# physics

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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!

• physics -

.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
r-Rearth
the distance above ground, not the bigger radius from earth center

• physics -

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..

• physics -

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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