Posted by **lissa** on Sunday, December 19, 2010 at 6:28pm.

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 -
**Damon**, Sunday, December 19, 2010 at 6:45pm
.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 -
**lissa**, Sunday, December 19, 2010 at 7:12pm
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 -
**Ventz**, Monday, December 5, 2011 at 10:52pm
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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