An interstellar cloud is found to be rotating such that a velocity shift of \(1 \mathrm{km} / \mathrm{s} / \mathrm{pc}\) is observed across it. (a) What is the angular speed (rad/s)? (b) What is the rotation period?

Angular speed helps us understand how fast an object, like an interstellar cloud, rotates around an axis. It expresses the angle through which the object turns in a certain period. Angular speed is crucial in astrophysics to describe rotational dynamics.

To calculate the angular speed (\(\omega\)), we use the formula: \[\omega = \frac{v} { r }\]. Here \(\omega\) is angular speed in rad/s, \(\text{v}\) is linear velocity (in meters per second), and \(\text{r}\) is the radius (in meters).

In our exercise, the linear velocity (v) is 1 km/s, which converts to \(10^3\) m/s in SI units. The radius (r) is 1 parsec, which converts to \(3.086 \times 10^{ 16 }\) meters. Plug these into our equation: \[\omega = \frac{10^3}{3.086 \times 10^{ 16 }} \approx 3.24 \times 10^{ -14 } \text{ rad/s }\].

This shows how slowly the interstellar cloud rotates around its axis.

The rotation period tells us how long it takes for the interstellar cloud to complete one full rotation. It’s the inverse of the angular speed and is expressed in seconds or other units like years for cosmic objects.

To find the rotation period (T), we use the formula: \[ T = \frac { 2 \pi }{ \omega } \.\](Here, \(\omega\) is the angular speed we previously calculated: \(3.24 \times 10^{ -14 }\) rad/s.

Let's calculate T: \[ T = \frac{ 2 \pi } { 3.24 \times 10^{ -14 } } = 1.94 \times 10^{ 14 } \text{ s } \.\]

To make it more intuitive, we convert seconds to years. Using the conversion factor (1 year \approx 3.154 \times 10^7 s), we get:

\[ T = \frac { 1.94 \times 10^{ 14 } }{ 3.154 \times 10^{ 7 } } \approx 6.15 \times 10^{ 6 } \text { years } \.\]

So, the cloud takes around 6.15 million years to make one full spin!

The velocity shift is the change in velocity observed across different parts of the interstellar cloud. In our exercise, the velocity shift is given as 1 km/s per parsec. This measures how the cloud's rotation speed varies with distance from its center.

To work in compatible units, we convert km/s to m/s. Since 1 km = 1000 m, a velocity shift of 1 km/s becomes \(10^3\) m/s.

Understanding the velocity shift in astronomical objects helps scientists learn about their structure and dynamics. It can reveal how mass is distributed within the cloud and how it might interact with other celestial bodies.

A parsec (pc) is a unit of distance used in astronomy to measure large scales like the space between stars. One parsec is approximately \(3.086 \times 10^{13}\) kilometers. Converting this to meters involves multiplying by 1000: \[ 1 \text{pc} = 3.086 \times 10^{ 13 } \times 10^3 \approx 3.086 \times 10^{ 16 } \text{ m } \].

This conversion is vital for calculations in astrophysics. Earth's atmosphere and space missions use SI units like meters and seconds for precise measurements and calculations. We converted parsecs to meters to calculate angular speed and the rotation period accurately in the initial exercise.

Knowing how to convert these units properly is essential for translating astronomical observations into meaningful data.