Question

Acceleration due to gravity and the mean density of the earth $'\rho '$ are related by which of the following relations where G is the gravitational constant and ${R_e}$is the radius of the earth.
A.$\left( {\dfrac{g}{G}} \right)\dfrac{{4\pi }}{3}{R_e}^3 = \rho$
B. $\dfrac{{\left( {\dfrac{g}{G}} \right)}}{{\left( {\dfrac{{4\pi }}{3}{R_e}^3} \right)}} = \rho$
C.$\dfrac{g}{G}\dfrac{{4\pi }}{3}{R_e}^2 = \rho$
D.$\dfrac{{\left( {\dfrac{g}{G}} \right)}}{{\left( {\dfrac{{4\pi }}{3}{R_e}^3} \right)}}$

Answer 1

$G$ is the universal constant of gravitation. And $g$ is the acceleration due to gravity of earth. You can compare the force of attraction using two formulas, one that involve $G$ and the one that involve $g$. And then equate them to find the relation between them. To include $\rho$ in the comparison. You should use the fact that density is the ratio of mass per unit volume and the earth is spherical.

Complete step by step answer:
Let us consider the radius of the earth is ${R_e}$ and the density of the earth is $e$
Then the mass of the earth is$M$
Therefore$M = \dfrac{4}{3}\pi \operatorname{R} _e^3$.
According to the law of motion of gravitation any particle of matter in the universe attracts any other with a force varying in diversity as the product.
In this question, the force on the body of mass$M$, near the surface of the earth is given by
$F = \dfrac{{G(mM)}}{{\operatorname{R} _e^2}}$ . . . . (1)
Where, $G \to$gravitational constant
$\operatorname{Re} \to$Radius of earth
$M \to$Mass of earth
$m \to$ Mass of body
$F \to$Force
If the magnitude of this, is the line joining the center of the earth and the center of the body of mass $M$,
Then the type of force produced on acceleration$g$in the body of mass$M$
And,$F = mg$ . . . (1)
By (1) and (2), we get
$\dfrac{{GMm}}{{\operatorname{R} _e^2}} = mg$
$\Rightarrow \dfrac{{GM}}{{\operatorname{R} _e^2}} = g$
$\Rightarrow GM = g \times \operatorname{R} _e^2$
$\Rightarrow M = \dfrac{{g \times \operatorname{R} _e^2}}{G}$
Substituting the value of$M$in the given equation.
$\dfrac{4}{{3\pi }}\operatorname{R} _e^3 = \dfrac{{g \times \operatorname{R} _e^2}}{G}$
$\Rightarrow \dfrac{4}{3}\pi \operatorname{R} _e^3eG = g \times \operatorname{R} _e^2$
$\Rightarrow \dfrac{4}{3}\pi {\operatorname{R} _e}eG = g$
$\Rightarrow \rho = \dfrac{{3g}}{{4\pi G\operatorname{Re} }}$
$\therefore$ $\dfrac{{\left( {\dfrac{g}{G}} \right)}}{{\left( {\dfrac{{4\pi }}{3}{R_e}^3} \right)}} = \rho$

So, the correct answer is “Option B”.

Note:
$G$ is the universal constant of gravitation. So its value, $G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$ is the same everywhere in the universe. But $g$ is the acceleration due to gravity of earth. So its value changes on different planets. Also, on the earth as well, the value of $g$ changes with height and depth.