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Physics

Mechanics

Easy

Question

The acceleration due to gravity g and density of the earth $rho$ are related by which of the following relations? (where G is the gravitational constant and $R subscript E$ is the radius of the earth)

$ρ = \frac{4 π G R_{\underline{g}}}{3 g}$
$ρ = \frac{3 g}{4 π G R_{E}}$
$ρ = \frac{3 G}{4 π g R_{E}}$
$ρ = \frac{4 π g R_{E}}{3 G}$

The correct answer is: $rho equals fraction numerator 3 g over denominator 4 pi G R subscript E end fraction$

Acceleration due to gravity on earth is $g equals fraction numerator G M subscript E over denominator R subscript E superscript 2 end fraction$
As $rho equals fraction numerator M subscript E over denominator 4 over 3 pi R subscript E superscript 3 end fraction rightwards double arrow M subscript E equals rho 4 over 3 pi R subscript capital sigma superscript 3$
$g equals fraction numerator G open parentheses rho 4 over 3 pi R subscript E superscript 3 close parentheses over denominator R subscript E superscript 2 end fraction equals 4 over 3 pi rho G R subscript E text or end text rho equals fraction numerator 3 g over denominator 4 pi G R subscript E end fraction$
Substituting this value in Eq. (i), we get

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