The gravitational pull of the Moon is partially responsible for the tides of the
ID: 1308256 • Letter: T
Question
The gravitational pull of the Moon is partially responsible for the tides of the sea. The Moon pulls on you, too, so if you are on a diet it is better to weigh yourself when this heavenly body is directly overhead! If you have a mass of 85.0 kg, how much less do you weigh if you factor in the force exerted by the Moon when it is directly overhead (compared to when it is just rising or setting)? Use the values 7.35 x 1022 kg for the mass of the moon, and 3.76 x 108 m for its distance above the surface of the Earth. (For comparison, the difference in your weight would be about the weight of a small candy wrapper. And speaking of candy...)
Explanation / Answer
Summing forces in the vertical direction with up being positive
N - m*g - Fm = 0
where N = reaction force due to your weight
m*g = weight
Fm = gravitational pull of the moon
Fm = G * m1 * m2 / r^2
where G = the gravitational constant
m1 = mass of the moon
m2 = mass of you
r = distance between surface of earth (where you are) and moon
Fm = 6.67428 x 10^-11 * 7.35 x 10^22 * 85 / (3.76 x 10^8)^2 = 2.9494 x 10^-3 N
N = 85 * 9.81 + 2.9494 x 10^-3 = 833.8529494 N
If the moon wasn't overhead then the equation becomes
N - mg = 0
and N = 833.85
The difference is 2.9494 x 10^-3 N
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