Question

If g is acceleration due to gravity on the surface of the Earth, G is universal gravitational constant and R is radius of the Earth then mean density $\rho$ of the Earth can be expressed as

• A. $\frac{3G}{4\pi gR}$
• B. $\frac{3g}{4\pi GR}$
• C. $\frac{4g}{3\pi GR}$
• D. $\frac{4G}{3\pi gR}$

Answer 1

Answer: (B)

(2) $\frac{3g}{4\pi GR}$

Answer 2

The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM}{R^2}$, where $M$ is the mass of the Earth, $G$ is the universal gravitational constant, and $R$ is the radius of the Earth. The volume of the Earth, assuming it to be a sphere, is $V = \frac{4}{3}\pi R^3$. The mean density $\rho$ of the Earth is defined as $\rho = \frac{M}{V}$. From the expression for $g$, we can express the mass $M$ as $M = \frac{gR^2}{G}$. Substituting this expression for $M$ into the density formula: $\rho = \frac{\frac{gR^2}{G}}{\frac{4}{3}\pi R^3}$ $\rho = \frac{gR^2}{G} \times \frac{3}{4\pi R^3}$ $\rho = \frac{3g}{4\pi GR}$