Question

At what height above the surface does the acceleration due to gravity reduce to 64%

Answer 1

Use the formula g{1-2h/r}

Answer 2

g=F/m=kMp/r²=kMp/(R+h)²

h=0=>go=kMp/Rp²

g=64%*go=64go/100=16go/25=>kMp/(Rp+h)²=kMp*16/25Rp²<=>1/(Rp+h)²=16/25Rp²<=>1/(Rp²+h²+2Rph)=16/25Rp²<=>16Rp²+16h²+32Rph=25Rp²<=>16h²+32Rph-9Rp²=0

Δ=32²Rp²-4*16*(-9Rp²)=1024Rp²+576Rp²=1600Rp²

h₁=(-32Rp+40Rp)/2=8Rp/2=4Rp=4*6371=25484 km

h₂ does not interest us because it is negative