The star HD 3651 shown in Figure 17-13 has a mass of \(0.79 \mathrm{M}_{\odot}\). Its brown dwarf companion, HD \(3651 \mathrm{~B}\), has about 40 times the mass of Jupiter. The average distance between the two stars is about \(480 \mathrm{AU}\). How long does it take the two stars to complete one orbit around each other?

In astronomy, the vast distances in space are often measured in astronomical units (AU), where one AU is the average distance from the Earth to the Sun, approximately \(1.496 \times 10^{11}\) meters. When dealing with orbital dynamics, scientists frequently convert this measure into meters to utilize in equations and calculations that are based on the metric system.

To convert from AU to meters, we simply multiple the distance in AU by \(1.496 \times 10^{11}\). As in our exercise example, for a distance of 480 AU between two stars, the conversion to meters would be \(480 \times 1.496 \times 10^{11}\) meters. This conversion is critical to ensure that all units are consistent when applying Kepler's Third Law or any other physical law, which allows us to accurately calculate the various properties of celestial bodies' orbits, such as their periods and velocities.

Understanding AU and its conversion to meters is vital for comprehending the scale of the solar system and beyond, and it's the first step in solving many astronomical problems.

The gravitational constant, represented by \(G\), is a key figure in the universe of physics, particularly in gravitational interactions. It appears in Newton's law of universal gravitation and is used to calculate the gravitational attraction between two masses.

With a value of \(6.67408 \times 10^{-11} m^3 kg^{-1} s^{-2}\), \(G\) is incredibly small, signifying the weakness of gravity compared to other fundamental forces. However, within Kepler's Third Law and in our exercise, we compute a system's gravitational constant, \(\mu\), which is the product of \(G\) and the system's total mass: \(\mu = G \times m\).

By determining \(\mu\), this allows astronomers to calculate the orbital period of celestial bodies with respect to their mutual center of mass. The gravitational constant is indispensable in orbital mechanics, defining the character of the force that maintains planets in orbit and leading to the development of celestial bodies' trajectories over time.

The orbital period is the time a celestial body takes to complete one full orbit around another object. According to Kepler's Third Law, the square of the orbital period is proportional to the cube of the semi-major axis of its orbit. Mathematically, this is expressed in the formula \(T = 2\pi \sqrt{\frac{d^3}{\mu}}\), where \(T\) is the orbital period, \(d\) is the distance between the orbiting bodies (after converting AU to meters), and \(\mu\) is the gravitational constant for the system calculated earlier.

In the context of our exercise, once we have \(d\) and \(\mu\), we plug them into this equation to solve for \(T\), the period in seconds. To make the result more comprehensible, we might convert this period from seconds to years by dividing by the number of seconds in one year, \(31536000\). This final conversion communicates the timescale in units that are more meaningful to our everyday experience and makes the vast temporal scopes involved in astronomical events more relatable.