Question
How high does a rocket have to go above the
Earth’s surface so that its weight is reduced
to 33.7 % of its weight at the Earth’s surface? The radius of the Earth is 6380 km
and the universal gravitational constant is
6.67 × 10
−11
N · m
2
/kg
2
.
Answer in units of km
Earth’s surface so that its weight is reduced
to 33.7 % of its weight at the Earth’s surface? The radius of the Earth is 6380 km
and the universal gravitational constant is
6.67 × 10
−11
N · m
2
/kg
2
.
Answer in units of km
Answers
Elena
mg=GmM/R²
g=GM/R²
g´=GM/(R+h)²
g´= 0.337g
GM/(R+h)² =0.337• GM/R²
R²/0.337=(R+h)²= R²+2Rh+h²
h²+2Rh - 1.96R² = 0
h=- R±sqrt{R²+1.96 R²}=
=-R±1.72R
h=0.72R=0.72•6.38•10⁶=4.59•10⁶ m
g=GM/R²
g´=GM/(R+h)²
g´= 0.337g
GM/(R+h)² =0.337• GM/R²
R²/0.337=(R+h)²= R²+2Rh+h²
h²+2Rh - 1.96R² = 0
h=- R±sqrt{R²+1.96 R²}=
=-R±1.72R
h=0.72R=0.72•6.38•10⁶=4.59•10⁶ m