Question

Acceleration due to gravity 'g' and the mean density of the earth ${}^{\prime }{\rho }^{\prime }$ are related by which of the following relations where G is the gravitational constant and $R_e$ is the radius of the earth

• A. $\left(\frac{g}{G}\right)\frac{4\pi }{3}{R}_e^3=\rho$
• B. $\frac{\left(\frac{g}{G}\right)}{\left(\frac{4\pi }{3}{R}_e\right)}=\rho$
• C. $\frac{g}{G}\frac{4\pi }{3}{R}_e^2=\rho$
• D. $\frac{\left(\frac{g}{G}\right)}{\left(\frac{4\pi }{3}{R}_e^3\right)}=\rho$

Answer 1

The correct option is B $\frac{\left(\frac{g}{G}\right)}{\left(\frac{4\pi }{3}{R}_e\right)}=\rho$

The gravitational acceleration on the earth's surface is $g=\frac{GM}{R_e^2}$

Now, Density $\rho =$ mass/volume $=\frac{M}{V}$

or $M=\rho V=\rho \times \left(4/3\right)\pi R_e^3$

Thus, $g=\frac{G\rho \left(4/3\right)\pi R_e^3}{R_e^2}$

or $\rho =\frac{\left(g/G\right)}{\left(4/3\right)\pi R_e}$