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8_murik_8 [283]
3 years ago
11

Why gravitational force at the center og earth is zero

Physics
2 answers:
KengaRu [80]3 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]3 years ago
4 0

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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\huge{\blue{\mathbb{HELLO\:BUDDY}}}

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❣️ ❣️ Follow me ❣️ Follow me

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

\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).

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

\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}πr^{3}

This makes

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

g = \frac{4πρG}{3}

So, at center of Earth

since, r = 0

so, g = 0.

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\boxed{\underline{\green{\star{HOPE \: YOU\:GOT \:THE \:ANSWER.}}}}

\boxed{\underline{\pink{\star{MARK \: ME  \: BRAINLIEST.}}}}

\Large{\mathcal{\green{FOLLOW \: ME}}}

Thankyou

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If the mass of one object is doubled, the force between these objects will also double.

Force refers to an influence on a body which can change its state of rest or motion.

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