Gravitational force due to the earth is given as
[tex]F=\frac{GM_em}{(r_e)^2}[/tex]
Here, G is the gravitational constant whose value is
[tex]G=6.67\times10^{-11}\text{ N m}^2\text{ /kg}^2[/tex]
The mass of the earth is
[tex]M_e\text{ = 5.97}\times10^{24}\text{ kg}[/tex]
The radius of the earth is
[tex]r_e=6371\times10^3\text{ m}[/tex]
Let m be the mass of any object.
Suppose the object is falling on the surface of the earth, so it will experience a force of mg.
Here, g is the acceleration due to gravity.
On equating both the forces,
[tex]\begin{gathered} mg=\frac{GM_em\text{ }}{(r_e)^2} \\ g=\frac{GM_e}{(r_e)^2} \end{gathered}[/tex]
On substituting the values, the value of g will be
[tex]\begin{gathered} g=\frac{6.67\times10^{-11}\times5.97\times10^{24}}{(6371\times10^3)^2} \\ =9.81\text{ m/s}^2 \end{gathered}[/tex]
