Problem 40 - Astronomy: At Play in the Cosmos

Question

Calculate the orbital velocity, in kilometers per second, of a star orbiting 15,000 pc from the center of a galaxy whose mass is 130 billion \(M_{\mathrm{Sun}}.\)

Step-by-Step Solution

Answer

The orbital velocity is approximately 193.1 km/s.

Step 1: Identify Formula for Orbital Velocity

To calculate the orbital velocity of a star, we use the formula for circular orbital velocity: \( v = \sqrt{\frac{GM}{r}} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), \( M \) is the mass of the galaxy, and \( r \) is the distance from the center of the galaxy.

Step 2: Convert Units appropriately

The given distance is 15,000 parsecs (pc). Convert this to meters: \(1 \, \text{pc} = 3.086 \times 10^{16} \, \text{m}\). Thus, \( r = 15,000 \times 3.086 \times 10^{16} = 4.629 \times 10^{20} \, \text{m}\). Also, convert the galaxy mass from solar masses: \( 1 \, M_{\text{Sun}} = 1.989 \times 10^{30} \text{ kg}\), so \( M = 130 \times 10^9 \times 1.989 \times 10^{30} = 2.5867 \times 10^{41} \, \text{kg}\).

Step 3: Insert Values into Formula

Substitute the known values for \( G \), \( M \), and \( r \) into the formula: \[ v = \sqrt{\frac{6.674 \times 10^{-11} \times 2.5867 \times 10^{41}}{4.629 \times 10^{20}}} \]

Step 4: Solve the Equation

Calculate the numerical value: - First, find \( GM = 6.674 \times 10^{-11} \times 2.5867 \times 10^{41} = 1.726 \times 10^{31} \).- Then divide by \( r \): \[ \frac{1.726 \times 10^{31}}{4.629 \times 10^{20}} \approx 3.728 \times 10^{10} \].- Finally, take the square root: \[ v = \sqrt{3.728 \times 10^{10}} \approx 1.931 \times 10^5 \, \text{m/s} \].

Step 5: Convert Orbital Velocity to Kilometers per Second

Convert the velocity from meters per second to kilometers per second by dividing by 1,000: \[ v \approx 193.1 \, \text{km/s} \].

Key Concepts

Gravitational Constant

The gravitational constant, often denoted as \( G \), plays a crucial role in understanding how objects in space interact through gravitational forces. It represents the force of attraction between two masses separated by a unit distance. Albert Einstein stated that gravity is the warping of spacetime, which helps in understanding concepts involving \( G \). The value of the gravitational constant is \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \). This seemingly small value signifies how relatively weak gravity is compared to other forces like electromagnetism.

• \( G \) is a universal constant, meaning its value is the same everywhere in the universe.
• It helps in calculating the gravitational force between two massive objects, such as planets or stars.
• In the orbital velocity formula \( v = \sqrt{\frac{GM}{r}} \), \( G \) allows us to determine the velocity required for a star or object to maintain its orbit at a specific distance \( r \) from a massive body of mass \( M \).

Understanding \( G \) is vital for solving astrophysical problems, such as calculating the dynamics of stars in galaxies.

Galactic Mass

Galactic mass is a measure of the total amount of matter contained within a galaxy. This includes stars, planets, interstellar gas, dark matter, and other celestial bodies. In our exercise, we consider the galactic mass to be 130 billion times the solar mass \( M_{\mathrm{Sun}} \), emphasizing the sheer scale of these cosmic structures compared to our Solar System.

• One solar mass \( M_{\mathrm{Sun}} \) is approximately \( 1.989 \times 10^{30} \, \text{kg} \).
• The conversion of galactic mass in terms of solar masses simplifies complex astronomical calculations.
• In calculating orbital velocities, knowing the mass of a galaxy is crucial because it directly affects the gravitational pull exerted on orbiting bodies.

By combining the galactic mass with the gravitational constant, we can explore the dynamics of stars orbiting the galaxy's center.

Parallax Distance

Parallax distance is an astronomical technique used to measure the distances to celestial objects. It relies on observing the apparent motion of a nearby star against the background of distant stars from two different positions of Earth's orbit six months apart. This method provides a way to calculate distance in parsecs (pc), where 1 parsec equals approximately \( 3.086 \times 10^{16} \, \text{m} \).

• Parallax provides one of the most precise distance measurement methods in astronomy.
• It is particularly helpful for measuring distances to stars within a few thousand parsecs of Earth.
• Distance measured in parsecs can be easily converted to meters if needed for calculations involving orbital dynamics.

Understanding parallax distance helps astronomers calibrate their observations and determine the scale of space around us.

Unit Conversion

Unit conversion is a fundamental skill in physics and astronomy, ensuring that all quantities in equations are expressed in compatible units. In our calculation, we converted distances from parsecs to meters and masses from solar masses to kilograms. This ensures precision when using the gravitational formula.

• 1 parsec (pc) is equivalent to \( 3.086 \times 10^{16} \, ext{m} \).
• 1 solar mass \( M_{\mathrm{Sun}} \) translates to \( 1.989 \times 10^{30} \, ext{kg} \).
• In our final step, converting velocity from meters per second to kilometers per second (1 km = 1000 m) simplified the interpretation of the result.

Effective unit conversion bridges the gap between abstract mathematical formulas and practical, understandable results, making it easier to communicate scientific findings globally.