Problem 107 - Chemistry Matter and Change

Question

Astronomy The black hole in the M82 galaxy has a mass about 500 times the mass of the Sun. It has about the same volume as the Moon. What is the density of this black hole? $$\begin{array}{c}{\text { mass of the Sun }=1.9891 \times 10^{30} \mathrm{kg}} \\ {\text { volume of the Moon }=2.1968 \times 10^{10} \mathrm{km}^{3}}\end{array}$$

Step-by-Step Solution

Answer

The density of the black hole is approximately \(4.5251 \times 10^{13}\) kg/m\(^3\).

Step 1: Calculate the Mass of the Black Hole

The mass of the black hole is given as 500 times the mass of the Sun. Therefore, we can calculate the mass of the black hole as follows:\[ \text{Mass of the Black Hole} = 500 \times 1.9891 \times 10^{30} \, \text{kg} \]Calculating, we get:\[ \text{Mass of the Black Hole} = 9.9455 \times 10^{32} \, \text{kg} \]

Step 2: Convert the Volume of the Black Hole to Cubic Meters

The volume of the Moon is given in cubic kilometers. We need to convert this volume into cubic meters for our density calculation. 1 km\(^3\) = \(10^9\) m\(^3\).Thus, the volume of the black hole in cubic meters is:\[ \text{Volume in } \text{m}^3 = 2.1968 \times 10^{10} \times 10^9 \, \text{m}^3 \]\[ \text{Volume in } \text{m}^3 = 2.1968 \times 10^{19} \, \text{m}^3 \]

Step 3: Calculate the Density of the Black Hole

Density is defined as mass per unit volume. Using the mass from Step 1 and the volume from Step 2, the density \( \rho \) is calculated as follows:\[ \rho = \frac{\text{Mass of the Black Hole}}{\text{Volume of the Black Hole}} \]\[ \rho = \frac{9.9455 \times 10^{32} \, \text{kg}}{2.1968 \times 10^{19} \, \text{m}^3} \]\[ \rho \approx 4.5251 \times 10^{13} \, \text{kg/m}^3 \]

Step 4: Conclusion

The density of the black hole is approximately \(4.5251 \times 10^{13}\) kg/m\(^3\). This reflects the incredibly high density of black holes, concentrated into a very small volume compared to their mass.

Key Concepts

mass calculation

Understanding how to calculate mass is crucial in various fields of science, including astronomy. Mass refers to the amount of matter within an object and is typically measured in kilograms (kg). In the given exercise, we encounter a situation where the black hole's mass is described as a multiple of the Sun's mass. To determine the black hole's mass, we simply multiply the known mass of the Sun by 500. This allows us to express the mass of the black hole in similar units. Here’s the calculation step by step:

• Start with the mass of the Sun, given as \(1.9891 \times 10^{30} \, \text{kg}\).
• Multiply this by 500 since the black hole is 500 times more massive than the Sun.
• The result is \(9.9455 \times 10^{32} \, \text{kg}\).

This highlights the immense mass of black holes, often measured in solar masses due to their astronomical size.

volume conversion

Volume conversion is an important step when working across different unit systems, such as from cubic kilometers to cubic meters. This is especially crucial in scientific problems where consistency of units is vital for correct calculations. In the problem, the black hole's volume is comparable to that of the Moon, provided in cubic kilometers (km³).
To convert volume from cubic kilometers to cubic meters, you need to recognize that one km equals 1,000 meters. Thus, one km³ is \(10^9\) m³.

• The given volume of the Moon is \(2.1968 \times 10^{10} \, \text{km}^3\).
• To convert this to cubic meters, multiply by \(10^9\).
• The converted volume is \(2.1968 \times 10^{19} \, \text{m}^3\).

This volume conversion is essential for computing the density, ensuring all measurements are in compatible units.

density formula

Density is an intrinsic property of matter, defined as mass per unit volume. It helps us understand how compact or concentrated the mass is in a given space. The formula for density (\( \rho \)), used in the problem, is as follows:
\(\rho = \frac{\text{Mass}}{\text{Volume}} \)
By using this formula, we can connect diverse physical properties such as mass and volume to better understand astronomical objects. For the black hole:

• We calculated its mass to be \(9.9455 \times 10^{32} \, \text{kg}\).
• The volume was converted to \(2.1968 \times 10^{19} \, \text{m}^3\).
• Substituting these into the density formula gives a density of approximately \(4.5251 \times 10^{13} \, \text{kg/m}^3\).

This extremely high density is a hallmark feature of black holes and highlights how they pack immense mass into a relatively small volume.

astronomy problem solving

Astronomy often involves solving complex problems using physics and mathematics, as demonstrated in the exercise. Tackling an astronomy problem usually follows systematic steps, from understanding the problem to applying known formulas and calculations.
In this exercise, we practiced several key skills:

• Mass calculation using multiplication and unit recognition.
• Volume conversion for compatible unit usage.
• Applying the density formula to relate mass and volume.

Each step requires careful consideration of units and mathematical operations, ensuring data remains accurate through conversions. Such problem-solving in astronomy ties together many disciplines and sharpens analytical skills. Through these techniques, astronomical phenomena, like the density of a black hole, can be understood quantitatively, providing insight into their extreme nature.