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I know the answer (hence, the title). But, why, i do not understand?

3 Answers3

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This is due Newtons shell theorem which says that for the particle $m$ all the mass outside the blue shell of radius $r_1$ cancels out:

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while the mass inside a shell of radius=0 equals also 0 so there is no gravitational acceleration in the gravitational center of the earth (or any planet).

Yukterez
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Because as you go down, the mass above you has a gravitational pull on you that resists the pull of the mass below you. This continues until you reach the center and the pulls all cancel out, making you weightless.

When you go up, the force of gravity on you decrease because of the equation $$F = G\frac{m_1m_2}{r^2}.$$ As the distance increases, its square does even more so, making the force of gravity weaker as you leave Earth i.e. go up.

Jimmy360
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Just as there is a Gauss's Law for the Electric Field, there is similarly a law for the Gravitational Field. It states the following: $$ -4\pi GM_{enclosed} = \oint \! g \cdot(dA) $$

When you go deeper into the earth, since the mass enclosed by a sphere of a certain radius is directly proportional to the volume enclosed (assuming constant density), the enclosed mass decreases faster than the surface area of the enclosed region decreases, causing a smaller gravitational field.

When you go into the atmosphere and further, since you've already exceeded the radius of the earth, the enclosed mass stays the same while the surface area of the gaussian sphere increases, causing a smaller gravitational field.