Chapter 1: Problem 67
A star is estimated to have a mass of \(2 \times 10^{36} \mathrm{~kg}\). Assuming it to be a sphere of average radius \(7.0 \times 10^{5} \mathrm{~km}\), calculate the average density of the star in units of grams per cubic centimeter.
Short Answer
Expert verified
The average density of the star is approximately \( 1.39 \times 10^{6} \mathrm{~g/cm^3} \).
Step by step solution
01
(Convert kg to g)
We are given the mass of the star in kilograms, and we need to convert it to grams.
1 kg = 1000 g
So our conversion will be:
\( (2 \times 10^{36} \mathrm{~kg}) * (1000 \mathrm{~g/kg}) = 2 \times 10^{39} \mathrm{~g} \)
#Step 2: Convert volume of the star to cubic centimeters#
02
(Convert km to cm)
We need to convert the radius of the star from kilometers to centimeters in order to calculate the volume in cubic centimeters.
1 km = 100,000 cm
So our conversion will be:
\( (7 \times 10^{5} \mathrm{~km}) * (10^{5} \mathrm{~cm/km}) = 7 \times 10^{10} \mathrm{~cm} \)
#Step 3: Calculate the volume of the star#
03
(Calculate volume of sphere)
Now that we have the radius of the star in centimeters, we can calculate its volume using the formula for the volume of a sphere:
\( V = \frac{4}{3} \pi r^{3} \)
where \( V \) is the volume and \( r \) is the radius. Plug in the values we just found:
\( V = \frac{4}{3} \pi (7 \times 10^{10} \mathrm{~cm})^{3} \)
Calculate the volume:
\( V \approx 1.44 \times 10^{33} \mathrm{~cm^3} \)
#Step 4: Calculate the average density of the star#
04
(Calculate density)
Now we have the mass of the star in grams, and its volume in cubic centimeters. We can calculate the average density using the formula:
\( \rho = \frac{m}{V} \)
where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume. Plug in the values we found:
\( \rho = \frac{2 \times 10^{39} \mathrm{~g}}{1.44 \times 10^{33} \mathrm{~cm^3}} \)
Calculate the average density:
\( \rho \approx 1.39 \times 10^{6} \mathrm{~g/cm^3} \)
The average density of the star is approximately \( 1.39 \times 10^{6} \) grams per cubic centimeter.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mass Conversion
Converting mass from kilograms to grams is a straightforward process but crucial when you want to calculate density in specific units like grams per cubic centimeter. Mass is often measured in kilograms (kg) in scientific contexts. However, smaller units like grams (g) can be more practical for detailed calculations, like density. To convert kilograms to grams, you multiply the mass value by 1000 since 1 kg equals 1000 g. For instance, in this exercise, the mass of the star is given as \(2 \times 10^{36} \mathrm{~kg}\). To convert it to grams, multiply by 1000:
- \((2 \times 10^{36} \mathrm{~kg}) \times (1000 \mathrm{~g/kg}) = 2 \times 10^{39} \mathrm{~g}\)
Volume Calculation
Calculating the volume of an object requires knowing the shape of the object and the formula that describes it. When working with stars, which are often modeled as spheres, we use the sphere volume formula. First, ensure that the radius is in the correct units. Once the radius is converted (as shown in the next section), use the volume formula for a sphere:
- \(V = \frac{4}{3} \pi r^3\)
- \(V = \frac{4}{3} \pi (7 \times 10^{10} \mathrm{~cm})^{3}\)
Unit Conversion
Unit conversion is critical in scientific calculations to ensure all measurements are in compatible units. In this problem, the radius of the star is initially given in kilometers but needs to be in centimeters for volume calculations in cubic centimeters. It's important to remember:
- 1 kilometer = 100,000 centimeters (or \(1 \mathrm{~km} = 10^5 \mathrm{~cm}\))
- \((7 \times 10^{5} \mathrm{~km}) \times (10^{5} \mathrm{~cm/km}) = 7 \times 10^{10} \mathrm{~cm}\)
Sphere Volume Formula
The sphere volume formula is essential when dealing with spherical objects, like stars in this context. The formula \(V = \frac{4}{3} \pi r^3\) allows us to compute the volume once the radius is known. Breaking it down:
- \(\frac{4}{3}\) is a constant that comes from integrating over a circular base.
- \(\pi\) is approximately 3.14159, a constant factor due to the circle in the sphere's cross-section.
- \(r^3\) means you're using the radius triply to scale space three-dimensionally, as volume is a 3D measurement.