Regarding the value of g(earth)

Hello guys(and gals) I needed a little clarification regarding the value of g as we go up and also as we go towards the center of the earth. Also is the value of g infinite or 0 at the center of the earth?

Thanks in advance.

Easy Math Editor

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Comments

The value of g will approach zero as we get closer and closer to the centre, but it will never reach zero.

This is because, g=(G*M/r^2) So if r=0, the equation would not make sence.

SABAB AHAD - 7 years, 7 months ago

When $r \to 0^+$ , $g \to \infty$ instead; a small denominator leads to a large value.

Ivan Koswara - 7 years, 7 months ago

thnx and that means as we go up it shall increase?

Jun Das - 7 years, 7 months ago

Well, I feel the simplest way to think about it is that there is mass all around you at the center of the Earth so you get an equal gravitational "pull" from all directions. The pulls cancel out so you get no net gravitational force or simply you wouldn't experience any force at all practically,i.e. $F=W=mg=0$ ,here $m$ is constant.Since, $F=0$ ,therefore, $g$ at the centre of the earth must equal $0$ or in one word, you would feel "weightless" there(same feeling as one feels it in space).Also, as we go up, value of $g$ decreases from the surface of the earth since value of $R+h$ increases,where $h$ is the height, and consequently resultant $g$ at that height decreases.Hope that clear things up!

Bhargav Das - 7 years, 7 months ago

@Bhargav Das – yea it did man thnx!

Jun Das - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$