The semi-major axis of Jupiter's orbit is \(5.20 \mathrm{AU}\). Find its orbital period in years, and its mean (time-averaged) orbital speed. (Mean orbital speed of Earth \(\left.=29.8 \mathrm{~km} \mathrm{~s}^{-1} .\right)\)

Kepler's Third Law states that \(\frac{T^2}{a^3} = \text{constant}\), where \(T\) is the orbital period and \(a\) is the semi-major axis of a planet's orbit. For our case, we will consider Jupiter and Earth. Let \(T_J\) and \(a_J\) represent Jupiter's orbital period and semi-major axis, respectively, and \(T_E\) and \(a_E\) represent Earth's orbital period and semi-major axis. Then we have: \(\frac{T_J^2}{a_J^3} = \frac{T_E^2}{a_E^3}\) We know that for Earth, \(T_E = 1 \text{ year}\) (by definition) and \(a_E = 1 \text{ AU}\) (by definition). We are given that \(a_J = 5.20 \text{ AU}\), and we need to find \(T_J.\)

Using the expression from step 1 and plugging in the known values, we can solve for \(T_J\): \(\frac{T_J^2}{5.20^3} = \frac{1^2}{1^3}\) Solving for \(T_J\), we get: \(T_J^2 = 5.20^3\) \(T_J = \sqrt{5.20^3}\) \(T_J \approx 11.86 \text{ years}\) Thus, Jupiter's orbital period is approximately 11.86 years.

To calculate Jupiter's mean orbital speed, we will first determine the distance covered in one orbit and then divide that by the orbital period. The circumference of an ellipse (which is an approximate shape of an orbit) is given by the formula: \(C \approx 2 \pi a \left[1 + \frac{1}{4} e^2 \right]\) Here, \(a\) is the semi-major axis, and \(e\) is the eccentricity of the orbit. Since Jupiter's orbit is nearly circular, its eccentricity is close to zero and we can use the approximation: \(C \approx 2 \pi a\) (for a nearly circular orbit) We can then find Jupiter's orbital speed, \(v_J\), using the distance covered in one orbit (C) and the orbital period, \(T_J\): \(v_J = \frac{C}{T_J}\) Given that the semi-major axis of Jupiter's orbit is \(a_J = 5.20 \mathrm{AU}\), we can convert it to kilometers: \(a_J = 5.20 \mathrm{AU} \times 149.6 \times 10^6 \mathrm{km} \mathrm{AU}^{-1} \approx 777.9 \times 10^6 \mathrm{km}\) Now, we can calculate Jupiter's mean orbital speed: \(v_J = \frac{2 \pi (777.9 \times 10^6 \mathrm{km})}{11.86 \text{ years}}\)

To convert Jupiter's mean orbital speed from km/year to km/s, we need to multiply it by the conversion factor: \(\text{Conversion factor} = \frac{1 \text{ year}}{3.154\times10^7 \text{ s}}\) \(v_J = \frac{2 \pi (777.9 \times 10^6 \mathrm{km})}{11.86 \text{ years}} \times \frac{1 \text{ year}}{3.154\times10^7 \text{ s}} \approx 13.1 \mathrm{~km} \mathrm{~s}^{-1}\) Hence, Jupiter's mean orbital speed is approximately 13.1 km/s.