Question

How far above the surface of the earth would you have to be before your weight is reduced by 13.63%?

Answer 1

a = GM / r^2

Variables and constants

a = Acceleration
G = Universal Gravitational Constant
M = Mass of Earth
r = Distance from Center of Earth
Re = Radius of Earth
g = Earth Surface Gravity

Values

g = 9.81 m/s^2
G = 6.67259E-11 m^3/kg-s^2
M = 5.9736E+24 kg
Re = 6,371,000 m

So you need 13.63% of Earth-g

Turn 13.63% into a decimal: 13.63 / 100 = 0.1363

Find how much that is of normal surface gravity: 0.1363 * 9.81 m/s^2 = 1.3371 m/s^2

That leaves "remaining gravitational acceleration" of: 9.81 m/s^2 - 1.3371 m/s^2 = 8.4729 m/s^2

Take the equation given above and solve for "r" gives

r = SQRT { [GM] / a }

r = SQRT { [ (6.67259E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (8.4729 m/s^2) }

r = 6,858,771 m

That's as measured from the center of the Earth. So subtract Earth's radius to surface to get altitude over the surface:

Alt. = r - Re
Alt. = (6,858,771 m) - (6,371,000 m)
Alt = 487,771.3 m