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
rRearth
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
=]