Answer:
1
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
[tex]M_e[/tex] = Mass of Earth
[tex]M_m[/tex] = Mass of Mercury
[tex]R_e[/tex] = Radius of Earth
[tex]R_m[/tex] = Radius of Mercury = [tex]0.375R_e[/tex]
Mass is given by
[tex]M=\rho v\\\Rightarrow M=\rho \frac{4}{3}\pi R^3[/tex]
Acceleration due to gravity on Earth
[tex]g_e=\frac{GM_e}{R_e^2}\\\Rightarrow g_e=\frac{G\rho_e \frac{4}{3}\pi R_e^3}{R_e^2}[/tex]
Acceleration due to gravity on Mercury
[tex]g_m=\frac{GM_m}{R_m^2}\\\Rightarrow g_m=\frac{G\rho_e \frac{4}{3}\pi (0.375R_e)^3}{(0.375R_e)^2}[/tex]
According to the question
[tex]\frac{g_m}{g_e}=\frac{3}{8}[/tex]
So,
[tex]\frac{3}{8}=\dfrac{\frac{G\rho_e \frac{4}{3}\pi (0.375R_e)^3}{(0.375R_e)^2}}{\frac{G\rho_e \frac{4}{3}\pi R_e^3}{R_e^2}}\\\Rightarrow \frac{3}{8}=\frac{\rho_m0.375}{\rho_e}\\\Rightarrow \frac{\rho_m}{\rho_e}=\frac{3}{8\times 0.375}\\\Rightarrow \frac{\rho_m}{\rho_e}=1[/tex]
Hence, the average density ratio of Mercury to Earth is 1
