Question

The density of the earth in terms of acceleration due to gravity$\left(g\right)$, the radius of the earth$\left(R\right)$ and universal gravitational constant $\left(G\right)$ is

• A. $\frac{4\pi RG}{3g}$
• B. $\frac{3\pi RG}{3g}$
• C. $\frac{4g}{3\pi RG}$
• D. $\frac{3g}{4\pi RG}$

Answer 1

The correct option is D

$\frac{3g}{4\pi RG}$

The explanation for the correct option

In the case of option (D)

Solution:

Step 1: Used the formula of gravitational

Know that the formula of gravitational

$g=\frac{GM}{R^2}$

Therefore,

$Desnsity\rho =Mass\left(M\right)/Volume\left(V\right)$

and

$$\begin{gathered} M=\rho \times V \\ M=\rho \times \left(\frac{4}{3}\right)\pi R^3 \end{gathered}$$

Therefore substitute the value of $M$ in the gravity formula,

$$\begin{gathered} g=\frac{G\left(\rho \times {\displaystyle \frac{4}{3}}\right)\pi R^3}{R^2} \\ \rho =\frac{3g}{4\pi RG} \end{gathered}$$

Therefore the density of the earth $\rho =\frac{3g}{4\pi RG}$.

The explanation for the incorrect option

In the case of option (A)

1. $\frac{4\pi RG}{3g}$is not correct for the density of the earth the correct density of the earth is $\rho =\frac{3g}{4\pi RG}$.

In the case of option (B)

1. $\frac{3\pi RG}{3g}$is not correct for the density of the earth the correct density of the earth is $\rho =\frac{3g}{4\pi RG}$.

In the case of option (C)

1. $\frac{4g}{3\pi RG}$is not correct for the density of the earth the correct density of the earth is $\rho =\frac{3g}{4\pi RG}$.

Hence, the correct option is (D).