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2 years ago

Why gravitational force at the center og earth is zero

Physics

2 answers:

KengaRu [80]2 years ago

8 0

F = G Mm/r²
mg = G Mm/r²
g = GM/r²
At centre of earth, r=0
g = GM/0
g =0

Anon25 [30]2 years ago

4 0

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$\huge{\green{\tt{\underline{\underline{QUESTION:-}}}}}$

Why value of 'g' is zero at center of Earth?

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$\huge{\red{\tt{\underline{\underline{EXPLANATION:-}}}}}$

When we move towards centre of earth, the mass is equally distributed in all directions.

The mass beneath you = Mass in front of you = Mass behind you

Thus, all the gravitational forces applied cancel each other and acceleration due to gravity (g) at centre of earth on centre of earth becomes zero (0).

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$\huge{\green{\tt{\underline{\underline{PROOF:-}}}}}$

The general equation of acceleration due to gravity is;-

g = $\frac{GM}{r^2}$

Assuming earth to be a perfect sphere and considering its uniform density.

We make the following adjustments in the following equation.

g = $\frac{GM}{r^2}$

Multiplying and dividing RHS by volume 'V'.

g = $\frac{GM}{V} × V × \frac{1}{r^2}$

We know that;-

$\frac{M}{V}$ = Density, ρ

Therefore,

g = $\frac{ρGV}{r^2}$

For sphere;-

V = $\frac{4}{3}$ π $r^{3}$

This makes

g = $\frac{4}{3}$ π $r^{3}$ × $\frac{ρG}{r^2}$

g = $\frac{4πρG}{3}$

So, at center of Earth

since, r = 0

so, g = 0.

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In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

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borishaifa [10]

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