Question

Two satellites are in circular orbits around Jupiter. One, with orbital radius \(r,\) makes one revolution every \(16 \mathrm{h}\) The other satellite has orbital radius \(4.0 r .\) How long does the second satellite take to make one revolution around Jupiter?

Short answer

Answer: The period of the second satellite is 128 hours.

Step-by-step solution

Write down the given information

The first satellite has an orbital radius r and a period of 16 hours. The second satellite has an orbital radius of 4.0r. We want to find the period (T) of the second satellite.

Use Kepler's Third Law

Kepler's Third Law states that \(\frac{T^2_1}{T^2_2} = \frac{r^3_1}{r^3_2}\).

Substitute the known values

Substitute the values for the first satellite into the equation: \(\frac{(16 \ \text{h})^2}{T^2_2} = \frac{(r)^3}{(4.0r)^3}\)

Simplify the equation

Simplify the equation to find the period of the second satellite: \(\frac{256 \ \text{h}^2}{T^2_2} = \frac{r^3}{64r^3}\)

Solve for the period of the second satellite

Cancel out the \(r^3\) on both sides of the equation and solve for \(T_2\): $\frac{256 \ \text{h}^2}{T^2_2} = \frac{1}{64} \\ T^2_2 = 256 \ \text{h}^2 \times 64 \\ T^2_2 = 16384 \ \text{h}^2 \\ T_2 = \sqrt{16384 \ \text{h}^2} \\ T_2 = 128 \ \text{h}$

State the answer

The second satellite takes 128 hours to make one revolution around Jupiter.