A 75 kg man weighs 735 N on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" 15% of his body weight?

Answer 1

m g = G M m /r^2
so m does not matter if we only want ration

G M is constant
r = earth radius (any units) but be consistent. You might use 6.38*10^6 kg

R = r + d

.85 G M/r^2 = G M/(r+d)^2

r^2 = .85 (r+d)^2
r = .922 (r+d)
r - .922 r = .922 d
d = r (1-.922)/.922 = .0846 r

Answer 2

The value of acceleration due to gravity on surface of the earth is
g=weight of the person/mass of the
person=735/75=9.8ms^-2
The man loses 15% of weight,
g'=g85/100
Required height =g'=g*R^2/(R+H)^2

Answer 3

only 6 years late

Answer 4

Thanks