Asked by lizbeth

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%?
Answered by Chanz
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.
Answered by Chanz
Oh sorry. And you need to subtract the radius of the earth from your answer to get how high above the surface.
Answered by lizbeth
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
?
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