Answer
(a) $h = 1434~km$
(b) $h = 4670~km$
Work Step by Step
(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 $h$:
$\frac{GM}{(R_E+h)^2} = \frac{1}{3}~\frac{GM}{R_E^2}$
$(R_E+h)^2 = 3R_E^2$
$R_E+h = \sqrt{3}~R_E$
$h = R_E~(\sqrt{3}-1)$
$h = (6380~km)~(\sqrt{3}-1)$
$h = 4670~km$
