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Chapter 14: Problem 46

Assume that the Sun's mass is about 300,000 Earth masses and that its radius is about 100 times that of Earth. The density of Earth is about \(5,500 \mathrm{kg} / \mathrm{m}^{3}\). a. What is the average density of the Sun? b. How does this compare with the density of water?

Short Answer

Expert verified
The average density of the Sun is 1650 kg/m³, which is 1.65 times the density of water.

Step by step solution

01

- Understand Given Values

The Sun's mass is 300,000 times the mass of Earth. The Sun's radius is 100 times the radius of Earth. The density of Earth is 5500 kg/m³.
02

- Volume of the Earth

Calculate the volume of the Earth using the formula for the volume of a sphere: \[ V_{\text{Earth}} = \frac{4}{3} \pi R_{\text{Earth}}^3 \] where \( R_{\text{Earth}} \) is the Earth's radius.
03

- Volume of the Sun

Using the volume formula for a sphere again but with the radius of the Sun: \[ V_{\text{Sun}} = \frac{4}{3} \pi (100 \cdot R_{\text{Earth}})^3 \]. This simplifies to: \[ V_{\text{Sun}} = \frac{4}{3} \pi (10^6 R_{\text{Earth}}^3) = 10^6 \cdot V_{\text{Earth}} \].
04

- Mass of the Sun

Given the mass of the Sun is 300,000 times that of Earth: \[ M_{\text{Sun}} = 300,000 \cdot M_{\text{Earth}} \].
05

- Calculate Density of the Sun

Density is mass divided by volume. Using the mass and volume of the Sun: \[ \rho_{\text{Sun}} = \frac{M_{\text{Sun}}}{V_{\text{Sun}}} = \frac{300,000 \cdot M_{\text{Earth}}}{10^6 \cdot V_{\text{Earth}}} \].Simplify to find: \[ \rho_{\text{Sun}} = \frac{300,000}{10^6} \cdot \frac{M_{\text{Earth}}}{V_{\text{Earth}}} = 0.3 \cdot \rho_{\text{Earth}} \].Using Earth's density: \[ \rho_{\text{Sun}} = 0.3 \cdot 5500 \text{ kg/m}^3 = 1650 \text{ kg/m}^3 \].
06

- Compare to Density of Water

The density of water is 1000 kg/m³. Compare it with the Sun's density: \[ \rho_{\text{Sun}} = 1650 \text{ kg/m}^3 \]. Thus, the Sun's density is 1.65 times the density of water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sun's Mass and Radius
To understand the density calculation of the Sun, it’s important to know its mass and radius first.
The problem states that the Sun's mass is 300,000 times the mass of Earth. This gives us a sense of how much matter the Sun contains compared to our planet.
Similarly, the Sun’s radius is mentioned to be about 100 times that of Earth. Knowing the radius is crucial because we need it to calculate the volume of the Sun. This larger radius gives us an insight into how much space the Sun occupies.
These values set the stage for further calculations regarding the Sun’s volume and density.
Density of Earth
The density of any object is defined as its mass divided by its volume. For Earth, this value is given as 5500 kg/m³.
Understanding Earth's density is important for two reasons:
  • It provides a baseline for comparing the density of the Sun.
  • We will use Earth's density to calculate the mass and volume needed in our formulas.
Knowing the density of Earth helps us in computing the density of the Sun by comparison.
Volume Calculation
To find the volume of the Sun and Earth, we use the formula for the volume of a sphere: \( V = \frac{4}{3} \pi R^3 \).
  • First, we calculate the volume of Earth (\(V_{\text{Earth}}\)) using Earth's radius (\(R_{\text{Earth}}\)).
  • Next, we calculate the volume of the Sun (\(V_{\text{Sun}}\)) using the fact that the Sun's radius is 100 times Earth's radius.
As a result, the formula for the Sun's volume simplifies to: \( V_{\text{Sun}} = 10^6 \times V_{\text{Earth}} \).
So, the Sun's volume is a million times that of Earth, showing how expansive the Sun is.
Density Comparison
Finally, we compare the densities of the Sun and Earth.
To find the density of the Sun (\( \rho_{\text{Sun}} \)), we use the simplified formula: \( \rho_{\text{Sun}} = \frac{M_{\text{Sun}}}{V_{\text{Sun}}} = 0.3 \times \rho_{\text{Earth}} = 0.3 \times 5500 = 1650 \text{ kg/m}^3 \).
This means the Sun's density is about 1650 kg/m³.
When we compare it to the density of water (1000 kg/m³), we see that the Sun's density is 1.65 times the density of water.

Understanding these comparisons highlights the differences in composition and structure between the Sun and Earth, despite the Sun's much larger size.

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Most popular questions from this chapter

In the proton-proton chain, the mass of four protons is slightly greater than the mass of a helium nucleus. The "lost"" mass a. binds the particles of the nucleus together. b. cancels the positive charge of the two protons that turned into neutrons. c. escapes as light, positrons, or neutrinos. d. is reabsorbed instantly, turning helium back into hydrogen.

Explain how hydrostatic equilibrium acts as a safety valve to keep the Sun at its constant size, temperature, and luminosity.

Which of the following describe sunspots? (Select all that apply.) a. brighter than the surrounding photosphere b. fainter than the surrounding photosphere c. hotter than the surrounding photosphere d. cooler than the surrounding photosphere e. round f. darker in the middle than at the edges g. caused by magnetic fields h. caused by convection

The solar wind pushes on the magnetosphere of Earth, changing its shape, because a. the solar wind is so dense. b. the magnetosphere is so weak. c. the solar wind contains charged particles. d. the solar wind is so fast.

Imagine that the source of energy in the interior of the Sun changed abruptly. a. How long would it take before a neutrino telescope detected the event? b. When would a visible-light telescope see evidence of the change?
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