g = GM/r^2
so r = sqrt(GM/g)
g = 9.8 so .2% reduced = 9.8 - (.002*9.8)
G = 6.67e-11 and M = 5.98e24 for Earth.
Lather rinse repeat.
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%?
3 answers
Oh sorry. And you need to subtract the radius of the earth from your answer to get how high above the surface.
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
6386044.4-6371000=15044 m or aprx 15 km
?
for .2%
g=9.8-(.002*9.8)
.0196 =9.7804
r=sqrt(GM/g)
sqrt(3.9886*10^14)/9.7804
6386044.4
6386044.4-6371000=15044 m or aprx 15 km
?