Question

If the value of g = 9.8m/second square, radius of earth = 6.37×10 the power 6 meter and the gravitational constant G =6.67×10 the power minus eleven meter cube per kg sec square then the density of earth is.

Answer 1

[Symbols have their usual meanings]

We know,
Acceleration due to gravity (g) = $\frac{GM}{ R^{2} }$
So, M = $\frac{gR^2}{G}$

Given values of g = 9.8 $m/ s^{2}$
R =$6.37 * 10^{6}$ 
G = $6.67 * 10^{-11}$

Hence, Mass of Earth (M) = $\frac{9.8 * (6.37*10^6)^{2} }{6.67*10^{-11}}$ = $5.96 * 10^{24}$

Volume of Earth (V) = $\frac{4}{3} \pi R^{3}$ = $\frac{4}{3} \pi (6.37*10^{6})^{3}$ = 1.08 * $10^{21}$

The density of Earth = M / V = $\frac{5.96 * 10^{24}}{1.08 * 10^{21}}$ = 5518.52 kg/$m^3$

[ANSWERED]