# Gravitational pull on Earth

March 28, 2021 Where G if Newton's gravitational constant, is the mass of the object being pulled on by the Moon, MMoon is the mass of the Moon, and rMoon is the distance between the Moon and the object. A similar force acts between the Earth and various objects, except that we use the distance to the Earth, rEarth, and the mass of the Earth, MEarth, in place of the lunar values. Since the Earth's force on something is the object's weight, W, we can write

MMoon / MEarth is about 1/80, and for an object at the surface of the Earth rEarth is about 4000 miles, while rMoon is about 240, 000 miles, or 60 times greater; so at the surface of the Earth the pull of the Moon is 80 times smaller than the object's weight because of its lesser mass, and another 3600 (= 60 squared) times smaller because of its greater distance. Combining these two effects, the Moon's pull on objects near the Earth is only 1/300, 000th of their Earth weight. So if something weighs 150 pounds due to the pull of the Earth, the pull of the Moon on that object would be about 150/300000, or 1/2000th of a pound. This is a very small force but it produces a number of interesting effects because it acts on every object near the Earth, including the Earth itself.

What does this force do?

(most of what follows will be considerably revised, being too incomplete to leave as-is)

(2) Tidal Effects
(incomplete first draft)
Gravitational Interactions of the Earth and Moon: Barycentric Motion describes the effect of the Moon's gravity on the Earth as a whole. But there is a second effect, due to the fact that if you are on the "front" side of the Earth (the side where the Moon is up), you are closer to the Moon than if you are on the "back" side of the Earth (the side where the Moon is down), and given the inverse square nature of the Law of Gravity, this means that you are pulled on harder when the Moon is up than when it is down. This produces what are referred to as differential forces, or because of their observable effect, tidal forces.
On the side of the Earth facing the Moon you are as much as 1/60th closer to the Moon than if you were in the middle of the Earth, producing a force which is 1/30th larger than average. On the side of the Earth facing away from the Moon, you are as much as 1/60th further away from the Moon than if you were in the middle of the Earth, producing a force which is 1/30th smaller than average.
So on the near side, you are pulled by a force which is 1/300, 000th of your weight, which moves you around the barycenter every month, AND by an additional force equal to 1/30th of this, or 1/10, 000, 000th of your weight, which tends to pull you away from the rest of the Earth. And on the far side, you are pulled on this much less than the rest of the Earth, which tends to pull the rest of the Earth away from you.
KEEP IN MIND that at the Equator, where things seem to weigh 1/3% less than at the Poles because of the Coriolis effect of the Earth's rotation, the Earth bulges out by about 1/3%. In other words, it bulges out by a fraction of its radius approximately equal to the apparent reduction in weight. IF THE SAME THING happened with the differential force = tidal force of the Moon, the 1/10, 000, 000th difference in force on different parts of the Earth would make various parts of the Earth deviate from their normal position by 1/10, 000, 000th of the radius of the Earth, which is about half a meter, or 1 1/2 feet. And on the average, that is exactly what happens in some places; but the actual tides involve a number of other factors, so they can differ from this simple result by a considerable amount.

Source: cseligman.com
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