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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}}}

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

❣️ ❣️ Follow me ❣️ Follow me

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

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

━━━━━━━━━━━━━✦✗✦━━━━━━━━━━━━━━

\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.

━━━━━━━━━━━━━✦✗✦━━━━━━━━━━━━━━

\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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Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on
qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

  • Δx=760 m
  • V₀=87 m/s
  • t=13.6 s
  • a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

760=(1183.2)(cos\theta)

cos\theta=\frac{760}{1183.2}

\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}

3 0
3 years ago
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