Question

A planet has two small satellites in circular orbits around the planet. the first satellite has a period 18.0 hours and an orbital radius 2.00 × 107 m. the second planet has an orbital radius 3.00 × 107 m. what is the period of the second satellite? a planet has two small satellites in circular orbits around the planet. the first satellite has a period 18.0 hours and an orbital radius 2.00 × 107 m. the second planet has an orbital radius 3.00 × 107 m. what is the period of the second satellite? 60.8 h 12.0 h 33.1 h 9.80 h 27.0 h

Answer 1

The two satellites orbit around the same planet, so we can use Kepler's third law, which states that the ratio between the cube of the radius of the orbit and the orbital period is constant for the two satellites:
$\frac{r_1^3}{T_1^2}= \frac{r_2^3}{T_2^2}$
where
$r_1$ is the orbital radius of the first satellite
$r_2$ is the orbital radius of the second satellite
$T_1$ is the orbital period of the first satellite
$T_2$ is the orbital period of the second satellite

If we use the data of the problem and we re-arrange the equation, we can calculate the orbital period of the second satellite:
$T_2 = \sqrt{T_1^2 ( \frac{r_2}{r_1} )^3} = \sqrt{(18.0 h)^2 ( \frac{3\cdot 10^7 m}{2 \cdot 10^7 m} )^3} = 33.1 h$