Question

At what height above Earth's surface is the gravitational acceleration reduced from its sea-level value by 0.20%?
At what height above Earth's surface is the gravitational acceleration reduced from its sea-level value by 2.0%?
At what height above Earth's surface is the gravitational acceleration reduced from its sea-level value by 20%?

What am I doing wrong...
for .2%
g=9.8-(.002*9.8)
.0196 =9.7804
r=sqrt(GM/g)
sqrt(3.9886*10^14)/9.7804
6386044.4
6386044.4-6370000=16044 m or aprx 16 km

am i not using the right equation?

Answer 1

g'=g*(re/(re+r))^2

so if you want g' to be say 1percetn less, then g'=.99g
or .99=(re/(re+r))^2
re/(re+r)= sqrt .99

then solve for r, which is the height above the Earth's surface

Answer 2

Your logic is perfectly sound. The problem is that using rounded figures like 5.98e24 for mass and 6.34e6 for radius gets you an initial value for earth's gravity as 9.83. Using less rounded figures like mass as 5.972kg gets closer to the truth (9.81).
But again we're dealing with really big numbers so getting .2% from these big numbers is subject to error. Using a little more exact figures your way gets a figure of about 11800m. But your way of doing it is completely correct.