Question:
A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is $6400 \mathrm{~km}$.)
Solution:
$F_{h}=\frac{1}{2} F_{e}$
$\frac{G M m}{(R+h)^{2}}=\frac{1}{2} \frac{G M m}{R^{2}}$
$(R+h)^{2}=2 R^{2}$
$h^{2}+2 R h-R^{2}=0$
Solving equation,
$h=\frac{-2 R \pm \sqrt{4 R^{2}+4 R^{2}}}{2}$
$h=-R \pm \sqrt{2} R$
$h=R(-1 \pm \sqrt{2})$
$h=R(\sqrt{2}-1)$
$h=6400 \times(1.414-1)$
$h=2649.6^{\approx} 2650 \mathrm{~km}$