Acceleration due to gravity g and the mean density ρ of the Earth are related by the relation: [G is the gravitational constant and Re is the radius of the Earth]

Question

Acceleration due to gravity g and the mean density ρ of the Earth are related by the relation: [G is the gravitational constant and Re is the radius of the Earth] (A) (g/G) (4 π/3) Re3=ρ (B) ((g/G)/(4 π)3 Re)=ρ (C) (g/G) (4 π/3) Re2=ρ (D) ((g/G)/(4 π)3 Re3)=ρ

Tardigrade Admin · Accepted Answer

g=(G M/Re2)=(G/Re2)((4/3) π Re3 ⋅ ρ) g=(4/3) π Re ⋅ G ⋅ ρ ⇒ ρ=((g/G)/(4 π)3 ⋅ Re)

Q. Acceleration due to gravity $g$ and the mean density $ρ$ of the Earth are related by the relation:
$[G$ is the gravitational constant and $R_{e}$ is the radius of the Earth]

$\frac{g}{G} \frac{4 π}{3} R_{e}^{3} = ρ$

$\frac{\frac{g}{G}}{\frac{4 π}{3} R _{e}} = ρ$

$\frac{g}{G} \frac{4 π}{3} R_{e}^{2} = ρ$

$\frac{\frac{g}{G}}{\frac{4 π}{3} R _{e}^{3}} = ρ$

$g = \frac{GM}{R _{e}^{2}} = \frac{G}{R _{e}^{2}} (\frac{4}{3} π R_{e}^{3} \cdot ρ)$
$g = \frac{4}{3} π R_{e} \cdot G \cdot ρ \Rightarrow ρ = \frac{\frac{g}{G}}{\frac{4 π}{3} \cdot R _{e}}$

Q. Acceleration due to gravity g and the mean density ρ of the Earth are related by the relation: [G is the gravitational constant and Re​ is the radius of the Earth]

Gg​34π​Re3​=ρ

34π​Re​Gg​​=ρ

Gg​34π​Re2​=ρ

34π​Re3​Gg​​=ρ

g=Re2​GM​=Re2​G​(34​πRe3​⋅ρ) g=34​πRe​⋅G⋅ρ⇒ρ=34π​⋅Re​Gg​​