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### Theory:

In the previous section, we learned about acceleration due to gravity ($$g$$) and its relation with universal gravitational constant ($$G$$). In this section, we will compute the mass of the earth.

We know that,

$\begin{array}{l}g\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{G×M}{{R}^{2}}\phantom{\rule{0.147em}{0ex}}\\ \\ \mathit{Where},\\ g\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\mathit{Acceleration}\phantom{\rule{0.147em}{0ex}}\mathit{due}\phantom{\rule{0.147em}{0ex}}\mathit{to}\phantom{\rule{0.147em}{0ex}}\mathit{gravity}\\ \\ G\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\mathit{Universal}\phantom{\rule{0.147em}{0ex}}\mathit{gravitational}\phantom{\rule{0.147em}{0ex}}\mathit{constant}\\ \\ M\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{Earth}\\ \\ R\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}\mathit{Radius}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{Earth}\end{array}$

Rearranging the above equation, the mass of the Earth ($$M$$ is obtained as follows:

$\begin{array}{l}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{Earth}\phantom{\rule{0.147em}{0ex}}\left(M\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{g×{R}^{2}}{G}\\ \\ \mathit{We}\phantom{\rule{0.147em}{0ex}}\mathit{know}\phantom{\rule{0.147em}{0ex}}\mathit{that},\\ \\ \mathit{Acceleration}\phantom{\rule{0.147em}{0ex}}\mathit{due}\phantom{\rule{0.147em}{0ex}}\mathit{to}\phantom{\rule{0.147em}{0ex}}\mathit{gravity}\phantom{\rule{0.147em}{0ex}}\left(g\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}9.8\phantom{\rule{0.147em}{0ex}}m{s}^{-1}\\ \\ \mathit{Universal}\phantom{\rule{0.147em}{0ex}}\mathit{gravitational}\phantom{\rule{0.147em}{0ex}}\mathit{constant}\phantom{\rule{0.147em}{0ex}}\left(G\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}6.674\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}{10}^{-11}\phantom{\rule{0.147em}{0ex}}N{m}^{2}{\mathit{kg}}^{-2}\\ \\ \mathit{Radius}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{Earth}\phantom{\rule{0.147em}{0ex}}\left(R\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}6378\phantom{\rule{0.147em}{0ex}}\mathit{km}\end{array}$

Substituting the known values of $$g$$, $$R$$ and $$G$$, you can calculate the mass of the Earth.

Variation of acceleration due to gravity ($$g$$):

We know that the acceleration due to gravity ($$g$$) is inversely proportional to the square of the geometric radius of the Earth ($g\phantom{\rule{0.147em}{0ex}}\propto \phantom{\rule{0.147em}{0ex}}\frac{1}{{R}^{2}}$); its value changes from one place to another on the surface of the Earth.

In the equatorial region, the geometric radius of the Earth is maximum, and in the polar region, it is minimum. So, the '$$g$$' value is maximum in the polar region and minimum in the equatorial region.

When you go to a higher altitude from the surface of the Earth, the value of '$$g$$' decreases. In the same way, when you go deep below the surface of the Earth, the value of '$$g$$' decreases. The value of $$g$$ is zero at the centre of the Earth.