What is the Schwarzschild radius (in km) of a 6Msun black hole? What fraction of the Earth's radius is this? What percent of the speed of light (2.998 x 10° m/s) is the escape velocity at the Schwarzschild radius?

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**Title: Understanding the Schwarzschild Radius for a Black Hole**

**Introduction**

This educational resource explains how to calculate the Schwarzschild radius of a black hole, specifically for a mass of \(6 M_{\text{Sun}}\), and explores its relationship to the Earth's radius and the speed of light.

**Key Questions**

1. What is the Schwarzschild radius (in km) of a \(6 M_{\text{Sun}}\) black hole?
2. What fraction of the Earth's radius is this?
3. What percent of the speed of light (\(2.998 \times 10^8\) m/s) is the escape velocity at the Schwarzschild radius?

**Derivation of Schwarzschild Radius**

The Schwarzschild radius (\(R_s\)) of a black hole is determined by the formula:

\[ R_s = \frac{2GM}{c^2} \]

For a black hole with mass \(6 M_{\text{Sun}}\), the formula becomes:

\[ R_s = \frac{2G(6)(M_{\text{Sun}})}{c^2} \]

where:
- \( M_{\text{Sun}} = 1.99 \times 10^{30} \) kg (mass of the Sun)
- \( G \) is the gravitational constant
- \( c \) is the speed of light

**Unit Conversion**

To express \(R_s\) in kilometers, note that \(1 \text{ km} = 1,000 \text{ m}\).

**Visual Explanation**

There are no graphs or diagrams in this section, but the explanation is focused on mathematical derivation and understanding.

**Summary**

This piece allows students to explore the concept of Schwarzschild radius and its implications for understanding the physics of black holes, utilizing basic physics concepts like gravitational force and the speed of light.
Transcribed Image Text:**Title: Understanding the Schwarzschild Radius for a Black Hole** **Introduction** This educational resource explains how to calculate the Schwarzschild radius of a black hole, specifically for a mass of \(6 M_{\text{Sun}}\), and explores its relationship to the Earth's radius and the speed of light. **Key Questions** 1. What is the Schwarzschild radius (in km) of a \(6 M_{\text{Sun}}\) black hole? 2. What fraction of the Earth's radius is this? 3. What percent of the speed of light (\(2.998 \times 10^8\) m/s) is the escape velocity at the Schwarzschild radius? **Derivation of Schwarzschild Radius** The Schwarzschild radius (\(R_s\)) of a black hole is determined by the formula: \[ R_s = \frac{2GM}{c^2} \] For a black hole with mass \(6 M_{\text{Sun}}\), the formula becomes: \[ R_s = \frac{2G(6)(M_{\text{Sun}})}{c^2} \] where: - \( M_{\text{Sun}} = 1.99 \times 10^{30} \) kg (mass of the Sun) - \( G \) is the gravitational constant - \( c \) is the speed of light **Unit Conversion** To express \(R_s\) in kilometers, note that \(1 \text{ km} = 1,000 \text{ m}\). **Visual Explanation** There are no graphs or diagrams in this section, but the explanation is focused on mathematical derivation and understanding. **Summary** This piece allows students to explore the concept of Schwarzschild radius and its implications for understanding the physics of black holes, utilizing basic physics concepts like gravitational force and the speed of light.
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