Question

expalin the value of g is zero at the centre of the earth

Answer 1

Answer:

Explanation:

Let us assume that the Earth is a homogenous uniform sphere of uniform density and the mean density of it is ρ in the whole answer.

a) F = mg

F = $\frac{GmM}{R^{2}}$

⇒ mg = $\frac{GmM}{R^{2}}$

∴ g = $\frac{GM}{R^{2}}$

$(\frac{4}{3} \pi pg)R$

b) As Volume of a sphere = $\frac{4}{3} \pi r^{3}$

M = $(\frac{4}{3} \pi R^{3})$ ρ

[∵ ρ = m/v]

⇒ $\frac{gR^{2}}{G}$ = $(\frac{4}{3} \pi R^{2})$ ρ

∴ g = $(\frac{4}{3}\pi pG) R$

c) Let us consider mass 'm' be placed at a depth 'd' below the surface of the Earth as shown in the figure.

Now, the gravitational force on m will be only due to the inner solid sphere of radius (R - d). The outer shell of the thickness d exterior to the mass exerts no resultant force on the mass. Hence, the mass of the Earth is :

M₁ = $\frac{4}{3} \pi (R - d)^{3}p$

Also now, r = R - d.

∴ $g_{d}$ = $\frac{GM_{1} }{(R - d)^{2}}$ = $\frac{G }{(R - d)^{2}} M_{1}$

⇒ $g_{d}$ = $\frac{G }{(R - d)^{2}} \frac{4}{3} \pi (R - d)^{3}p$

∴ $g_{d}$ =  $(\frac{4}{3}\pi pG)(R - d)$

As g = $(\frac{4}{3}\pi pG) R$

⇒ $g_{d}$/g = $\frac{R - d}{R}$

∴ $g_{d}$ = g$(1-\frac{d}{R})$

d) As d is the depth measured from the surface of the Earth, d at the centre of the Earth is R.

⇒ $g_{d}$ = g(1-1) = 0

∴ g at the centre of the Earth is 0.

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