A star is observed in a circular orbit about a black hole with an orbital radius of \(1.5 \times 10^{11}\) kilometers and an average speed of \(2,000 \mathrm{km} / \mathrm{s}\). What is the mass of this black hole in solar masses?

Orbital velocity is the speed at which an object travels along its circular orbit around a larger body. The equation for orbital velocity is: \[ v = \sqrt{\frac{GM}{r}} \] Here, v represents the orbital velocity, G is the gravitational constant, M is the mass of the larger body (in this case, the black hole), and r is the radius of the orbit.
In the context of our exercise, the star's average speed of 2,000 km/s is its orbital velocity. To find the mass of the black hole, we use this speed along with the gravitational constant and the orbital radius.
Orbital velocity is crucial for calculating many astronomical parameters, including the mass of celestial bodies. A higher orbital velocity at a given distance signifies a larger mass of the central body.

The gravitational constant, commonly denoted as G, is a fundamental constant in physics that appears in Newton's law of universal gravitation. Its value is approximately \( 6.67 \times 10^{-11} \) \( m^3 kg^{-1} s^{-2} \) and it quantifies the strength of the gravitational force between two masses.
In our exercise, we use G to relate the orbital velocity and radius of the star to the mass of the black hole. The gravitational constant allows us to understand how massive celestial bodies influence the motion of objects around them.
By substituting the known values into the orbital velocity formula, G helps us compute the mass of the black hole, showcasing the power of this fundamental constant in gravitational calculations.

Solar masses are a standard unit of mass used in astronomy to express the mass of stars and other astronomical objects. One solar mass is equivalent to the mass of our Sun, approximately \( 1.989 \times 10^{30} \text{kg} \).
In our exercise, after calculating the mass of the black hole in kilograms, we convert it to solar masses for easier comparison with other astronomical objects. To perform this conversion, we divide the black hole's mass by the mass of the Sun.
Using solar masses simplifies our understanding and communication of the sizes and masses of celestial bodies. For instance, a mass calculation of \( 4.52 \text{solar masses} \) means that the black hole is about 4.52 times the mass of the Sun.

A circular orbit is one in which an object moves around a larger body in a path that forms a perfect circle. The radius of this orbit remains constant, meaning the distance between the orbiting object and the central body does not change.
In our example, the star moves around the black hole in a circular orbit with a radius of \( 1.5 \times 10^{14} \text{ meters} \). This distance is crucial for applying the orbital velocity formula to determine the black hole's mass.
Circular orbits simplify calculations because the gravitational force acts as the centripetal force that keeps the object in its path. Therefore, the relationship between velocity, mass, and radius can be directly applied.

Unit conversion is an essential step in solving many physics problems. It ensures that all the values used in equations are in the correct and compatible units. In our exercise, we first convert the star's orbital radius from kilometers to meters, because the gravitational constant G is expressed in SI units.
Given that \( 1 \text{ kilometer} \ = 10^3 \text{ meters} (1 \text{ km} = 10^3\text{ m}) \), we converted the orbital radius: \( 1.5 \times 10^{11} \text{ kilometers} = 1.5 \times 10^{14} \text{ meters} \).
Additionally, unit conversion helps when expressing the final mass of the black hole in solar masses, a more meaningful unit in astrophysics.
Always pay attention to units and make sure all measurements are consistent to avoid errors in calculations.