The orbit of a spacecraft about the Sun has a perihelion distance of \(0.1 \mathrm{AU}\) and an aphelion distance of \(0.4 \mathrm{AU}\). What is the semimajor axis of the spacecraft's orbit? What is its orbital period?

: We are given the perihelion distance, which is the closest distance between the spacecraft and the Sun, and the aphelion distance, which is the farthest distance between the spacecraft and the Sun. The perihelion distance is \(0.1 \mathrm{AU}\) and the aphelion distance is \(0.4 \mathrm{AU}\).

: The semimajor axis is the average of the distance of perihelion and aphelion, given by the formula: \(a =\frac{Dist_{peri} + Dist_{aphel}}{2}\). Plugging in the given values, we get: \(a =\frac{0.1 \mathrm{AU} + 0.4 \mathrm{AU}}{2} =\frac{0.5 \mathrm{AU}}{2}=0.25 \mathrm{AU}\) So, the semimajor axis of the spacecraft's orbit is \(0.25 \mathrm{AU}\).

: Kepler's Third Law states that the square of the orbital period (\(T^2\)) of a planet is directly proportional to the cube of the semimajor axis (\(a^3\)) of its orbit. The formula for the constant of proportionality is: \(T^2 = k \cdot a^3\), where \(T\) is the orbital period in Earth years, \(a\) is the semimajor axis in astronomical units (AU), and \(k\) is the constant of proportionality, which is approximately equal to 1 for objects orbiting the Sun. We already found the semimajor axis to be \(0.25 \mathrm{AU}\). Now, we need to solve for the orbital period \(T\). From Kepler's Third Law, we have: \(T^2 = k \cdot (0.25 \mathrm{AU})^3\) Since \(k \approx 1\), we have: \(T^2 = (0.25 \mathrm{AU})^3 = 0.015625\) To find the orbital period \(T\), take the square root of both sides: \(T = \sqrt{0.015625} \approx 0.125 \mathrm{years}\) So, the orbital period of the spacecraft is approximately \(0.125 \mathrm{years}\) or about \(45.625 \mathrm{days}\).