## Physics: Principles with Applications (7th Edition)

$\frac{1}{2} \times G\frac{m}{r^2} = G\frac{m}{(\sqrt{2}~r)^2}$ We can see that the acceleration due to gravity will be half when the distance from the center of the Earth is $\sqrt{2}~r$. $\sqrt{2}~r = \sqrt{2}\times (6380 ~km) = 9020 ~km$ The height above the Earth's surface is 9020 km - 6380 km which is 2640 km. At a height of 2640 km above the Earth's surface, the acceleration due to gravity is half the value of the acceleration due to gravity at the Earth's surface.