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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
N · m
Answer in units of km

  • physics -

    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}=
    h=0.72R=0.72•6.38•10⁶=4.59•10⁶ m

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