From Mechanics

Use the theory of gravitation equation

\displaystyle F=\frac{Gm_{1}m_{2}}{{d}^{\ 2}}

where,

\displaystyle G=6\textrm{.}67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-2}}

In this case

\displaystyle {{m}_{1}} is the mass of the earth and \displaystyle {{m}_{2}} is the mass of the man.

\displaystyle d is the distance of the man from the centre of the earth and thus is the sum of the radius of the earth and his height above sea level.

Thus

\displaystyle {{m}_{1}}= \displaystyle \text{5}\textrm{.}\text{98}\times \text{1}0^{\text{24}}\text{ }\text{kg}

\displaystyle {{m}_{2}}=80\ \text{kg}

The radius of the earth is

\displaystyle \text{6}\textrm{.}\text{37}\times \text{1}0^{\text{6}}\text{ m }

and the height of the man above sea level is \displaystyle 5000\ \text{m}

which gives

\displaystyle d=\displaystyle \text{13}\textrm{.}\text{37}\times \text{1}0^{\text{6}}\text{ m }

Inserting all these quantities in the gravitation equation

\displaystyle \begin{align} & F=\frac{\left( 6\textrm{.}67\times {{10}^{-11}} \right)\times \left( 5\textrm{.}98\times {{10}^{24}} \right)\times 80}{{{\left( 6\textrm{.}375\times {{10}^{6}} \right)}^{2}}} \\ & =\frac{319\textrm{.}10\times {{10}^{14}}}{40\textrm{.}64\times {{10}^{12}}}=7\textrm{.}85\times {{10}^{2}}\ =785\ \text{N} \\ \end{align}