I got curious and wondered how much weight we lose while a solar eclipse is in effect so I calculated it here are my calculations please correct me if I'm wrong/made mistakes; Earth's g=9.81N/Kg       Moon's gravitational pull acting on a man standing at the surface of the earth = ((6.67x^-11)x(7.35x10^22))/(384.4x10^6)^2   where 6.67x10^-11 is the gravitational constant,  7.35x10^22 is the mass of the moon, 384.4x10^6 is the distance from the moon to earth therefore moon's g on us is 3.318x10^-5 N/Kg.  Now the sun; ((6.67x10^-11)x(1.989x10^30))/(149.6x10^9)^2 therefore sun's g on us = 5.928x10^-3N/Kg. so 9.81/9.81+(3.318x10^-5)+(5.928x10^-3) = 0.999393 so x100 = 99.94% of g therefore you are 0.06% lighter during a solar eclipse good time to measure your weight if you are on a diet

 

We're in free fall with respect to the sun, as we are in orbit.

 

15 minutes ago, swansont said:

We're in free fall with respect to the sun, as we are in orbit.

 

That is true, however we are continuously accelerating towards its centre therefore a force is constantly applied on us by its gravity. But yes I do agree I shouldn't have factored in the sun now that I look at it since if the force is constant when we jump onto a weighing scale it will still have the sun's force included before or after a solar eclipse.   

1 hour ago, Encryptor said:

That is true, however we are continuously accelerating towards its centre therefore a force is constantly applied on us by its gravity. But yes I do agree I shouldn't have factored in the sun now that I look at it since if the force is constant when we jump onto a weighing scale it will still have the sun's force included before or after a solar eclipse.   

If you were in orbit around the sun (and away from the earth), and strapped a scale to your feet it would read zero. It has no direct effect on the weight here on earth. (There's a differential effect stemming from your location which are the tides.)

7 hours ago, Encryptor said:

I got curious and wondered how much weight we lose while a solar eclipse is in effect so I calculated it here are my calculations please correct me if I'm wrong/made mistakes; Earth's g=9.81N/Kg       Moon's gravitational pull acting on a man standing at the surface of the earth = ((6.67x^-11)x(7.35x10^22))/(384.4x10^6)^2   where 6.67x10^-11 is the gravitational constant,  7.35x10^22 is the mass of the moon, 384.4x10^6 is the distance from the moon to earth therefore moon's g on us is 3.318x10^-5 N/Kg.  Now the sun; ((6.67x10^-11)x(1.989x10^30))/(149.6x10^9)^2 therefore sun's g on us = 5.928x10^-3N/Kg. so 9.81/9.81+(3.318x10^-5)+(5.928x10^-3) = 0.999393 so x100 = 99.94% of g therefore you are 0.06% lighter during a solar eclipse good time to measure your weight if you are on a diet

 

You can't just use the gravitational pull of the Moon or the Sun on the person to work this out.  The Earth (as a whole) is in free fall with respect to each.  So, if like in Swansont's example you were to replace the Earth with a man standing on a scale, but located where the center of the Earth is, he would weigh nothing, even though the gravitational pull on him from Both the Sun and Moon are equal to what you calculated.  In order to get the effect caused by the Sun and Moon being directly overhead compared to no Sun and Moon,  you have to take the difference between the gravitational pull they have on a mass at the distance that the center of the Earth is and that on the mass if located 1 Earth radius closer.

2 hours ago, GLekter said:

These are very interesting calculations. It would be cool to weigh yourself right before the eclipse and after. And show your friend on the phone that you lost weight in 4 minutes. The difficulty is only syncing with the phone.

1 hour ago, adrickhauan said:

It seems that it is pretty easy to synchronize the information from the scales.

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