If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

The correct option is 3g / 4πRG.

To find the mean density of the Earth, we start by relating the acceleration due to gravity on the Earth's surface, $g$, with the Earth's mass, $M$, and its radius, $R$.

The acceleration due to gravity $g$ on the surface of the Earth is given by Newton's law of universal gravitation:
$g=\frac{GM}{R^2}$
where $G$ is the universal gravitational constant.

Assuming the Earth is a perfect sphere of radius $R$, its volume $V$ is:
$V=\frac{4}{3}\pi R^3$

Let $\rho$ represent the mean density of the Earth. Since density is defined as mass divided by volume ($\rho =\frac{M}{V}$), we can express the mass $M$ of the Earth as:
$M=\rho ·V=\rho ·\left(\frac{4}{3}\pi R^3\right)$

Now, substitute this expression for mass $M$ back into the equation for $g$:
$g=\frac{G}{R^2}·\left(\frac{4}{3}\pi R^3\rho \right)$

Simplify the equation by canceling the $R^2$ term in the denominator with the $R^3$ in the numerator:
$g=\frac{4}{3}\pi R\rho G$

To find the mean density $\rho$, rearrange the terms to solve for $\rho$:
$\rho =\frac{3g}{4\pi RG}$

Thus, the mean density of the Earth is $\frac{3g}{4\pi RG}$, which corresponds to the option 3g / 4πRG.