## University Physics with Modern Physics (14th Edition)

At a distance of 13,800 km above the earth's surface, the acceleration due to gravity is $0.98~m/s^2$.
Let $M$ be the mass of the earth. Let $R$ be the radius of the earth. (equation 1): $\frac{G~M}{R^2} = 9.80~m/s^2$ Let $h$ be the distance above the earth's surface where the acceleration due to gravity is $0.98~m/s^2$. (equation 2):$\frac{G~M}{(R+h)^2} = 0.98~m/s^2$ We can divide equation 1 by equation 2. $\frac{(R+h)^2}{R^2} = 10$ $R+h = \sqrt{10}~R$ $h = (\sqrt{10}-1)~R$ $h = (\sqrt{10}-1)(6380~km)$ $h = 13,800~km$ At a distance of 13,800 km above the earth's surface, the acceleration due to gravity is $0.98~m/s^2$.