Which of the following is the value of gravitational force on a body of mass $80\mathit{ kg}$ lying on the surface of the earth? (Given gravitational constant $\mathit{G }$is $6.67\times {10}^{-11}N{m}^2{\mathit{kg}}^{-2}$, Mass of the earth is $6\times {10}^{24}\mathit{ kg}$, radius of the earth is $6.4\times {10}^6\mathit{ m}$).

The value of gravitational force between the body of mass $80\mathit{ kg}$ and the earth is $781.6\mathit{ N}$ .
As per Newton's universal law of gravitation, we know that force of attraction between 2 bodies is given by $\mathrm{F}=\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}$ 

Where, $G$ is the gravitational constant 
$m_1$is the mass of the first body 
$m_2$ is the mass of the second body 
$r$ is the distance between two bodies.
Mass of earth  $M=6\times {10}^{24}\mathit{kg}$ Mass of the boy $m=80\mathit{ kg}$
The radius of the earth  $r=6.4\times {10}^{6}m$ 

Hence, the force of attraction between the body and the earth can be written as 
$\mathrm{F}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2} .......\left(\mathrm{i}\right)$ 

Substituting the values in (i)
$\therefore \mathrm{F}=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\mathrm{ }\times 80\mathrm{ }}{{(6.4\times {10}^6\mathrm{ })}^2}$ 

$\Rightarrow \mathrm{F} =\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\mathrm{ }\times 80\mathrm{ }}{{(6.4\times {10}^6\mathrm{ })}^2}$ 

$\Rightarrow \mathrm{F} =\frac{3201.6\times {{10}^{-11}\times 10}^{24}\mathrm{ }}{{40.96\times {(10}^6\mathrm{ })}^2}$ 

$\Rightarrow \mathrm{F}=78.16\times 10 \mathrm{N}$ 

Thus, the value of gravitational force between the body and the earth is $781.6\mathit{ N}$.