Question

Consider earth to be a uniform sphere of radius, $6.4\times {10}^6$$m$ and its density is $5500kg{m}^{-3}$. Find the value of acceleration due to gravity on its surface.

Answer 1

Step 1: Given

Radius of earth $R=6.4\times {10}^{6}m$

Density of Earth $=5500kg{m}^{-3}$

Step 2: Finding the mass of Earth

Density $\left(e\right)=\frac{mass}{volume}$ (volume sphere $=\frac{4}{3}{\mathrm{\pi r}}^3$)

$5500=\frac{Mass}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}$

$Mass=5500\times \frac{4}{3}\times \mathrm{\pi }\times {(6.4\times {10}^6)}^3$

The gravitational constant is specified and given as

$G=6.7\times {10}^{-11}N{m}^{2}k{g}^2$

Step 3: Finding the value of acceleration due to gravity

$g=\frac{GM}{R^2}$

Substituting the values into given formula

$g=\frac{6.7\times {10}^{-11}\times 5500\times {\displaystyle \frac{4}{3}}\times \mathrm{\pi }\times {(6.4\times {10}^6)}^3}{{(6.4\times {10}^6)}^2}$

$=6.7\times {10}^{-11}\times 5.5\times {10}^3\times \frac{4}{3}\times \mathrm{\pi }\times 6.4\times {10}^6$

$g=9.83m{s}^{-2}$

Therefore, the value of acceleration due to gravity is surface is $9.83m{s}^2$.