Question

If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight to drop to 800 N

Answer 1

To calculate weight we use formula:
$F=m*g=882N$

When we have two bodies we use general gravity formula:
$F=G* \frac{ M_{p}*M_{E} }{ r^{2} }$
Where:
G = gravity constant = $6.67* 10^{-11} m^{3} kg^{-1} s^{-2}$
Mp = mass of person = 882 / 9.81= 89.9kg
ME = mass of Earth = $5.97* 10^{24} kg$
r = distance between Earth and person

From these two equations we find that left sides are equal so the right sides must be equal too.
$m*g=G* \frac{ M_{p}*M_{E} }{ r^{2} }$

We solve this for r:
$r= \sqrt{G* \frac{ M_{p}*M_{E}}{M_{p}*g}$
r=6371116m = 6371.116km

This is distance from center of Earth. Radius of Earth is 6370km and height above surface is 6371.116 - 6370 = 1.116km or 1116m.