physics
posted by lissa .
A 70 kg man weighs 686 N on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" 14% of his body weight?
i don't even know where to start with this question... i need help!

.86 = Rearth^2/r^2
because
W = G m M/R^2
and G m M does not change
remember that they really ask for
rRearth
the distance above ground, not the bigger radius from earth center 
could try and explain more in depth? like wat does W stand for? his weight? and i don't really get how your solving it..

F=(G*m1*m2)/r^2
F force
G gravitational constant
m1 mass on earth
m2 mass of object
r radius on earth plus given distance over it
.86*686=589.96 to find final force
plug in everything..
589.96=(6.67*10^11*70*5.98*10^24)/ (h+6,370,000)^2
Solve for h!
h=509411
=]
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