## College Physics (4th Edition)

(a) $h = 1434~km$ (b) $h = 4670~km$
(a) The gravitational field strength is $\frac{GM}{R^2}$, where $M$ is the Earth's mass, and $R$ is the distance from the center of the Earth. Let $R_E$ be the radius of the Earth (which is 6380 km). Let $h$ be the height above the surface where the gravitational field strength is two-thirds of its value at the surface. We can find $h$: $\frac{GM}{(R_E+h)^2} = \frac{2}{3}~\frac{GM}{R_E^2}$ $2(R_E+h)^2 = 3R_E^2$ $R_E+h = \sqrt{\frac{3}{2}}~R_E$ $h = R_E~(\sqrt{\frac{3}{2}}-1)$ $h = (6380~km)~(\sqrt{\frac{3}{2}}-1)$ $h = 1434~km$ (b) The gravitational field strength is $\frac{GM}{R^2}$, where $M$ is the Earth's mass, and $R$ is the distance from the center of the Earth. Let $R_E$ be the radius of the Earth (which is 6380 km). Let $h$ be the height above the surface where the gravitational field strength is one-third of its value at the surface. We can find $h$: $\frac{GM}{(R_E+h)^2} = \frac{1}{3}~\frac{GM}{R_E^2}$ $(R_E+h)^2 = 3R_E^2$ $R_E+h = \sqrt{3}~R_E$ $h = R_E~(\sqrt{3}-1)$ $h = (6380~km)~(\sqrt{3}-1)$ $h = 4670~km$