Asked by Anonymous
A man has a mass of 65kg on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" 13% of his body weight?
Answers
Answered by
Damon
You are asking how far up before g = .87 ge where ge is g at earth surface.
gravitational force is inversely proportional to distance squared.
we are at r radius, Re is earth radius.
so (r/Re)^2 = 1/.87 = 1.1494
r/Re = 1.07 or r = 1.0721 Re
then
r-Re = height above earth
=.0721 Re but Re = 6.4*10^6 meters approx
so
.0721*6.4*10^6 = .46*10^6 = 4.6*10^5 =460,000 meters
gravitational force is inversely proportional to distance squared.
we are at r radius, Re is earth radius.
so (r/Re)^2 = 1/.87 = 1.1494
r/Re = 1.07 or r = 1.0721 Re
then
r-Re = height above earth
=.0721 Re but Re = 6.4*10^6 meters approx
so
.0721*6.4*10^6 = .46*10^6 = 4.6*10^5 =460,000 meters
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