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) 60.8 h B) 12.0 h C) 33.1 h D) 9.80 h E) 27.0 h

Question

Quizplus · Accepted Answer

We can use Kepler's third law to solve this problem, which states:

$$\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3}$$

where $T$ is the period of the satellite and $R$ is its orbital radius.

Plugging in the values given in the problem, we get:

$$\frac{(18.0\ \mathrm{hours})^2}{(2.00	imes10^{17}\ \mathrm{m})^3} = \frac{T_2^2}{(3.00	imes10^{17}\ \mathrm{m})^3}$$

Solving for $T_2$, we get:

$$T_2 = \sqrt{\frac{(18.0\ \mathrm{hours})^2	imes(3.00	imes10^{17}\ \mathrm{m})^3}{(2.00	imes10^{17}\ \mathrm{m})^3}} \approx 33.1\ \mathrm{hours}$$

Therefore, the answer is C.

A Planet Has Two Small Satellites in Circular Orbits Around