A space probe is studying a newly discovered spherical planet of...

The gravitational acceleration at a distance $r$ from the center of a planet is given by $g = \dfrac{GM}{r^2}$.

For the first orbit, the height is $h_1 = \dfrac{R}{2}$, so the distance from the center is $r_1 = R + \dfrac{R}{2} = \dfrac{3R}{2}$.

The acceleration is $a_1 = \dfrac{GM}{\left(\dfrac{3R}{2}\right)^2} = \dfrac{4GM}{9R^2}$.

For the second orbit, the height is $h_2 = 2R$, so the distance from the center is $r_2 = R + 2R = 3R$.

The acceleration is $a_2 = \dfrac{GM}{(3R)^2} = \dfrac{GM}{9R^2}$.

The ratio of the accelerations is $\dfrac{a_1}{a_2} = \dfrac{\dfrac{4GM}{9R^2}}{\dfrac{GM}{9R^2}} = 4$.

Thus, the ratio is $4 : 1$.

Answer: $4 : 1$