Question

Assuming earth to be a solid sphere of uniform density $5800kg{m}^{-3}$and radius $6\times {10}^{6}m$, calculate the value of ‘g’ on its surface.

Answer 1

Step 1: Given data:

$$\begin{gathered} DensityofEarth,\rho =5800kg{m}^{-3} \\ RadiusofEarth,R=6\times {10}^{6}m \\ Accelerationduetogravity,g=? \end{gathered}$$

Step 2: Writing the formula for the acceleration due to gravity in terms of the density and radius of the earth

$$\begin{gathered} g=G\times \frac{M}{R^2} \\ =G\times \frac{\rho \times V}{R^2} \\ =G\times \frac{\rho \times \frac{4}{3}\pi R^3}{R^2} \\ =\frac{4}{3}\times G\times \rho \times \pi \times R \end{gathered}$$

Step 3: Substituting the given values in the expression

$$\begin{gathered} g=\frac{4}{3}\times G\times \rho \times \pi \times R \\ =\frac{4}{3}\times \left(6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}\right)\times \left(5800kg{m}^{-3}\right)\times \left(3.14\right)\times \left(6\times {10}^{6}m\right) \\ =9.72m{s}^{-2} \end{gathered}$$

Hence, the value of acceleration due to gravity is $9.72m{s}^{-2}$.