about 0.3 to 0.4 grams of change due to the moon. Which is why the sensitivity of the scale needs to be very high. Measuring over a full rotation of earth should give a variability in the weight of about double, meaning close to 0.7 or 0.8g. The scale would have to measure 100.0000kg to show these changes. It would go from 100.0003kg to 99.9997kg over the max and min readings.
The moon gets closer and farther than the mean distance, so additional calculations can be done to get more accurate readings.
Yes.
If you could measure down to 0.001 pounds and stood there, without respiration, food, or sweating for the experiment.
Another option would be to get a very large scale, - the more sensitive the better - and find something very close to max weight there too.
Think a 2 ton scale that might go down to 0.1 or even 0.01 pounds, and putting just under 2 tons on it.
I have no idea if such things exist or how much they would cost, but it would be a lot. A big balloon of water suspended on a tree might work on a still day, but any variables such as wind or animals might cause weird readings.
I don't know what could be done with standard household equipment.
Load cells are pretty good devices. they convert forces (loads) into electrical signals which can then be measured. This is how all digital scales work.
The key to a good load cell measurement is to get the proper range, and sensitivity, along with electronic conditioning (amplification and filtering) to make the output workable for your situation.
People are notoriously bad experiment subjects as everything we do - including movement - typically affects the readings. Better to use a solid object on a "step on" scale, and anything that can be done with suspension is usually good with water, since it is readily available and relatively heavy.
I might do the calculations to see how much of an impact the moon has on weight, but I don't suspect it is much on the whole.
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